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MT EDUCARE LTD. GEOMETRY

SCHOOL SECTION 1

P

RC QD

Al

mSB

PROPERTY - I

PROPERTIES OF AREAS OF TRIANGLES

The ratio of areas of two triangles is equal to the ratio of the productof their bases and corresponding heights.In ABC,seg AD is the height andseg BC is the base.In PQR,seg PS is the height andseg QR is the base.

A ( ABC) BC AD

A ( PQR) QR PS

To learn the next property, we have to first understand the meaning ofTriangles with equal heights.

In theorems and problems we will come across three situations where two

or more triangles have equal heights.

1. In the adjoining figure,

seg AD and seg PS are the heights

of ABC and PQR respectively.

If AD = PS then ABC and PQR

are said to have equal heights.


line l || line m

ABC and PQR lie between

the same two parallel lines l

and m, they are said to have

equal heights.


ABD, ADC and ABC have common

vertex A and their bases BD, DC and BC

lie on the same line BC.

Also seg AE line BC.

seg AE is the height of ABD, ADC and ABC.

These three triangles have same height.

A

CB D QS

P

R

A

E DB C

Similarity1.

QS

P

R

A

CB D

GEOMETRY MT EDUCARE LTD.

SCHOOL SECTION2

PROPERTY - IV

PROPERTY - III

D

QB

C

A P

PROPERTY - II

A

B Q

P

C R

l

m

A

DB C

B C

A

D

The ratio of areas of two triangles having equal heights, is equal to theratio of their corresponding bases.

1. In the adjoining figure, ABC

and PQR lie between the same

two parallel lines l and m.

Their heights are equal.

A ( ABC) BC

A ( PQR) QR

2. In the adjoining figure, ABD, ADC

and ABC have common vertex A, and their

bases BD, CD and BC lie on the same line BC.

Hence they have equal heights. Considering two

triangles at a time we get the following results.

(i) A ( ABD) BD

A ( ADC) CD

(ii)

A ( ABD) BD

A ( ABC) BC

(iii)

A ( ADC) DC

A ( ABC) BC

The ratio of areas of two triangles having equal bases, is equal to theratio of their corresponding heights.In the adjoining figure,ABC and ABD havethe same base AB.

A ( ABC) CP

A ( ABD) DQ

Areas of two triangles having equal bases and equal heights, are equal.In the adjoining figure,

ABD and ACD have common vertex A and

their bases BD and CD lie on the same line BC.


Also, D is the midpoint of seg BC.

BD = CD.

Their bases are equal.

A (ABD) = A (ACD)


SCHOOL SECTION 3

EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5)

1. In the adjoining figure,seg BE seg AB andseg BA seg AD.If BE = 6 and AD = 9 (1 mark)

find A ( ABE)A ( ABD)

Sol.A ( ABE)

A ( ABD)

=

BE

AD[Triangles with common base]

A ( ABE)

A ( ABD)

=

6

9

A ( ABE)

A ( ABD)

=

2

3


2. In the adjoining figure,seg SP side YK andseg YT seg SK.If SP = 6, YK = 13, YT = 5and TK = 12 then find :A (SYK) : A (YTK). (2 marks)

Sol.A ( SYK)

A ( YTK)

=

YK SP

TK YT

[ The ratio of the areas of two triangles is

equal to the ratio of the products of their

bases and corresponding heights ]

A ( SYK)

A ( YTK)

=

13 6

12 5

[Given]

A ( SYK)

A ( YTK)

=

13

10


3. In the adjoining figure,RP : PK = 3 : 2 thenfind the values of :(i) A (TRP) : A (TPK)(ii) A (TRK) : A (TPK)(iii) A (TRP) : A (TRK) (3 marks)

Sol.RP

PK=

3

2[Given]

Let the common multiple be x RP = 3x and PK = 2x RK = RP + PK [ R - P - K] RK = 3x + 2x RK = 5x

TRP, TPK and TRK have a common vertex T and their bases RP,PK and RK lie on the same line RK

Their heights are equal

BE

A D

S

P Y

T

K

T

R P K


SCHOOL SECTION4

(i)A ( TRP)

A ( TPK)

=

RP

PK[Triangles having equal heights]

A ( TRP)

A ( TPK)

=

3x

2x

A ( TRP)

A ( TPK)

=

3

2

Similarly,

(ii)A ( TRK)

A ( TPK)

=

RK

PK

A ( TRK)

A ( TPK)

=

5x

2x

A ( TRK)

A ( TPK)

=

5

2

A

P

B R S T C

Q

(iii)A ( TRP)

A ( TRK)

=

RP

RK

A ( TRP)

A ( TRK)

=

3x

5x

A ( TRP)

A ( TRK)

=

3

5


4. The ratio of the areas of two triangles with the common base is 6 : 5.Height of the larger triangle is 9 cm. Then find the corresponding heightof the smaller triangle. (2 marks)

Sol. Let the areas of the larger and the smaller triangle be A1 and A2 respectively.Let their heights be h1 and h2 respectively.A

A1

2=

6

5 and h1 = 9 cm [Given]

The two triangles have a common base [Given]

A

A1

2=

h

h1

2[Triangles with common base]

6

5=

9

h2

h2 =5 9

6

h2 =15

2 h2 = 7.5

The corresponding height of the smaller triangle is 7.5 cm.


5. In the adjoining figure,seg PR seg BC, seg AS seg BC andseg QT seg BC. Find the following ratios :

(i)

A ( ABC)A ( PBC) (ii)

A ( ABS)A ( ASC)

(iii)

A ( PRC)A ( BQT) (iv)

A ( BPR)A ( CQT) (3 marks)

Sol. (i)A ( ABC)

A ( PBC)

=

AS

PR[Triangles with common base]

(ii)A ( ABS)

A ( ASC)

=

BS

SC[Triangles having equal heights]


SCHOOL SECTION 5

(iii)A ( PRC)

A ( BQT)

=

RC PR

BT QT

[ The ratio of the areas of two triangles

equal to the ratio of the products of their

bases and corresponding heights ]Similarly,

(iv)A ( BPR)

A ( CQT)

=

BR PR

CT QT


6. In the adjoining figure,seg DH seg EF and seg GK seg EF.If DH = 12 cm, GK = 20 cm andA (DEF) = 300 cm2 then find(i) EF (ii) A (GEF) (iii) A (DFGE). (3 marks)

Sol. (i) Area of triangle =1

2 × base × height

A (DEF) =1

2 × EF × DH

300 =1

2 × EF × 12

300 = EF × 6

EF =300

6

EF = 50 cm

(ii)A ( DEF)

A ( GEF)

=

DH

GK[Triangles with common base]

300

A ( GEF) =12

20

A (GEF) =300 20

12

A (GEF) = 500 cm2

(iii) A (DFGE) = A (DEF) + A (GEF) [Area addition property]= 300 + 500

A (DFGE) = 800 cm2


7. In the adjoining figure,seg ST || side QRFind the following ratios :

(i)

A ( PST)A ( QST) (ii)

A ( PST)A ( RST) (iii)

A ( QST)A ( RST) (3 marks)

Sol. PST and QST have a common vertex Tand their bases PS and QS lie on the same line PQ.

A ( PST)

A ( QST)

=

PS

QS [Triangles having equal heights]

D

E HK F

G

P

S T

Q R


SCHOOL SECTION6

A

CB

D E

(ii) PST and RST have a common vertex S and their bases PT and RTlie on the same line PR.

A ( PST)

A ( RST)

=

PT

RT[Triangles having equal heights]

(iii)seg ST || side QR [Given]QST and RST lie between the same two parallel lines ST and QR

Their heights are equalAlso, they have a common base ST

A (QST) = A (RST) [Areas of two triangles having equal basesand equal heights are equal]

A ( QST)

A ( RST)

= 1

BASIC PROPORTIONALITY THEOREM (B.P.T.)

Statement : If a line parallel to a side of a triangle intersects other sides intwo distinct points, then the line divides those sides in proportion. (4 marks)Given : In ABC,

(i) line DE || side BC

(ii) Line DE intersects sides AB and AC

at points D and E respectively.

To Prove : AD

DB =

AE

ECConstruction : Draw seg BE and seg CD.

Proof : ADE and BDE have a common vertex E and

their bases AD and BD lie on the same line AB.


A( ADE)

A( BDE)

=

AD

DB.......(i) [Triangles having equal heights]

ADE and CDE have a common vertex D andtheir bases AE and EC lie on the same line AC.


A( ADE)

A( CDE)

=

AE

CE.......(ii) [Triangles having equal heights]

line DE || side BC [Given]BDE and CDE are between the same two parallel lines DE and BC.

Their heights are equal.Also, they have same base DE.

A(BDE) = A(CDE) ......(iii)[ Areas of two triangles having equal

basesand equal heights are equal ]

A( ADE)

A( BDE)

=

A( ADE)

A( CDE)

......(iv) [From (i), (ii) and (iii)]

ADDB

= AEEC

[From (i), (ii) and (iv)]


SCHOOL SECTION 7


1. Find the values of x in the following figures, if line l is parallel to one ofthe sides of the given triangles : (2 marks)

Sol. (i) In ABC,line l || side BC [Given]

AP

PB=

AY

YC[By B.P.T.]

3

6=

5

x

x =6 5

3

x = 10

(ii) In STR,

line l || side TR [Given]

SP

TP=

SQ

RQ [By B.P.T.]

x

4.5=

1.3

3.9

x =4.5 1.3

3.9

x =4.5

3

x = 1.5

(iii)In LMN,

line l || side LN [Given]

MP

LP=

MQ

NQ [By B.P.T.]

8

2=

x

3

x =8 3

2

x = 12


6. In the adjoining figure,ML || BC and NL || DC.

Then prove that : AM AN

=AB AD

. (3 marks)

Proof : In ABC,seg ML || side BC [Given]

B

P

A 5 Y x C

6

3

l

Sx

4.5

T

R

1.3

3.9

lP Q

B

M

AL C

D

N

L

N

2

8

Mx

3l

P

Q


SCHOOL SECTION8

D

C

A

B

E

F

l

BM

AM=

CL

AL[By B. P. T]

BM AM

AM

=

CL AL

AL

[By Componendo]

AB

AM=

AC

AL[ A - M - B and A - L - C]

AM

AB=

AL

AC......(i) [By Invertendo]

In ADC,seg NL || side DC [Given]

DN

AN=

CL

AL[By B. P. T.]

DN AN

AN

=

CL AL

AL

[By Componendo]

AD

AN=

AC

AL[ A - N - D and A - L - C]

AN

AD=

AL

AC......(ii) [By Invertendo]

AMMB

=ANAD

[From (i) and (ii)]

CONVERSE OF BASIC PROPORTIONALITY THEOREMStatement : If a line divides two sides of a triangle in the same ratio, then theline is parallel to the third side. (4 marks)

Given :In ABC,line l intersects sides AB and AC at pointsD and E respectively such thatAD

DB =

AE

EC To prove : Line l || side BC Proof : (Indirect method)

Let us suppose that line DE is not parallel to side BC We can draw a line DF parallel to side BC, such that A-F-C.

In ABC,line DF || side BC

AD

DB=

AF

FC.........(i) [By B.P.T.]

But, AD

DB=

AE

EC.........(ii) [Given]

AF

FC=

AE

EC[From (i) and (ii)]

AF FC

FC

=

AE EC

EC

[By componendo]

AC

FC=

AC

EC[ A - F - C, A - E - C]

FC = EC F and E are not two different points. Line DF and line DE coincide line DE || side BC line l || side BC


SCHOOL SECTION 9


2. E and F are points on the side PQ and PR respectively of PQR. Foreach of the following cases, state whether EF || QR.(i) PE = 3.9 cm, EQ = 1.3 cm, PF = 3.6 cm and FR = 2.4 cm.(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.

Sol. (i)PE

EQ =3.9

1.3

PE

EQ =3

1......(i)

PF

FR =

3.6

2.4

PF

FR=

3

2.....(ii)

In PQR,

PE

EQ PF

FR[From (i) and (ii)]

line EF is not parallel to side QR.

(ii)PE

EQ =4

4.5

PE

EQ =4 10

4.5 10

PE

EQ =40

45

PE

EQ =8

9......(i)

PF

FR=

8

9......(ii)

In PQR,PE

EQ = PF


line EF || side QR [By converse of B.P.T.]

(iii) PQ = PE + EQ [ P - E - Q] 1.28 = 0.18 + EQ EQ = 1.28 – 0.18 EQ = 1.10 cm

PR = PF + FR [ P - F - R] 2.56 = 0.36 + FR FR = 2.56 – 0.36 FR = 2.20 cm

PE

EQ =0.18

1.10

PE

EQ =0.18 100

1.10 100

P

E F

Q R

3.9 3.6

2.41.3

P

E F

Q R

4 8

94.5

P

0.18

E F

0.36

Q R

1.28 2.56


SCHOOL SECTION10

PE

EQ =18

110

PE

EQ =9

55......(i)

PF

FR =

0.36

2.20

PF

FR =

0.36 ×100

2.20 ×100

PF

FR =

36

220

PF

FR =

9

55.....(ii)

In PQR,

PE

EQ =PF


line EF || side QR [By converse of B.P.T.]

PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190)

15. D and E are the points onsides AB and AC such thatAB = 5.6, AD = 1.4, AC = 7.2and AE = 1.8.Show that DE || BE. (3 marks)

Proof : AB = AD + DB [ A - D - B] 5.6 = 1.4 + DB DB = 5.6 – 1.4 DB = 4.2 units

AC = AE + EC [ A - E - C] 7.2 = 1.8 + EC EC = 7.2 – 1.8 EC = 5.4 units

AD

DB=

1.4

4.2

AD

DB=

1.4 ×10

4.2 ×10

AD

DB=

14

42

AD

DB=

1

3.......(i)

AE

EC=

1.8

5.4

AE

EC=

1.8 ×10

5.4 ×10

B

D

AE C


SCHOOL SECTION 11

AE

EC=

18

54

AE

EC=

1

3.......(ii)

In ABC,

AD

DB=

AE

EC[From (i) and (ii)]

seg DE || side BC [By converse of B.P.T.]


9. In the adjoining figure,points A, B and C are onseg OP, seg OQ and seg OR respectivelysuch that AB || PQ and AC || PR.Then show that BC || QR. (3 marks)

Proof : In POQ,seg AB || side PQ [Given]

OA

AP =

OB

BQ ......(i) [By B.P.T.]

In POR,seg AC || side PR [Given]

OA

AP =

OC

CR.....(ii) [By B.P.T.]

In OQR,

OB

BQ = OC

CR[From (i) and (ii)]

seg BC || side QR [By converse of B.P.T.]


7. In the adjoining figure,seg ED || seg QO andseg DF || seg OR.Prove that seg EF || side QR. (3 marks)

Proof : In PQO,seg ED || side QO [Given]

PE

EQ =PD

DO......(i) [By B.P.T.]

In PRO,seg DF || seg OR [Given]

PF

FR=

PD

DO.....(ii) [By B.P.T.]

In PQR,

PE

EQ =PF


seg EF || side QR [By converse of B.P.T.]

Q

P

R

B

A

C

O

P

DE F

O

Q R


SCHOOL SECTION12


19. Two triangles ABC and DBC lieon the same side of the base BC.From a point P on BC, PQ || ABand PR || BD are drawn. Theyintersect AC at Q and DC at R.Prove that QR || AD. (3 marks)

Proof : In ABC,seg PQ || side AB [Given]

CP

BP =

CQ

AQ .......(i) [By B.P.T.]

In BCD,seg PR || side BD [Given]

CP

BP =

CR

DR.......(ii) [By B.P.T.]

In ACD,

CQ

AQ = CR

DR[From (i) and (ii)]

seg QR || side AD [By converse of B.P.T.]


21. In a ABC, if D is a point on BC such that BD AB

=DC AC

, prove that AD is the

bisector of A. (Hint : Produce BA to E such that AE = AC. Join EC).(4 marks)

Construction : Take a point E on ray BA,such that AE = AC.Draw seg EC

Proof : In AEC,seg AE seg AC ......(i) [Construction]

AEC ACE .....(ii) [Isosceles triangle theorem]

BD

DC =

AB

AC[Given]

BD

DC =

AB

AE.....(iii) [From (i)]

In BEC,

BD

DC =

AB

AE[From (iii)]

seg AD ll side EC [By converse of B.P.T.]On transversal AC,DAC ACE .....(iv) [Converse of alternate angles test]On transversal BE,BAD AEC ......(v) [Converse of corresponding

angles test] DAC BAD [From (ii), (iv) and (v)] Ray AD is the bisector of A.

DA

Q R

P CB

D

A

E

CB


SCHOOL SECTION 13

PROPERTY OF INTERCEPTS MADE BY THREE PARALLEL LINES

x

E

GR

Q F

P

y

lm

n

p q

R

TC

B S

Al

m

n

8

10

The ratio of the intercepts made on a transversal by three parallellines is equal to the ratio of the corresponding intercepts made onany other transversal by the same parallel lines.line l || line m || line nand lines l, m and n cut the transversal xin points P, Q and R respectively.and lines l, m and n cut the transversal yin point E,F and G respectively.

PQ

QR = EF

FG

EXAMPLE :In the adjoining figure,line l || line m ||line n.Lines p and q are transversals.From given informationfind ST. (2 marks)

Sol. line l || line m || line n [Given]

On transversals p and q,

AB

BC=

RS

ST

[By Property of Intercepts made

by threeparallel lines]

8

10=

12

ST[Given]

ST =12 ×10

8

ST = 15 units

PROPERTY OF AN ANGLE BISECTOR OF A TRIANGLE Statement : In a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides. (4 marks) Given : In ABC,

ray AD is the bisector of BACsuch that B - D - C.

To Prove :BD

DC =

AB

AC Construction : Draw a line passing through C, parallel to line AD and intersecting line BA at point E, B - A - E. Proof : In BEC,

line AD || side CE [Construction]

BD

DC=

AB

AE.........(i) [By B.P.T.]

line CE || line AD [Construction]

On transversal BE,

BADAEC ........(ii) [Converse of corresponding angles test]

E

D

A

B C

xx


SCHOOL SECTION14

Also, On transversal AC,

DAC ACE ........(iii) [Converse of alternate angles test]

But, BAD DAC ........(iv) [ ray AD bisects BAC]

In AEC,

AEC ACE [From (ii), (iii) and (iv)]

seg AC seg AE [Converse of Isosceles triangle theorem]

AC = AE ........(v)

BDDC

=ABAC

[From (i) and (v)]


5. Ray PT is the anglebisector of QPR.Find the value of x andthe perimeter of PQR. (2 marks)

Sol. In PQR,ray PT bisects QPR [Given]

PQ

PR=

QT

TR[Property of angle bisector of a triangle]

5.6

x=

4

5

x =5 5.6

4

x = 7 PR = 7 cm

QR = QT + TR [ Q - T - R] QR = 4 + 5 QR = 9 cm

Perimeter of PQR = PQ + QR + PR= 3.6 + 9 + 7

Perimeter of PQR = 19.6 cm


3. Point Q is on the side MP such thatMQ = 2 and MP = 5.5.Ray NQ is the angle bisectorMNP of MNP.Find MN : NP. (2 marks)

Sol. MP = MQ + QP [M - Q - P] 5.5 = 2 + QP QP = 5.5 – 2 QP = 3.5 units

In MNP,ray NQ bisects MNP [Given]

P

QT R

4 cm5 cm

5.6 cmx

• •

N

M Q P

× ×


SCHOOL SECTION 15

MN

NP=

MQ

QP [Property of angle bisector of a triangle]

MN

NP=

2

3.5

MN

NP=

2 10

3.5 10

MN

NP=

20

35

MN

NP=

4

7

MN : NP = 4 : 7


4. Ray YM is the angle bisectorof XYZ, where XY = YZ.Find the relation between XM and MZ. (2 marks)

Sol.In XYZ,

ray YM bisects XYZ [Given]

XY

YZ=

XM

MZ[Property of angle bisector of a triangle]

1 =XM

MZ[ XY = YZ]

XM = MZ


16. In PQR, if QS is the angle bisectorof Q then, show that

A ( PQS)A ( QRS) =

PQQR (3 marks)

Proof :In PQRray QS bisects PQR [Given]

PQ

QR = PS

SR......(i)

[By property of angle bisec tor

of a triangle]PQS and QRS have a common vertex Q and their bases PS andSR lie on the same line PR.


A ( PQS)

A ( QRS)

=

PS

SR

[ The ratio of the areas of

two triangles having equal

heights is equal to the ratio

of their corresponding bases ]

A ( PQS)A ( QRS)

=

PQQR [From (i)]

X

M

Y Z• •

P

Q

S

R


SCHOOL SECTION16

Q R

P

2

3

1

•

×

C

A

4 2

•B 6

×

Q

R

×P

C

A×

B

Definition : For a given one - to - one correspondence between the verticesof two triangles, if(i) their corresponding angles are congruent.(ii) their corresponding sides are in proportion,then the correspondence is known as Similarity and the triangles aresaid to be Similar Triangles.

In above figure, for correspondence ABC PQR

(i) A P, B Q, C R and

(ii)AB

PQ = BC

QR = AC

PR =

2

1Hence ABC and PQR are similar triangles.

ABC is similar to PQR under ABC PQR, this statement is writtensymbolically as ABC ~ PQR.

NOTE If two triangles are similar, theni) their corresponding angles are congruentii) their corresponding sides are in proportion.If ABC ~ PQR, theni) A P, B Q, C R

ii)AB BC AC

PQ QR PR

TEST OF SIMILARITYWhen two triangles are similar, then three pairs of corresponding angles

are congruent and three pairs of corresponding sides are in proportion.

But to prove that two triangles are similar, we select only threeconditions taken in proper order. These conditions are called Tests ofsimilarity. There are 3 tests of similarity :

1. A - A - A test (A - A test) :For a given one - to - one correspondence between the vertices of two

triangles, if the corresponding angles are congruent, then the two trianglesare similar.

In fig. ABC PQR,

if, A P, B Q and C R,

then ABC ~ PQR

We know that sum of measures of three angles of a triangle is 180º. Becauseof this if two pairs of corresponding angles of two given triangles are

SIMILARITY OF TRIANGLES


SCHOOL SECTION 17

congruent then remaining pair is also congruent, and thus the trianglesbecome similar triangles. This is known as A - A test.

A - A Test : For a given one-one correspondence between the vertices oftwo triangles, if two angles of one triangle are congruent with thecorresponding two angles of other triangle, then the two triangles are similar.

2. S - A - S Test :For a given one-one correspondence between the vertices of two

triangles, if two sides of one triangle are proportional to the correspondingsides of the other triangle and angles included by them are congruent,then the two triangles are similar.

In the adjoining figure,under the correspondenceABC PQR,AB

PQ = BC

QR = 2

1and B Q,then ABC ~ PQR by S - A - S test of similarity.

3. S - S - S Test :For a given one-one correspondence between the vertices two triangles,

if three sides of one triangle are proportional to the three correspondingsides of other triangle, then the two triangles are similar.In the adjoining figure,under the correspondenceABC PQR,AB

PQ = BC

QR = AC

PR =

2

1

then ABC ~ PQR by S - S - S test of similarity.


1. In each of the following figures you find two triangles. Indicate whether thetriangles are similar. Give reasons in support of your answer. (2 marks)

Sol. (i)AB

PQ =4.6

2.3

AB

PQ =2

1......(i)

BC

QR =10

5

BC

QR =2

1......(ii)

AC

PR=

8

4

AC

PR=

2

1......(iii)

In ABC and PQR,AB

PQ =BC

QR = AC

PR[From (i), (ii) and (iii)]

ABC PQR [By SSS test of similarity]

RQ

P

5 4.5

4

A

CB

10 9

8

P

RQ40º

3

2.5

A

B C40º

6

5

A

P

Q R

B C

4.62.3 4

8

10

5


SCHOOL SECTION18

(ii)

AO

CO=

5

3......(i)

BO

DO=

5

3......(ii)

In AOB and COD,

AO

CO =

BO

DO[From (i) and (ii)]

AOB COD [Vertically opposite angles]

AOB COD [By SAS test of similarity]

(iii)m ABC = m ADE = 60º [Given]

ABC ADE ......(i)

In ABC and ADE,

ABC ADE [From (i)]

BAC DAE [Common angle]

ABC ADE [By AA test of similarity]

(iv)AB

PQ =24

12

AB

PQ =2

1......(i)

BC

QR =7

5......(ii)

AC

PR=

25

13.....(iii)

In ABC and PQR,

AB

PQ BC

QR AC

PR[From (i), (ii) and (iii)]

ABC is not similar to PQR.

(v)AB

BC=

2.5

2.5

AB

BC= 1 ......(i)

AD

CD=

2.5

2.5

AD

CD= 1 ......(ii)

BD = BD [Common side]

BD

BD= 1 ......(iv)

In ABD and CBD,

AB

BC =

BD

BD =

AD

CD[From (i), (ii) and (iii)]

ABD CBD [By SSS test of similarity]

5

A B

C D

O

5

33

A

B C

D E60º

60º

B

A

C

P

Q R

24 25

7

1213

5

A

B C

D2.5

2.5

2.5

2.5


SCHOOL SECTION 19

(vi) In AOB and DOC,AOB DOC [Common angles]mBAO = 52ºm CDO = 42º

BAO is not congruent to CDO AOB is not similar to DOC.


1. Study the following figures and find out in each case whether thetriangles are similar. Give reason. (2 marks)(i) MN = MT + TN [ M - T - N] MN = 2 + 4 MN = 6 units

MK = MP + PK [ M - P - K] MK = 3 + 6 MK = 9 units

MN

MT=

6

2

MN

MT=

3

1......(i)

MK

MP=

9

3

MK

MP=

3

1.....(ii)

In MNK and MTP,

MN

MT=

MK

MP[From (i) and (ii)]

NMK TMP [Common angle]

MNK ~ MTP [By SAS test of similarity]

(ii) PX = PR + RX [ P - R - X] PX = a + 2a PX = 3a

PX

PR=

3a

a

PX

PR=

3

1......(i)

XS

RT=

3b

2b

XS

RT=

3

2.....(ii)

In PXS and PRT,

PX

PR

XS

RT[From (i) and (ii)]

PXS is not similar to PRT

A

B C

D

O

52º

42º

M

T P

N K

2 3

4 6

P

R T

X S

2b

3b


SCHOOL SECTION20

(iii)

m M = m Q = 55º ......(i)m N = m R = a .....(ii)In DMN and AQR,M Q [From (i)]N R [From (ii)]

DNM ~ AQR [By AA test of similarity]


9. Using the information given inthe adjoining figure, find F.

(3 marks)

Sol.AB

DE=

3.8

7.6

AB

DE=

1

2........(i)

BC

EF=

6

12

BC

EF=

1

2.......(ii)

AC

DF=

3 3

6 3

AC

DF=

1

2.......(iii)

In ABC and DEF,AB

DE=

BC

EF =

AC

DF[From (i), (ii) and (iii)]

ABC ~ DEF [By SSS test of similarity] C F .....(iv) [c.a.s.t]

In ABC,m A = 80º [Given]m B = 60º

m C = 40º ......(v) [Remaining angle]

m F = 40º [From (iv) and (v)]


5. A vertical pole of a length 6m casts a shadow of 4 m long on the ground.At the same time a tower casts a shadow 28 m long. Find the height ofthe tower. (2 marks)

Sol. In the adjoining figure,seg AB and seg PQrepresents the pole andtower respectively andseg BC and seg QRrepresents the shadowcast by them respectively.

D

A

R

NM

Q

70º

55º a

a55º

B C

A

6 m

4 m

P

Q R28 m

?

F E

D

12 cm

7.6

cm

6 3 cm

A

B C

3.8

cm 3 3 cm

6 cm

80º

60º


SCHOOL SECTION 21

Now, ABC ~ PQR,

AB

PQ =BC

QR [c.s.s.t.]

6

PQ =4

28[Given]

PQ =6 28

4

PQ = 42

The height of the tower is 42 m.


10. A vertical stick 12 m long casts a shadow 8 m long on the ground. Atthe same time a tower casts the shadow 40 m on the ground. Determinethe height of the tower. (2 marks)

Sol. In the adjoining figure,seg AB and seg PQrepresents the vertical stickand the tower respectivelyand seg BC and seg QRrepresents the shadowcast by them respectively.ABC ~ PQR

AB

PQ =BC

QR [c.s.s.t.]

12

PQ =8

40[Given]

PQ =40 ×12

8 PQ = 60

Height of the tower is 60 m.


2. A triangle ABC with sides AB = 6 cm, BC = 12 cm and AC = 8 cm isenlarged to PQR such that its longest side is 18 cm. Find the ratio andhence find the lengths of the remaining sides of PQR. (3 marks)

Sol.

ABC and the enlarged PQR are similar.

AB

PQ = BC

QR = AC

PR[c.s.s.t.]

6

PQ = 12

18 =

8

PR[Given]

P

Q R

?

40 m

A

B C8 m

12 m

P

Q R18 cm

? ?

A

B C

8 cm

12 cm

6 cm


SCHOOL SECTION22

6

PQ = 2

3 =

8

PR......(i)

6

PQ = 2

3[From (i)]

8

PR =

2

3[From (i)]

PQ = 6 3

2

PR =

8 3

2

PQ = 9 cm PR = 12 cm

The ratio is 2

3 and PQ = 9 cm, PR = 12 cm.


6. Triangle ABC has sides of length 5, 6 and 7 units while PQR hasperimeter of 360 units. If ABC is similar to PQR then find the sidesof PQR. (3 marks)

Sol. In ABC,AB = 5 unitsBC = 6 units [Given]AC = 7 unitsPerimeter of PQR = 360 units [Given]

PQ + QR + PR = 360 ......(i)ABC ~ PQR [Given]

AB

PQ =BC

QR = AC

PR[c.s.s.t.]

5

PQ =6

QR =7

PR

5

PQ =6

QR =7

PR=

5 6 7

PQ QR PR

[By theorem on equal ratios]

5

PQ =6

QR =7

PR=

18

360[From (i)]

5

PQ =6

QR =7

PR =

1

20.....(ii)

5

PQ = 1

20[From (ii)]

PQ = 100 units

6

QR = 1

20[From (ii)]

QR = 120 units

7

PR=

1

20[From (ii)]

PR = 140 units


SCHOOL SECTION 23


3. In the adjoining figure,E is a point on side CB producedof an isosceles ABC with AB = AC.If AD BC and EF AC,prove that ABD ~ ECF. (2 marks)

Proof : In ABC,seg AB seg AC [Given]

ABC ACB ......(i) [Isosceles triangle theorem]In ABD and ECF,ABD FCE [From (i) and B - D - C, A - F - C ,

C - B - E]ADB EFC [ each is 90º]

ABD ~ ECF [By AA test of similarity]


2. ABC is a right angled at B.D is any point on AB.DE AC. If AD = 6 cm,AB = 12 cm,AC = 18 cm, find AE. (2 marks)

Sol. In ABC and AED,BAC DAE [Common angle]ABC AED [ Each is 90º]

ABC ~ AED [By AA test of similarity]

AB

AE=

AC

AD[c.s.s.t.]

12

AE=

18

6[Given]

AE =12 × 6

18 AE = 4 units


8. In the adjoining figure,AB || DC.Using the information givenfind the value of x. (3 marks)

Sol. seg AB || seg DC [Given] On transversal BD,

CDB ABD [Converse of alternate angles test] CDO ABO .......(i) [D - O - B]

In DOC and BOA,CDO ABO [From (i)]DOC BOA [Vertically opposite angles]

DOC ~ BOA [By AA test of similarity]

A

EB C

F

D

B C

E

D

A

D C

A

O

B

3

x – 33x – 19

x – 5


SCHOOL SECTION24

DO

BO=

OC

OA[c.s.s.t.]

3

x – 3=

x – 5

3x –19 3(3x – 19) = (x – 5) (x – 3)

9x – 57 = x2 – 3x –5x + 15

9x – 57 = x2 – 8x + 15

x2 – 8x – 9x + 15 +57 = 0

x2 – 17x + 72 = 0

x2 – 9x – 8x + 72 = 0

x (x – 9) – 8 (x – 9) = 0

(x – 9) (x – 8) = 0

x – 9 = 0 or x – 8 = 0

x = 9 or x = 8


4. D is a point on side BC of ABC such that ADC = BAC.Show that AC2 = BC × DC. (2 marks)

Proof : In ABC and DAC,

BAC ADC [Given]

ACB ACD [Common angle]

ABC ~ DAC [By AA test of similarity]

AC

DC =

BC

AC[c.s.s.t.]

AC2 = BC × DC


8. ABCD is a trapezium in which AB || DC and its diagonals intersect

each other at the point O. Show that AO CO

=BO DO

. (3 marks)

Proof : ABCD is a trapeziumside AB || side DC [Given]

On transversal AC,BAC DCA [Converse of alternate angles test]

BAO DCO ......(i) [ A - O - C]In AOB and COD,BAO DCO [From (i)]AOB COD [Vertically opposite angles]

AOB ~ COD [By AA test of similarity]

AO

CO =

BO

DO[c.s.s.t.]

AO CO

=BO DO

[By Alternendo]

A B

O

D C

A

B DC


SCHOOL SECTION 25


10. Using the basic proportionality theorem, prove that the line joining themid points of any two sides of a triangle is parallel to the third side andhalf of it. (5 marks)

Given : In ABC,D and E are the midpoints of sides ABand AC respectively.

To prove : (a) seg DE || sideBC

(b) DE = 1

BC2

Proof :(a) Let us assume that seg DE is not parallel to side BC. We can draw a line passing through D parallel to side BC intersecting

side AC at F such that A - F - E - C.In ABC,line DF || side BC

AD

DB =

AF

FC........(i) [By B.P.T.]

But, AD = DB [ D is the midpoint of side AB]

AD

DB = 1 ......(ii)

AF

FC = 1 [From (i) and (ii)]

AF = FC F is the midpoint of side AC

But, E is also the midpoint of side AC [Given]But, a segment can have one and only midpoint

F and E are not different points. seg DE || side BC ......(iii)

(b) line DE || line BC [From (iii)] On transversal AB,

ADE ABC ......(iv) [Converse of corresponding angles test]In ADE and ABC,ADE ABC [From (iv)]DAE BAC [Common angle]

ADE ~ DABC [By AA test of similarity]

AD

AB =

DE

BC =

AE

AC.......(v) [c.s.s.t.]

But, AD = 1

AB2

[D is the midpoint of side AB]

AD

AB =

1

2......(vi)

DE

BC =

1

2[From (v) and (vi)]

DE = 1

BC2

A

D E

B C


SCHOOL SECTION26


7. In the adjoining figure,DE || BC and AD : DB = 5 : 4,Find (i) DE : BC

(ii) DO : DC (iii) A (DOE) : A (DCE) (5 marks)Sol. seg DE || seg BC [Given]

On transversal AB,ADE ABC ......(i) [By converse of corresponding

angles test]In ADE and ABC,ADE ABC [From (i)]DAE BAC [Common angle]

ADE ~ ABC [By AA test of similarity]

AD

AB=

DE

BC.....(ii) [c.s.s.t]

AD

DB=

5

4[Given]

DB

AD=

4

5[By Invertendo]

DB AD

AD

+=

4 5

5

+[By componendo]

AB

AD=

9

5[A - D - B]

AD

AB=

5

9......(iii)

DE

BC=

5

9[From (ii) and (iii)]

DE : BC = 5 : 9 ......(iv)

(ii) seg DE || seg BC [Given]On transversal DC,EDC BCD [Converse of alternate angles test]

EDO BCO .......(v) [D - O - C]In DOE and COB,EDO BCO [From (v)]DOE BOC [Vertically opposite angles]DOE ~ COB [By AA test of similarity]

DO

CO=

DE

BC[c.s.s.t]

DO

CO=

5

9[From (iv)]

CO

DO=

9

5[By Invertendo]

CO DO

DO

+=

9 5

5

+[By Componendo]

DC

DO=

14

5[ D - O - C]

DO

DC=

5

14[By Invertendo]

DO : DC = 5 : 14 ......(vi)

(iii)DOE and DCE have a common vertex E and their bases DO andDC lie on the same line DC.

A

B C

O

D E


SCHOOL SECTION 27


A ( DOE)

A ( DCE)

=

DO

DC[Triangles having equal heights]

A ( DOE)

A ( DCE)

=

5

14[From (vi)]


18. Let X be any point on side BC of ABC,XM and XN are drawn parallel to BAand CA. MN meets produced CB in T.Prove that TX2 = TB . TC. (5 marks)

Proof : seg AB || seg MX [Given]seg NB || seg MX [ A - N - B]

On transversal TX,TBN TXM ......(i) [Converse of corresponding angles test]In TBN and TXM,TBN TXM [From (i)]BTN XTM [Common angle]

TBN TXM [By AA test of similarity]

TB

TX =

TN

TM.....(ii) [c.s.s.t.]

seg NX ll seg AC [Given] seg NX ll seg MC [ A - M - C] On transversal TC,

TXN TCM ......(iii) [Converse of corresponding angles test]In TXN and TCM,TXN TCM [From (iii)]NTX MTC [Common angle]

TXN TCM [By AA test of similarity]

TX

TC =

TN

TM.....(iv) [c.s.s.t.]

TB

TX =

TX

TC[From (ii) and (iv)]

TX2 = TB × TC


25. In the adjoining figure,DEFG is a square and BAC = 90ºProve that :(i) AGF ~ DBG (ii) AGF ~ EFC(iii) DBG ~ EFC (iv) DE2 = BD . EC

Proof :(i) DEFG is a square [Given] DE = EF = GF = DG .......(i) [Sides of a square] seg GF || seg DE [Opposite side of a square] seg GF || seg BC [B - D - E - C) On transversal AB,

AGF ABC .......(ii) [Converse of corresponding angles test]On transversal AC,AFG ACB ......(iii)

A

N

M

CXBT

A

G F

B D E C


SCHOOL SECTION28

In AGF and DBG,AGF GBD [From (ii) and A - G - B, B - D - C]GAF BDG [ each is 90º]

AGF DBG ......(iv) [By AA test of similarity]

(ii) In AGF and EFC,AFG FCE [From (iii) and A - F - C, C - E - A]GAF FEC [ Each is 90º]

AGF ~ EFC ......(v) [By AA test of similarity]

(iii) DBG ~ EFC [From (iv) and (v)]

(iv) BD

EF =

DG

EC[c.s.s.t.]

EF × DG = BD × EC DE × DE = BD × EC [From (i)] DE2 = BD × EC


20. In the adjoining figure,ADB and CDB have the samebase DB. If AC and BD intersect at O

then prove that

A ( ADB) AO=

A ( CDB) CO . (4 marks)

Proof : ADB and CDB have a common base BD,

A( ADB)

A( CDB)

=

AN

CM....(i) [Triangles with common base]

In ANO and CMO,ANO CMO [ Each is 90º]AON COM [Vertically opposite angles]

ANO ~ CMO [By AA test of similarity]

AN

CM =

AO

CO......(ii) [c.s.s.t.]

A ( ADB)A ( CDB)

=

AOCO

[From (i) and (ii)]


24. Two poles of height ‘a’ meters and‘b’ meters are ‘p’ meters apart.Prove that the height ‘h’ drawn fromof the point of intersection N of thelines joining the top of each pole tothe foot of the opposite pole is

aba + b meters. (5 marks)

Proof : AB = AT + TB [ A - T - B] AB = (x + y) = p ......(i)

In ATN and ABR,TAN BAR [Common angle]ATN ABR [ each is 90º]

ATN ABR [By AA test of similarity]

AT

AB=

TN

BR[c.s.s.t]

A

D NO M B

C

S

N ab

h

x yTBA

R

p


SCHOOL SECTION 29

x

p =h

a......(ii) [Given and from (i)]

In BTN and BAS,TBN ABS [Common angle]BTN BAS [ each is 90º]

BTN BAS [By AA test of similarity]

BT

AB=

TN

AS[c.s.s.t]

y

p =h

b......(iii) [Given and from (i)]

Adding (ii) and (iii),x y

+p p =

h h+

a b

x + y

p = 1 1

h +a b

p

p =a + b

hab

[From (i)]

1 =a + b

hab

ab

a + b = h

h =ab

a +b metres


22. The bisector of interior Aof ABC meets BC in D.The bisector of exteriorA meets BC produced in E.

Prove that BD CD

=BE CE

.

[Hint : For the bisector of A which

is exterior of BAC, AB BE

=AC CE

] (5 marks)

Construction : Draw seg CF || seg AE such that B - F - A - KProof : In ABC,

ray AD bisects BAC [Given]

BD

CD =

AB

AC......(i) [By property of angle bisector of a triangle]

In ABE,seg CF || side AE [Construction]

BF

AF =

BC

CE[By B.P.T.]

BF + AF

AF =

BC + CE

CE[By Componendo]

AB

AF =

BE

CE......(ii) [B - F - A and B - C - E]

seg CF || seg AE [Construction] on transversal AF,

KAE AFC ......(iii) [Converse of corresponding angles test] on transversal AC,

EAC ACF ......(iv) [Converse of alternate angles test]

F

A

BD C

E

××••

K•


SCHOOL SECTION30

Q R

P

SB C

A

D

But,KAE EAC .......(v) [ ray AE is the bisector of KAC]

AFC ACF ......(vi) [From (iii), (iv) and (v)]In AFC,AFC ACF [From (vi)]

AF = AC ......(vii) [Converse of Isosceles triangle theorem]

AB

AC =

BE

CE......(viii) [From (ii) and (vii)]

BD

CD =

BE

CE[From (i) and (viii)]

BDBE

=CDCE

[By Alternendo]

AREAS OF SIMILAR TRIANGLES Statement : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (4 marks) Given : ABC ~ PQR.

To Prove :A ( ABC)

A ( PQR)

=

BC

QR

2

2 = AB

PQ

2

2 = AC

PR

2

2

Construction : (i) Draw seg AD side BC,

B - D - C (ii) Draw seg PS side QR,

Q - S - R

Proof :A ( ABC)

A ( PQR)

=

BC AD

QR PS

......(i)

[ The ratio of the areas of two triangles

is equal to ratio of the products of a

base and its corresponding height ]ABC ~ PQR [Given]

AB BC AC

PQ QR PR ......(ii) [c.s.s.t.]

Also, B Q .....(iii) [c.a.s.t.]In ADB and PSQ,ADB PSQ [Each is a right angle]B Q [From (ii)]

ADB ~ PSQ [By A-A test of similarity]

AD BD AB

PS QS PQ .....(iv) [c.s.s.t.]

AD BC

PS QR ......(v) [From (ii) and (iv)]

A ( ABC)

A ( PQR)

=

BC AD

QR PS [From (i)]

A ( ABC)

A ( PQR)

=

BC BC

QR QR [From (v)]

A ( ABC)

A ( PQR)

=

BC

QR

2

2 .......(vi)

A( ABC)A( PQR)

=

BC²QR² =

AB²PQ² =

AC²PR²

[From (ii) and (vi)]


SCHOOL SECTION 31


1. ABC ~ DEF(i) If A (ABC) = 9 cm2, A (DEF) = 64 cm2, DE = 5.6 cm then find AB.(ii) If A (ABC) : A (DEF) = 16 : 25, BC = 2.2 cm then find EF.(iii)If AB = 2.4 cm, DE = 1.6 cm, find the ratio of the area of ABC andDEF. (2 marks)

Sol. (i) ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

AB

DE

2

2 [Areas of similar triangles]

9

64=

AB

(5.6)

2

2 [Given]

3

8=

AB

5.6[Taking square roots]

AB =3 5.6

8

AB = 3 × 0.7

AB = 2.1 cm

(ii) ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

BC

EF

2


16

25=

(2.2)

EF

2

2 [Given]

4

5=

2.2

EF[Taking square roots]

EF =2.2 5

4

EF =5.5

2

EF = 2.75 cm

(iii)ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

AB

DE

2


A ( ABC)

A ( DEF)

=

(2.4)

(1.6)

2

2 [Given]

A ( ABC)

A ( DEF)

=

5.76

2.56

A ( ABC)

A ( DEF)

=

576

256

A ( ABC)

A ( DEF)

=

9

4

A (ABC) : A (DEF) = 9 : 4.


SCHOOL SECTION32


3. A model of a ship is made in ratio 1 : 200(i) The length of the model is 4 m. Calculate the length of the ship.(ii) The area of the deck of the ship is 16,0000 m2. Find the area of

the deck of the model. (2 marks)Sol. (i) A model of a ship is made in the ratio 1 : 200.

Length of the model is 4 m [Given] Length of the ship = 4 × 200

Length of the ship = 800 m

(ii) Area of the deck of the ship = 160000 m2

Area of the deck of the model

Area of the deck of the ship = (length of the model)

(length of the ship)

2

2


160000 =

1

200

2


160000 =

1

40000

Area of the deck of the model = 1

16000040000

Area of the deck of the model = 4 m2


4. In the adjoining figure,D is a point on BC such thatABD = CAD. If AB = 5 cm,AD = 4 cm and AC = 3 cm. Find(i) BC (ii) DC (iii) A (ACD) : A (BCA). (3 marks)

Sol. In ACD and BCA,CAD ABC [Given and B - D - C]ACD ACB [Common angle]

ACD BCA [By AA test of similarity]

AC

BC=

DC

AC =

AD

AB[c.s.s.t]

3

BC=

DC

3 =

4

5......(i) [Given]

(i)3

BC=

4

5[From (i)]

BC =3 × 5

4

BC =15

4

BC = 3.75 cm

(ii)DC

3=

4

5[From (i)]

DC =4 × 3

5

DC =12

5

DC = 2.4 cm

A

B CD

5 cm

4 cm

3 cm


SCHOOL SECTION 33

(iii) ACD BCA

A( ACD)

A( BCA)

=

AD

AB

2


A( ACD)

A( BCA)

=

4

5

2

2 [Given]

A( ACD)

A( BCA)

=

16

25

A(ACD) : A(BCA) = 16 : 25


5. In ABC, PQ is a line segment intersecting AB at P and AC at Q suchthat PQ || BC. If PQ divides ABC into two equal parts means equal in

area, find BPAB

. (5 marks)

Sol. seg PQ divides ABC into two parts of equal areas [Given]

A (APQ) = 1

2 A (ABC)

A ( APQ)

A ( ABC)

=

1

2......(i)

seg PQ || side BC [Given]On transversal AC,AQP ACB ......(ii) [Converse of corresponding angles test]In APQ and ABC,AQP ACB [From (ii)]PAQ BAC [Common angle]

APQ ~ ABC [By AA test of similarity]

A ( APQ)

A ( ABC)

=

AP

AB

2


1

2=

AP

AB

2

2 [From (i)]

1

2=

AP

AB[Taking square roots]

AP

AB=

1

2

AB – BP

AB=

1

2[ A - P - B]

AB BP

–AB AB

=1

2

BP

1 –AB

=1

2

1

1 –2

=BP

AB

BP

AB=

2 – 1

2

B

P

A QC


SCHOOL SECTION34


17. In the adjoining figure,XY || AC and XY divides thetriangular region ABC into twoequal areas.Determine AX : AB. (4 marks)

Sol. seg XY divides ABC in twoparts of equal areas

A(XYB) = 1

2 A(ACB)

A ( XYB)

A ( ACB)

=

1

2.......(i)

seg XY || side AC [Given]On transversal BC,XYB ACB .....(ii) [Converse of corresponding angles test]In XYB and ACB,XYB ACB [From (ii)]XBY ABC [Common angle]

XYB ~ ACB [By AA test of similarity]

A ( XYB)

A ( ACB)

=

XB

AB

2


1

2=

2

2

XB

AB[From (i)]

1

2=

XB

AB[Taking square roots]

XB

AB=

1

2

AB – AX

AB=

1

2[ A - X - B]

AB

AB –

AX

AB=

1

2

AX

1 –AB

=1

2

1

1 –2

=AX

AB

AX

AB=

2 – 1

2


23. In the adjoining figure,ABCD is a square. The BCE onside BC and ACF on the diagonal ACare similar to each other. Then show

that A (BCE) = 1

A ( ACF)2

(3 marks)

Proof : BCE ~ ACF [Given]

A ( BCE)

A ( ACF)

=

2

2

BC

AC......(i) [Areas of similar triangles]

ABCD is a square [Given]

F

CD

A B

E

A

C

X

BY


SCHOOL SECTION 35

AB = BC = CD = AD ......(ii) [Sides of a square]In ABC,m ABC = 90º [Angle of a square]

AC2 = AB2 + BC2 [By Pythagoras theorem] AC2 = BC2 + BC2 [From (i)] AC2 = 2BC2 .....(iii)

A ( BCE)

A ( ACF)

=

BC

2BC

2

2 [From (i) and (iii)]

A ( BCE)

A ( ACF)

=

1

2

A (BCE) =12

A (ACF)


4. In the adjoining figure, DE ||BC,(i) If DE = 4 cm, BC = 8 cm,

A (ADE) = 25 cm2, find A (ABC)(ii) If DE : BC = 3 : 5,then A (ADE) : A (DBCE). (5 marks)

Sol. seg DE || seg BC [Given]On transversal AC,AED ACB ......(i) [Converse of corresponding angles test]In ADE and ABC,AED ACB [From (i)]DAE BAC [Common angle]

ADE ~ ABC .....(ii) [By AA test of similarity]

A ( ADE)

A ( ABC)

=

DE

BC

2


25

A ( ABC) =4

8

2

2 [Given]

25

A ( ABC) = 16

64

A (ABC) =64 25

16

A (ABC) = 100 cm2

(ii) ADE ABC [From (ii)

A ( ADE)

A ( ABC)

=

DE

BC

2


A ( ADE)

A ( ABC)

=

DE

BC

2

A ( ADE)

A ( ABC)

=

3

5

2

[Given]

A ( ADE)

A ( ABC)

=

9

25

A ( ABC)

A ( ADE)

=

25

9[By Invertendo]

A ( ABC) – A ( ADE)

A ( ADE)

=

25 – 9

9[By Dividendo]

B

D

A E C


SCHOOL SECTION36

A ( DBCE)

A ( ADE)

=16

9[Area addition property]

A ( ADE)

A ( DBCE)

=

9

16[By Invertendo]

A (ADE) : A (DBCE) = 9 : 16


6. In ABC, seg DE || side BC. If 2A (ADE) = A (DBCE), find AB : AD and

show BC = 3 .DE (4 marks)

Sol. seg DE || side BC [Given]On transversal AB,ABC ADE .....(i) [Converse of corresponding angles test]In ABC and ADE,ABC ADE [From (i)]BAC DAE [Common angle]

ABC ~ ADE [By AA test of similarity]AB

AD =

BC

DE =

AC

AE.....(ii) [c.s.s.t.]

A ( ABC)

A ( ADE)

=

AB

AD

2

2 .....(iii) [Areas of similar triangles]

A (ABC) = A (ADE) + A (DBCE) [Area addition property]

= A (ADE) + 2A (ADE) [ A (DBCE) = 2A (ADE)]

A (ABC) = 3A (ADE)

A ( ABC)

A ( ADE)

=

3

1 .....(iv)

3

1=

AB

AD

2

2 [From (iii) and (iv)]

AB

AD=

3

1 ......(v) [Taking square roots]

AB : AD = 3 :1

BC

DE=

AB

AD[From (ii)]

BC

DE=

3

1[From (v)]

BC = 3 × DE


2. The corresponding altitudes of two similar triangles are 6 cm and9 cm respectively. Find the ratio of their areas. (3 marks)

Given : (i) In ABC, seg AD side BC(ii) In PQR, seg PS side QR(iii) ABC ~ PQR(iv) AD = 6 cm, PS = 9 cm

A

D E

B C

P

SQ RB D C

A


SCHOOL SECTION 37

To find :A ( ABC)

A ( PQR)

Sol. ABC ~ PQR [Given]

A ( ABC)

A ( PQR)

=

AD

PS

2

2 [Altitudes of similar triangles]

A ( ABC)

A ( PQR)

=

6

9

2

2 [Given]

A ( ABC)

A ( PQR)

=

6

9

2

A ( ABC)

A ( PQR)

=

2

3

2

A ( ABC)

A ( PQR)

=

4

9

A (ABC) : A (PQR) = 4 : 9


3. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively.Find the ratio of their corresponding heights. What is the ratio of theircorresponding medians ? (3 marks)

Given :(i) ABC ~ PQR(ii) In ABC, seg AD side BC and seg AE is the median(iii)In PQR, seg PS side QR and seg PT is the median(iv) A (ABC) = 81cm2, A (PQR) = 49 cm2

To find : (i) AD

PS (ii)

AE

PTSol. ABC ~ PQR [Given]

A ( ABC)

A ( PQR)

=

AD

PS

2

2 [Altitudes of similar triangles]

81

49=

AD

PS

2

2 [Given]

AD

PS=

9

7[Taking square roots]

AD : PS = 9 : 7ABC ~ PQR [Given]

A ( ABC)

A ( PQR)

=

AE

PT

2

2 [Medians of similar triangles]

81

49=

AE

PT

2

2 [Given]

AE

PT=

9


AE : PT = 9 : 7

Q S T R

PA

B D E C


SCHOOL SECTION38


6. In the adjoining figure, ABCDis a parallelogram whose diagonalsintersect at O. P is a point on thediagonal AC such that PA : AO = 1 : 2.BP meets DA produced at Q. Then find(i) PQ : QB (ii) A (PQA) : A (PBC)(iii) A (PQA) : A (QBCA)

[Hint : PA PA 1 PA 1 1 1

= = × = × =AC 2AO 2 AO 2 2 4

] (5 marks)

Sol. (i) ABCD is a parallelogram [Given] seg AD || seg BC ......(i) [By definition]

In PBC,seg QA || side BC [From (i) and Q - A - D]

PQ

QB = PA

AC....(ii) [By B.P.T.]

PA

AC =

PA

2AO

[ Diagonals of paralle log ram

bisec t each other ]

PA

AC =

1

2 ×

PA

AO

PA

AC =

1

2 ×

1

2

PA 1

AO 2

PA

AC =

1

4.......(iii)

PQ

QB = 1


PQ : QB = 1 : 4

(ii) seg QA || seg BC [From (i) and Q - A - D]On transversal PB,PQA PBC .......(iv) [Converse of corresponding angles test]In PQA and PBC,PQA PBC [From (iv)]QPA BPC [Common angles]

PQA ~ PBC [By AA test of similarity]

A ( PQA)

A ( PBC)

=

PA

PC

2


A ( PQA)

A ( PBC)

=

PA

PC

2

.......(v)

PA

AC=

1

4[From (iii)]

AC

PA=

4

1[By Invertendo]

AC PA

PA

+=

4 1

1

+[By Componendo]

PC

PA=

5

1[ P - A - C]

P

A

OD

CB

Q


SCHOOL SECTION 39

A

B E F

G

CD

CENTROID OF A TRIANGLE : A

B D C

GF E

PA

PC=

1

5.......(vi) [By Invertendo]

A ( PQA)

A ( PBC)

=

1

5

2

[From (v) and (vi)]

A ( PQA)

A ( PBC)

=

1

25

A (PQA) : A (PBC) = 1 : 25

(iii)A ( PQA)

A ( PBC)

=

1

25

A ( PBC)

A ( PQA)

=

25

1[By Invertendo]

A ( PBC) – A ( PQA)

A ( PQA)

=

25 –1

1[By Dividendo]

A ( QBCA)

A ( PQA)

=24


A ( PQA)

A ( QBCA)

=

1

24[By invertendo]

A (PQA) : A (QBCA) = 1 : 24

In ABC,seg AD, seg BE and seg CF are the medians

G is the centroid of ABC

AG =2

AD3

GD = 1

AD3

[ Centroid of a triangle trisects each

AG = 2GD median]

GD = 1

AG2


5. G is the centroid of ABC.GE and GF are drawn parallelto AB and AC respectively.Find A (GEF) : A (ABC).(Hint : Draw the median AD.Then GD : AD = 1 : 3.A (GED) : A (ABD) = 1 : 9) (5 marks)

Construction : Draw seg AG and let it intersect side BC at point D.Sol. seg GE ll side AB [Given]

On transversal BC, GEC ABC [Converse of corresponding

angles test] GED ABD ......(i) [E - D - C and B - D - C]


SCHOOL SECTION40

In GED and ABDGED ABD [From (i)]GDE ADB [Common angle]

GED ABD [By AA test of similarity]

A( GED)

A( ABD)

=

GD

AD

2


A( GED)

A( ABD)

=

GD

AD

2

......(ii)

But, GD =1

AD3

[ Centriod of a triangle

trisec ts each median]

GD

AD=

1

3.....(iii)

A( GED)

A( ΑΒD)

=

1

3

2

[From (ii) and (iii)]

A( GED)

A( ΑΒD)

=

1

9.....(iv)

Similarly we can get,

A( GFD)

A( ΑCD)

=

1

9.....(v)

A ( GED)

A ( ΑΒD)

=

A ( GFD)

A ( ΑCD)

=

1

9[From (iv) and (v)]

A ( GED) A ( GFD)

A ( ΑΒD) + A ( ΑCD)

=

1

9[Theorem on equal ratio]

A ( GEF)

A ( ΑBC)

=

1


A (GEF) : A (ABC) = 1 : 9

ALTERNATIVE METHOD :seg GE ll side AB [Given]On transversal BC,

GEC ABC ......(i) [Converse of corresponding angles test]seg GF ll side AC [Given]On transversal BC,

GFB ACB ......(ii) [Converse of corresponding angles test]In GEF and ABC,GEF ABC [From (i) and E - F - C]GFE ACB [From (ii) and E - F - C]

GEF ABC [By AA test of similarity]

A ( GEF)

A ( ABC)

=

GE

AB

2


A ( GEF)

A ( ABC)

=

GE

AB

2

......(iii)

In GED and ABD,GED ABD [From (i) and E - D - C]

GDE ADB [Common angle] GED ABD [By AA test of similarity]


SCHOOL SECTION 41

A

B C D

SIMILARITY IN RIGHT ANGLED TRIANGLES

A

B C

D

PROPERTY OF GEOMETRIC MEAN

P

Q R

S

GE

AB=

GD

AD......(iv) [c.s.s.t.]

But, GD =1

AD3

[ Centriod of a triangle

trisec ts each median]

GD

AD=

1

3......(v)

GE

AB=

1

3.....(vi) [From (iv) and (v)]

A ( GEF)

A ( ΑΒC)

=

1

3

2

[From (iii) and (vi)]

A ( GEF)

A ( ΑΒC)

=

1

9

A (GEF) : A (ABC) = 1 : 9

In a right angled triangle, if the perpendicular is drawn from the vertexof the right angle to the hypotenuse, then the triangles on either side ofthe perpendicular are similar to the original triangle and to each other.

In ABC,

m ABC = 90º

seg BD hypotenuse AC

ABC ~ ADB ~ BDC

In a right angled triangle, the length of perpendicular segment drawnon to the hypotenuse from the opposite vertex, is the geometric mean ofthe length of the segments into which the hypotenuse is divided.

In PQR,m PQR = 90ºseg QS hypotenuse PR

QS² = PS × RS

EXERCISE 1.5 (TEXT BOOK PAGE NO. 30)

2. ABD is a triangle in which A = 90ºand seg AC hypotenuse BD.Show that :(i) AB2 = BC . BD(ii) AD2 = BD . CD(iii)AC2 = BC . CD (3 marks)

Proof :(i) In ABD,m BAD = 90º [Given]


SCHOOL SECTION42

M

L R N

K

A

B C

D

seg AC hypotenuse BD [Given] BAD ~ BCA ~ ACD .....(i) [Similarity in right angled triangles]

BAD ~ BCA [From (i)]

AB

BC =

BD

AB[c.s.s.t.]

AB2 = BC × BD

(ii) BAD ~ ACD [From (i)]

AD

CD =

BD

AD[c.s.s.t.]

AD2 = BD × CD

(iii)BCA ~ ACD [From (i)]

AC

CD =

BC

AC[c.s.s.t.]

AC2 = BC × CD


4. In the adjoining figure,LMN = 90º and LKN = 90º,seg MK seg LN. Prove thatR is the midpoint of seg MK. (3 marks)

Proof : In LMN,m LMN = 90º [Given]seg MR hypotenuse LN [Given]

MR2 = LR × RN .....(i) [By property of geometric mean]In LKN,m LKN = 90º [Given]seg KR hypotenuse LN [Given]

KR2 = LR × RN .....(ii) [By property of geometric mean] MR2 = KR2 [From (i) and (ii)] MR = KR [Taking square roots] R is the midpoint of seg MK

THEOREM OF PYTHAGORAS

Statement : In a right angled triangle, the square of the hypotenuse is equalto the sum of the squares of the remaining two sides. (4 marks)Given : In ABC, m ABC = 90º

To Prove : AC² = AB² + BC² Construction : Draw seg BD side AC such that A - D - C. Proof : In ABC,

m ABC = 90º [Given]seg BD hypotenuse AC [Construction]

ABC ~ ADB ~ BDC .....(i) [Similarity in right angled triangles]ABC ~ ADB [From (i)]

AB

AD=

AC

AB[Corresponding sides of similar triangles]

AB² = AC × AD ........(ii)ABC ~ BDC [From (i)]

BC

DC=

AC

BC[Corresponding sides of similar triangles]

BC² = AC × DC ........(iii)


SCHOOL SECTION 43

A

B C

A

B C

Dx

y

z

4

5

Adding (ii) and (iii) we get,AB² + BC² = AC × AD + AC × DC

AB² + BC² = AC (AD + DC) AB² + BC² = AC × AC [ A - D - C] AB² + BC² = AC² AC² = AB² + BC²


3. A ladder 10 m long reaches a window 8 m above the ground. Find thedistance of the foot of the ladder from the base of the wall. (2 marks)

Sol. In the adjoining figure,seg AB represents the wallseg AC represents the ladderseg BC represents the distance of thefoot of the ladder from the base of the wallAC = 10 mAB = 8 mIn ABC,m ABC = 90º [Given]AC2 = AB2 + BC2

(10)2 = (8)2 + BC2

100 = 64 + BC2

BC2 = 100 – 64 BC2 = 36 [Taking square roots] BC = 6 m

The distance of the foot of the ladder from the base of the wall is 6 m.


11. In the adjoining figure,an altitude is drawn tothe hypotenuse. The lengthsof different segments are markedin each figure. Determine thevalues of x, y, z in each case. (3 marks)

Sol. (i) In ABC,mABC = 90º [Given]seg BD hypotenuse AC [Given]

BD2 = AD × DC [By property of geometric mean] y2 = 4 × 5 y = 4 × 5 [Taking square roots]

y = 2 5

In ADB,m ADB = 90º [Given]AB2 = AD2 + BD2 [By Pythagoras theorem]

AB2 = 42 + y2

x2 = 42 + (2 5 )2

x2 = 16 + 20 x2 = 36

x = 6 [Taking square roots]


SCHOOL SECTION44

In BDC,m BDC = 90º

BC2 = BD2 + CD2 [By Pythagoras theorem] z2 = y2 + 52

z2 = (2 5 )2 + 52

z2 = 20 + 25 z2 = 45

z = 9 × 5 [Taking square roots]

z = 3 5

(ii) In PSQ,m PSQ = 90º [Given]

PQ2 = PS2 + QS2 [By Pythagoras theorem] 62 = 42 + y2

36 = 16 + y2

y2 = 36 – 16 y2 = 20

y = 4 5 [Taking square roots]

y = 2 5In PQR,m PQR = 90º [Given]seg QS hypotenuse PR [Given]

QS2 = PS × SR [By property of geometric mean] y2 = 4 × x

( 2 5 )2 = 4 × x 20 = 4 × x

x =20

4 x = 5

In QSR,m QSR = 90º [Given]

QR2 = QS2 + SR2 [By Pythagoras theorem] z2 = y2 + x2

z2 = (2 5 )2 + (5)2

z2 = 20 + 25 z2 = 45

z = 9 × 5 [Taking square roots]

z = 3 5


2. Find the side of square whose diagonal is 16 2 cm. (2 marks)

Given : ABCD is a square.

AC = 16 2 cmTo find : Side of a squareSol. ABCD is a square [Given]

Let the sides of the square be x cmIn ABC,m ABC = 90º [Angle of a square]

P

Q R

S6

y

z

4

X

A D

B C

16 2 cm

x

x

x

x


SCHOOL SECTION 45

AC2 = AB2 + BC2 [By Pythagoras theorem]

16 22

= x2 + x2

256 × 2 = 2x2

x2 =256 2

2

x2 = 256 x = 16 [Taking square roots]

The side of a square is 16 cm.


4. Find the perimeter of an isosceles right triangle with each of itscongruent sides is 7 cm. (2 marks)

Given : In ABC,m ABC = 90ºAB = BC = 7 cm

To find : Perimeter of ABCSol. In ABC,

m ABC = 90º [Given] AC2 = AB2 + BC2 [By Pythagoras theorem] AC2 = (7)2 + (7)2

AC2 = 49 + 49 AC2 = 98 AC = 49 2 [Taking square roots]

AC = 7 2 cmPerimeter of ABC = AB + BC + AC

= 7 7 7 2

= 14 7 2

Perimeter of ABC = 7 2 2 cm


4. Prove that the sum of the squares of the sides of a rhombus is equal tothe sum of the squares of its diagonals. (4 marks)

Given : (i) ABCD is a rhombus(ii) Diagonals AC and BDintersect each other at point O.

To prove : AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Proof : ABCD is a rhombus [Given] AB = BC = CD = AD .....(i) [Sides of a rhombus] m AOB = 90º ....(ii)

[Diagonals of a rhombus areAO = 1

2 AC ....(iii)

perpendicular bisectors of each other]

BO = 1

2 BD .....(iv)

In AOB,m AOB = 90º [From (ii)]

AB2 = AO2 + BO2 [By Pythagoras theorem]

A

B C7 cm

7 cm

A D

B C

O


SCHOOL SECTION46

AB2 = 1 1

AC BD2 2

2 2

[From (iii) and (iv)]

AB2 = 1

4AC2 +

1

4BD2

Multiplying throughout by 4, 4AB2 = AC2 + BD2

AB2 + AB2 + AB2 + AB2 = AC2 + BD2

AB2 + BC2 + CD2 + AD2 = AC2 + BD2 [From (i)]


5. The perpendicular AD on thebase BC of ABC intersectsBC at D so that BD = 3 CD.Prove that 2AB2 = 2AC2 + BC2. (3 marks)

Proof : BC = BD + CD [B - D - C] BC = 3CD + CD [ BD = 3CD] BC = 4CD .....(i)

In ADB,m ADB = 90º [Given]

AB2 = AD2 + BD2 [By Pythagoras theorem] AB2 = AD2 + (3CD)2 [ BD = 3CD] AB2 = AD2 + 9CD2

AB2 = AD2 + CD2 + 8CD2 .....(ii)In ADC,m ADC = 90º [Given]

AC2 = AD2 + CD2 .....(iii) [By Pythagoras theorem]AB2 = AC2 + 8CD2 [From (ii) and (iii)]

AB2 = AC2 + 8BC

4

2

[From (i)]

AB2 = AC2 + 8 × BC

16

2

AB2 = AC2 + BC

2

2

2AB2 = 2AC2 + BC2 [Multiplying throughout by 2]


8. ABC is a triangle in which AB = ACand D is any point on BC.Prove that : AB2 – AD2 = BD. CD. (4 marks)

Proof : In AEB,m AEB = 90º [Given]

AB2 = AE2 + BE2 ......(i) [By Pythagoras theorem]In AED,m AED = 90º [Given]

AD2 = AE2 + DE2 .....(ii) [By Pythagoras theorem]Subtracting equation (ii) from (i),AB2 – AD2 = AE2 + BE2 – (AE2 + DE2)

AB2 – AD2 = AE2 + BE2 – AE2 – DE2

AB2 – AD2 = BE2 – DE2

AB2 – AD2 = (BE + DE) (BE – DE) AB2 – AD2 = (BE + DE) × BD .......(iii) [ B - D - E]

A

DB C

A

BD E

C


SCHOOL SECTION 47

In AEB and AEC,m AEB = m AEC = 90º [Given]hypotenuse AB hypotenuse AC [Given]seg AE seg AE [Common side]

AEB AEC [By hypotenuse side theorem] seg BE seg CE .....(iv) [c.s.c.t.] AB2 – AD2 = (CE + DE) × BD [From (iii) and (iv)] AB2 – AD2 = CD × BD [ D - E - C]


5. Seg AD is the median of ABC,and AM BC. Prove that :

(i) AC2 = AD2 + BC × DM +

2BC2

(ii) AB2 = AD2 – BC × DM +

2BC2

(4 marks)

Proof :(i) In AMD,m AMD = 90ºAD2 = AM2 + DM2 ......(i) [By Pythagoras theorem]In AMC,m AMC = 90º [Given]

AC2 = AM2 + MC2 [By Pythagoras theorem] AC2 = AM2 + (DM + DC)2 [ M - D - C] AC2 = AM2 + DM2 + 2DM × DC + DC2

AC2 = AM2 + DM2 + 2DC × DM + DC2

AC2 = AD2 + BC × DM +

2BC2

[From (i) and D is the midpoint

of side BC]

(ii) In AMB,m AMB = 90º [Given]

AB2 = AM2 + BM2 [By Pythagoras theorem] AB2 = AM2 + (BD – DM)2

AB2 = AM2 + BD2 – 2BD × DM + DM2

AB2 = AM2 + DM2 – 2BD × DM + BD2

AB2 = AD2 – BC × DM +

2BC2

[From (i) and D is the midpoint

of side BC]


6. ABC is a triangle where C = 90º. Let BC = a, CA = b, AB = c and let ‘p’be the length of the perpendicular from C on AB. Prove that (i) cp = ab,

(ii) 2 2 2

1 1 1= +

p a b (5 marks)

Proof : Area of a triangle =1

2 × base × height

A (ABC) =1

2 × AB × CD

A (ABC) =1

2 × c × p ......(i)

A

B M D C

A

BC

cDb

p

a


SCHOOL SECTION48

Also, A (ABC) =1

2 × BC × AC

A (ABC) =1

2 × a × b ......(ii)

From (i) and (ii) we get,

1

2 × c × p =

1

2 × a × b

cp = ab(ii) cp = ab

1

cp =1

ab[By Invertendo]

1

c p2 2 =1

a b2 2 [Squaring both sides]

1

p2 =c

a b

2

2 2 .....(iii)

In ACB,m ACB = 90º [Given]

AB2 = AC2 + BC2 [By Pythagoras theorem] c2 = b2 + a2 .....(iv)

1

p2 =b a

a b

2 2

2 2 [From (iii) and (iv)]

1

p2 = b a

a b a b

2 2

2 2 2 2

2

1p = 2 2

1 1+

a b


12. ABC is a right angled trianglewith A = 90º. A circle is inscribedin it. The lengths of the sidescontaining the right angle are 6 cmand 8 cm. Find the radius of the circle. (4 marks)

Construction : Let the sides AB, BC and ACtouch the circle at points L, M and Nrespectively. Draw seg OA, seg OB,seg OC, seg OL, seg OM and ON.

Sol. In ABC,m BAC = 90º [Given]

BC2 = AC2 + AB2 [By Pythagoras theorem] BC2 = 62 + 82

BC2 = 36 + 64 BC2 = 100 BC = 10 cm .......(i) [Taking square roots]

Let the radius of the circle be rOL = OM = ON = r .......(ii) [Radii of the same circle]seg OL side ABseg OM side BC [Radius is perpendicular to the tangent]seg ON side AC

C

A L B

M

ON

6 cm

8 cm


SCHOOL SECTION 49

Q

P

RB C

A

Area of a triangle =1

2 × base × height

A (AOB) =1

2 × AB × OL

A (AOB) =1

2 × 8 × r .....(iii) [From (ii)]

A (AOB) = 4rSimilarly,

A (BOC) = 5r .....(iv)A (AOC) = 3r .....(v)

Adding (iii), (iv) and (v),A (AOB) + A (BOC) + A (AOC) = 4r + 5r + 3r [From (iii), (iv) and (v)]

A (ABC) = 12r [Area addition property)

1

2 × AB × AC = 12r

1

2 × 8 × 6 = 12r

24 = 12r r = 2

The radius of the circle is 2 cm.

CONVERSE OF THEOREM OF PYTHAGORAS

Statement : In a triangle, if the square of one side is equal to the sum of thesquares of the remaining two sides, then the angle opposite to the first sideis a right angle and the triangle is right angled triangle. (4 marks)Given : In ABC,

AC² = AB² + BC²To Prove : m ABC = 90ºConstruction : Draw PQR such

that PQ = AB, QR = BC andm PQR = 90º

Proof : QR = BC .......(i)[Construction]

PQ = AB ......(ii)In PQR,m PQR = 900 ......(iii) [Construction]

PR² = PQ² + QR² [By Pythagoras theorem] PR² = AB² + BC² .......(iv) [From (i) and (ii)]

But,AC² = AB² + BC² .......(v) [Given]

PR² = AC² [From (iv) and (v)] PR = AC .......(vi) [Taking square roots]

In ABC and PQR,side AB side PQ [From (ii)]side BC side QR [From (i)]side AC side PR [From (vi)]

ABC PQR [SSS test of congruence] ABC PQR [c.a.c.t] m ABC = m PQR = 90º [From (iii)] m ABC = 90º


SCHOOL SECTION50


1. Sides of triangles are given below. Determine which of them are rightangled triangles. (2 marks)

(i) 8, 15, 17Sol. (17)2 = 289 ......(i)

(8)2 + (15)2 = 64 + 225= 289 ......(ii)

(17)2 = (8)2 + (15)2 [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of

Pythagoras theorem]

(ii) 9, 40, 41Sol. (41)2 = 1681 ......(i)

(9)2 + (40)2 = 81 + 1600= 1681 ......(ii)


Pythagoras theorem]

(iii) 40, 20, 30Sol. (40)2 = 1600 ......(i)

(20)2 + (30)2 = 400 + 900= 1300 ......(ii)

(40)2 (20)2 + (30)2 [From (i) and (ii)] The given sides do not form a [By Converse of

right angled triangle. Pythagoras theorem]

(iv) 11, 60, 61Sol. (61)2 = 3721 ......(i)

(11)2 + (60)2 = 121 + 3600= 3721 ......(ii)


Pythagoras theorem]

(v) 11, 12, 15Sol. (15)2 = 225 ......(i)

(11)2 + (12)2 = 121 + 144= 265 ......(ii)

(15)2 (11)2 + (12)2 [From (i) and (ii)] The given sides do not form a [By Converse of

right angled triangle. Pythagoras theorem]

(vi) 12, 35, 37Sol. (37)2 = 1369 ......(i)

(12)2 + (35)2 = 144 + 1225= 1369 ......(ii)


Pythagoras theorem]


SCHOOL SECTION 51

THEOREM OF 300 - 600 - 900 TRIANGLE

B C

A

300

600

THEOREM OF 450 - 450 - 900 TRIANGLE

A

B C

450

450

If the angles of a triangle are 300, 600 and 900, then the side oppositeto 300 is half of the hypotenuse and the side opposite to 600 is

32

times the hypotenuse.

In ABC,m A = 30º, m C = 60º and m B = 90º

BC = 1

2 AC and AB =

3

2AC.

If the angles of a triangle are 45º - 45º - 90º then the length of the

perpendicular sides are 1

2 times the hypotenuse.

In ABC,m A = 45ºm C = 45ºm B = 90º

AB = BC = 1

2 AC.


1. Find the length of the altitude of an equilateral triangle, each side measuring‘a’ units. (3 marks)

Given : ABC is an equilateral triangle.AB = BC = AC = aseg AD side BC

To find : ADSol. ABC is an equilateral triangle

AB = BC = AC = a ......(i) [Given]In ADB,m ADB = 90º [Given]m ABD = 60º [Angle of an equilateral triangle]

m BAD = 30º [Remaining angle] ADB is a 30º - 60° - 90º triangle By 30º - 60º - 90º triangle theorem,

AD = 3

2 × AB [Side opposite to 60º]

AD = 3

2 × a [From (i)]

AD = 3

2a units.

A

BD

C

aa

a


SCHOOL SECTION52


5. In the adjoining figure,PQR = 60º andray QT bisects PQR.seg BA ray QP andseg BC ray QR.If BC = 8 find theperimeter of ABCQ. (4 marks)

Sol. Ray QT is the angle bisector of PQR [Given]B lies on ray QTseg BA ray QP [Given]seg BC ray QR

BA = BC [Angle bisector theorem]But, BC = 8 units .......(i) [Given]

BA = 8 units ......(ii)m PQR = 60º [Given]

m PQT = m RQT = 1

2 × 60º [ Ray QT bisects QR]

m PQT = m RQT = 30º ......(iii)In BAQ,m BAQ = 90º [Given]m AQB = 30º [From (iii), P - A - Q and Q - B - T]

m ABQ = 60º [Remaining angle] BAQ is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem

AB =1

2BQ [Side opposite to 30º]

8 =1

2BQ [From (ii)]

BQ = 16 units

AQ =3

2 × BQ [Side opposite to 60º]

AQ =3

2 × 16

AQ = 8 3 units ......(iv)

Similarly, QC = 8 3 cm ......(v) Perimeter of ABCQ = AB + BC + QC + AQ

= 8 8 8 3 8 3 [From (i), (ii), (iv) and (v)]

= 16 16 3

Perimeter of ABCQ = 16 1 3 units



if LK = 6 3 find MK, ML, KN, MN

and the perimeter of MNKL. (4 marks)

Sol. In MLK,m MLK = 90º [Given]

P

T

RCQ

A

B

N

M

LK

45º

30º

6 3


SCHOOL SECTION 53

m MKL = 30º [Given] m LMK = 60º [Remaining angle] MLK is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem,

LK =3

2MK [Side opposite to 60º]

6 3 =3

2 × MK [Given]

MK =6 3 2

3

MK = 12 units ......(i)

ML =1

2 × MK [Side opposite to 30º]

ML =1

2 × 12

ML = 6 units ......(ii)In MKN,m MKN = 90º [Given]m MNK = 45º [Given]

m NMK = 45º [Remaining angle] MKN is a 45º - 45º - 90º triangle

By 45º - 45º - 90º triangle theorem,

MK = KN = 1

2 × MN ......(iii)

MK =1

2 × MN [From (iii)]

12 =1

2 × MN

MN = 12 2 units .....(iv)KN = 12 units ......(v) [From (i) and (iii)]

Perimeter of MNKL = MN + KN + KL +ML= 12 2 12 6 3 6

[From (ii), (iv) and (v) and given]= 18 12 2 6 3

Perimeter of MNKL = 6 3 2 2 3 units


7. In the adjoining figure,PQRV is a trapezium inwhich seg PQ || seg VR.SR = 4 and PQ = 6. Find VR.

(4 marks)Sol. In QSR,

m QSR = 90º [Given]m SQR = 45º [Given]

m QRS = 45º [Remaining angle] QSR is a 45º - 45º - 90º triangle. QS = SR [Congruent sides of 45º - 45º - 90º

triangle]

P Q

T SV R

45º60º

4

6


SCHOOL SECTION54

But, SR = 4 units ......(i) [Given] QS = 4 units .....(ii)

In PQTS,seg PQ || seg TS [Given and V - T - S - R]m T = m S = 90º [Given]

PQTS is a rectangle PQ = TS = 6 units .....(iii) [Opposite sides of a rectangle

QS = PT = 4 units .....(iv) and from (ii) and given]

In PTV,m PTV = 90º [Given]m VPT = 60º [Given]

m PVT = 30º [Remaining angle] PTV is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem,

PT =1

2 × PV [Side opposite to 30º]

4 =1

2 × PV [From (iv)]

PV = 8 units ......(v)

VT =3

2 PV [Side opposite to 60º]

VT =3

2 × 8 [From (v)]

VT = 4 3 units .....(vi)VR = VT + TS + SR [V - T - S - R]

VR = 4 3 6 4 [From (i), (iii) and (vi)]

VR = 4 3 10

VR = 2 2 3 5 units


7. Prove that three times the square of any side of an equilateral triangleis equal to four times the square of an altitude. (3 marks)

To prove : 3AB2 = 4AD2

Proof : ABC is an equilateral triangle [Given]In ADB,m ADB = 90º [Given]m ABD = 60º [Angle of an

equilateral triangle] m BAD = 30º [Remaining angle] ADB is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem,

AD = 3

2AB [Side opposite to 60º]

2AD = 3 AB

4AD2 = 3AB2 [Squaring both sides] 3AB2 = 4AD2

A

B CD


SCHOOL SECTION 55

A

B C

5 10

5 3

APPOLLONIUS THEOREM

A

CB D

A

B C

36 cm

18 3 cm

18 cm

CONVERSE OF 30º - 60º - 90º TRIANGLE THEOREM

Statement : In a right angled triangle, if the length of one side is 3

2 time

the hypotenuse, then the measure of the angle opposite to the side is 60º.In ABC,m ABC = 90º [Given]

AB = 1

2 AC

m C = 30º

BC = 3

2 AC

m A = 60º


3. If the sides of a triangle measure 18 cm, 18 3 cm and 36 cm, showthat it is a 30º - 60º - 90º triangle. (3 marks)

Given : In ABC,AC = 36 cm, AB = 18 cm, BC = 18 3 cm

To prove : ABC is a 30º - 60º - 90º triangle.Proof : AC2 = (36)2

AC2 = 1296 .......(i)

AB2 + BC2 = (18)2 + 18 32

= 324 + 972 AB2 + BC2 = 1296 ......(ii) AC2 = AB2 + BC2 [From (i) and (ii)] ABC is a right angled [Converse of Pythagoras

triangle at point B .....(iii) theorem]1

2 × AC =

1

2 × 36 [ AC = 36]

1

2 × AC = 18

1

2× AC = AB [ AB = 18]

m C = 30º .....(iv) [By converse of 30º - 60º - 90ºtriangle theorem]

In ABC,m B = 90º [From (iii)]m C = 30º [From (iv)]

m A = 60º [Remaining angle] ABC is a 30º - 60º - 90º triangle.

Appollonius theorem is a theorem relating the length of a median of atriangle to the lengths of its sides. Let us study how to make use ofAppollonius theorem in the given figure.

In ABC,seg AD is a median,

By Appollonius theorem,AB² + AC² = 2AD² + 2BD²

orAB² + AC² = 2ADB + 2DC²


SCHOOL SECTION56


1. In ABC, AP is a median.If AP = 7, AB2 + AC2 = 260then find BC. (2 marks)

Sol. In ABC,seg AP is the median [Given]

AB2 + AC2 = 2AP2 + 2BP2 [By Appollonius theorem] 260 = 2 (7)2 + 2BP2 [Given] 260 = 2 (49) + 2BP2

260 = 98 + 2BP2

260 – 98 = 2BP2

2BP2 = 162

BP2 =162

2 BP2 = 81 BP = 9 units [Taking square roots]

BP =1

2 BC [ P is the midpoint of seg BC]

9 =1

2BC

BC = 18 units


2. In the adjoining figure,AB2 + AC2 = 122, BC = 10.Find the length of the medianon side BC. (2 marks)

Sol. In ABC,seg AQ is the median [Given]

BQ = QC =1

2 × BC

BQ = QC =1

2 × 10 [Given]

BQ = QC = 5 units ......(i)AB2 + AC2 = 2AQ2 + 2BQ2 [By Appollonius theorem]

122 = 2AQ2 + 2 (5)2 [From (i) and given] 122 = 2AQ2 + 2 (25) 122 = 2AQ2 + 50 2AQ2 = 122 – 50 2AQ2 = 72 AQ2 = 36

AQ = 6 units [Taking square roots]


13. In PQR, seg PM is the median. If PM = 9 and PQ2 + PR2 = 290. Find QR.(2 marks)

Sol. In PQR,seg PM is the median [Given]

PQ2 + PR2 = 2PM2 + 2QM2 [By Appolloniustheorem]

290 = 2 (9)2 + 2QM2 [Given] 290 = 2 (81) + 2QM2

290 = 162 + 2QM2

A

B PC

A

B Q C

P

Q M R

9


SCHOOL SECTION 57

290 – 162 = 2QM2

128 = 2QM2

QM2 =128

2 QM2 = 64 QM = 8 units [Taking square roots]

QM =1

2 QR [ M is midpoint of side QR]

8 =1

2QR

8 × 2 = QR

QR = 16 units


14. From the information given in the figure show that PM = PN = 3 a .(3 marks)

Proof :

MQ = QR = a [Given] Q is the midpoint of seg MR .......(i)

In PMR,seg PQ is the median [From (i) and by definition]

PM2 + PR2 = 2PQ2 + 2MQ2 [By Appollonius theorem] PM2 + a2 = 2 (a)2 + 2 (a)2

PM2 + a2 = 2a2 + 2a2

PM2 + a2 = 4a2

PM2 = 4a2 – a2

PM2 = 3a2

PM = 3 a .......(ii) [Taking square roots]Similarly we can prove,

PN = 3 a ......(iii)

PM = PN = 3 a [From (ii) and (iii)]


3. Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length ofone of its diagonals is 26 cm. Find the length of the other. (3 marks)

Sol. ABCD is a parallelogram [Given]

OB = OD =1

2 × BD .....(i) [ Diagonals of paralle log ram

bisec t each other]

OB = OD =1

2 × 26 [Given]

OB = OD = 13 cm

P

a a

a aaM Q RS N

A D

B C

17 cm

11 cmO


SCHOOL SECTION58

In ADB,seg AO is the median [From (i) and by definition]

AB2 + AD2 = 2AO2 + 2OB2 [By Appollonius theorem] (11)2 + (17)2= 2AO2 + 2 (13)2

121 + 289 = 2AO2 + 2 (169) 410 = 2AO2 + 338 410 – 338 = 2AO2

72 = 2AO2

AO2 = 36 AO = 6 cm [Taking square roots]

AO =1

2 × AC [ Diagonals of paralle log ram

bisec t each other]

6 =1

2 × AC

AC = 12 cm

Length of other diagonal is 12 cm.


6. In the adjoining figure,PQR = 90º.T is the mid point of the side QR.Prove that PR2 = 4PT2 – 3PQ2. (3 marks)

Proof : In PQR,seg PT is the median [Given]

PQ2 + PR2 = 2PT2 + 2QT2 ......(i) [By Appollonius theorem]

In PQT,m PQT = 90º [Given]

PT2 = PQ2 + QT2 [By Pythagoras theorem] QT2 = PT2 – PQ2 .....(ii)

PQ2 + PR2 = 2PT2 + 2 (PT2 – PQ2) [From (i) and (ii)] PQ2 + PR2 = 2PT2 + 2PT2 – 2PQ2

PR2 = 4PT2 – 2PQ2 – PQ2

PR2 = 4PT2 – 3PQ2

ALTERNATIVE METHOD :In PQR,m PQR = 90º [Given]

PR2 = PQ2 + QR2 [By Pythagoras theorem] PR2 = PQ2 + (2QT)2 [ T is the midpoint of side QR] PR2 = PQ2 + 4QT2 .....(i)

In PQT,m PQT = 90º [Given]

PT2 = PQ2 + QT2 [By Pythagoras theorem] QT2 = PT2 – PQ2 .....(ii) PR2 = PQ2 + 4 (PT2 – PQ2) [From (i) and (ii)] PR2 = PQ2 + 4PT2 – 4PQ2

PR2 = 4PT2 – 3PQ2

P

QT

R


SCHOOL SECTION 59

HOTS PROBLEM(Problems for developing Higher Order Thinking Skill)

1. In ABCD, side BC || side AD.side AC and side BD intersect in

point Q. If AQ = 13

AC then

show that DQ = 12

BQ. (5 marks)

Proof : side AD || side BC [Given] On transversal BD

ADB CBD [Converse of alternate angles test] ADQ CBQ .........(i) [ B - Q - D]

In AQD and CQB,AQD CQB [Vertically opposite angles]ADQ CBQ [From (i)]

AQD ~ CQB [By A-A test of similarity]

AQ

QC =DQ

BQ .........(ii) [c.s.s.t.]

But, AQ =1

3AC [Given]

3AQ = AC 3AQ = AQ + QC [ A - Q - C] 3AQ – AQ = QC 2AQ = QC

AQ

QC =1

2.........(iii)

1

2=

DQ

BQ [From (ii) and (iii)]

DQ =12

BQ

2. A line cuts two sides AB andside AC of ABC in points P andQ respectively then show that

A ( APQ) AP × AQ=

A ( ABC) AB × AC (2 marks)

Construction : Draw seg BQProof : APQ and ABQ have a common vertex Q and their bases AP and AB lie

on the same line AB.

A ( APQ)

A ( ABQ)

=

AP

AB......(i) [Triangles having equal heights]

ABQ and ABC have a common vertex B and their bases AQ and AClie on the same line AC


A ( ABQ)

A ( ABC)

=

AQ

AC......(ii) [Triangles having equal heights]

Multiplying (i) and (ii),

A D

Q

B C

A

P

Q

B C


SCHOOL SECTION60

CE

DA B

A ( APQ)

A ( ABQ)

×

A ( ABQ)

A ( ABC)

=

AP

AB ×

AQ

AC

A ( APQ)A ( ABC)

=

AP × AQAB × AC

3. In the adjoining figure,AD is the bisector of theexterior A of ABC.seg AD intersects theside BC produced in D.

Prove that BD AB

=CD AC

. (5 marks)

Construction : Draw seg CE || seg AD such that B - E - A.Proof : In ABD,

seg EC || side AD [Construction]

BC

CD =

BE

EA[By B.P.T.]

BC + CD

CD =

BE + EA

EA[By Componendo]

BD

CD =

AB

AE......(i) [B - C - D and B - E - A]

seg EC || seg AD [Construction] on transversal EK,

KAD AEC ......(ii) [Converse of corresponding angles test] on transversal AC,

DAC ACE ......(iii) [Converse of alterante angles test]But,KAD DAC .......(iv) [ ray AD is the bisector of KAC]

AEC ACE ......(v) [From (ii), (iii) and (iv)]In AEC,

AEC ACE [From (v)] AE = AC ......(vi) [Converse of isosceles triangle theorem]

BDCD

= ABAC

[From (i) and (vi)]

4. In ABC, ACB = 90º,seg CD seg ABseg DE seg CBshow that : CD2 × AC = AD × AB × DE. (2 marks)

Proof : In ACB,m ACB = 90º [Given]seg CD hypotenuse AB

CD2 = AD × DB .....(i) [Property of geometric mean]In ACB and DEB,ACB DEB [ Each is 90º]

ABC DBE [Common angle] ACB DEB [By AA test of similarity]

AC

DE =

AB

DB[c.s.s.t.]

AC × DB = AB × DE

K

A

B CD

••

E


SCHOOL SECTION 61

AC = AB × DE

DB.....(ii)

Multiplying (i) and (ii),

CD2 × AC = AD × DB × AB × DE

DB CD2 × AC = AD × AB × DE

5. In an equilateral triangle ABC,the side BC is trisected at D. Prove that9AD2 = 7AB2. (Hint : Draw AE BC)

(2 marks)

Construction :Draw seg AE side BC,such that B - D - E - C.

Proof : ABC is an equilateral triangle. [Given] AB = BC = AC ........(i) [Sides of an

equilateral triangle]In AED,m AED = 90º [Construction]

AD² = AE² + DE² .......(ii) [By Pythagoras Theorem]In AEB,m AEB = 90º [Construction]m ABE = 60º [Angle of an equilateral triangle]

m BAE = 30º [Remaining angle] AEB is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem,

AE = 3

2AB ......(iii) [Side opposite to 60º]

BE = 1

2AB ......(iv) [Side opposite to 30º]

DE = BE – BD [ B - D - E]

DE = 1

2 AB –

1

3BC [From (iv) and Given]

DE = 1

2 AB –

1

3 AB [From (i)]

DE = 3AB 2AB

6

DE = 1

6AB .......(v)

AD² = 3

AB2

2

+ 1

AB6

2

[From (ii), (iii) and (v)]

AD² = 3

4AB² +

1

36AB²

AD² = 27AB + AB

36

² ²

AD² = 28AB

36

²

AD² = 7

9 AB²

9AD² = 7AB²

A

B D E C


SCHOOL SECTION62

6. In the adjoining figure,each of segments PA, QB, RC andSD is perpendicular to line l.If AB = 6, BC = 9, CD = 12 andPS = 36, then determine PQ, QRand RS.

Sol. seg PA line lseg QB line l [Given]seg RC line lseg SD line l

seg PA || seg QB || seg RC || seg SD ......(i)[Perpendiculars drawn to the sameline are parallel to each other]

AD = AB + BC + CD [A - B - C - D] AD = 6 + 9 + 12 AD = 27 units .....(ii)

seg PA || seg QB || seg SD [From (i)]On transversals AD and PS,AB

AD=

PQ

PS

[By property of intercepts made

by three parallel lines]

6

27=

PQ

36[From (ii) and given]

PQ =36 × 6

27 PQ = 8 units ......(iii)

seg PA|| seg QB || seg RC [From (i)]On transversals AD and PS,AB

BC=

PQ

QR

[By property of intercepts made


6

9=

8

QR [Given and from (iii)]

QR =9 × 8

6

QR = 12 units ......(iv)

PS = PQ + QR + RS [ P - Q - R - S] 36 = 8 + 12 + RS [From (iii), (iv) and given] 36 = 20 + RS RS = 36 – 20

RS = 16 units

7. In ABC, ABC = 135º. Prove that AC2 = AB2 + BC2 + 4A (ABC).Proof : m ABC + m ABD = 180º [Angles forming linear pair]

135º + m ABD = 180º m ABD = 180º – 135º m ABD = 45º .....(i)

In ADB,m ADB = 90º [Given]m ABD = 45º [From (i)]

m BAD = 45º ......(ii) [Remaining angle]In ABD,ABD BAD [From (i) and (ii)]

seg AD seg DB .....(iii) [Converse isosceles triangle theorem]

SRQ

P

l A B C D

A

D B C

135º


SCHOOL SECTION 63

In ADB,m ADB = 90º [Construction]

AB2 = AD2 + DB2 .....(iv) [By Pythagoras theorem]In ADC,m ADC = 90º [Construction]

AC2 = AD2 + DC2 [By Pythagoras theorem] AC2 = AD2 + (DB + BC)2 [ D - B - C] AC2 = AD2 + DB2 + 2 × DB × BC + BC2

AC2 = AB2 + BC2 + 2 × DB × BC[From (iv)]

AC2 = AB2 + BC2 + 2 × AD × BC.....(v) [From (iii)]

Area of triangle = 1

2 × base × height

A (ABC) = 1

2 × BC × AD

4A (ABC) = 4 × 1

2 × BC × AD [Multiplying throughout by 4]

4A (ABC) = 2 × AD × BC ......(vi) AC2 = AB2 + BC2 + 4A (ABC) [From (v) and (vi)]

8. PQR is a right triangle, right angledat Q such that QR = b anda = A (PQR). If QN PR then

show that QN = 4 2

2a . b

b + 4a

Proof : Area of triangle =1

2 × base × height

A (PQR) =1

2 × QR × PQ

a =1

2 × b × PQ [Given]

2a

b= PQ .....(i)

Also,

A (PQR) =1

2 × PR × QN

a =1

2 × PR × QN [Given]

QN =2a

PR.....(ii)

In PQR,m PQR = 90º [Given]

PR2 = PQ2 + QR2 [By Pythagoras theorem]

PR = PQ QR2 2 [Taking square roots]

PR = 2a

bb

22

[From (i) and given]

P

Q R

N


SCHOOL SECTION64

PR = 4a

bb

2

22

PR = 4a b

b

2 4

2

PR = b 4a

b

4 2

......(iii)

QN =

2a

b 4a

b

4 2[From (ii) and (iii)]

QN = 4 2

2a . b

b + 4a

15. ABCD is an quadrilateral M is the midpoint of diagonal AC and N is themidpoint of diagonal BD. Prove that : AB2 + BC2 + CD2 + DA2 = AC2 + BD2 +4MN2.

Proof :

In ABC,seg BM is the median [ M is midpoint of side AC]

AB2 + BC2 = 2BM2 + 2CM2 ......(i) [By Appollonius theorem]In ADC,seg DM is the median [ M is midpoint of side AC]

CD2 + DA2 = 2DM2 + 2CM2 ......(ii)[By Appollonius theorem]

Adding (i) and (ii),AB2 + BC2 + CD2 + DA2 = 2BM2 + 2CM2 + 2DM2 + 2CM2

AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + 4CM2

AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + 4 1

AC2

2

[ M is the midpoint of seg AC]

AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + 4 × 1

4AC2

AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + AC2 ......(iii)In BMD,seg MN is the median [ N is the midpoint of side BD]

BM2 + DM2 = 2MN2 + 2BN2 [By Appollonius theorem] 2BM2 + 2DM2 = 4MN2 + 4BN2 [Multiplying throughout by 2]

2BM2 + 2DM2 = 4MN2 + 4×1

BD2

2

[ N is the midpoint of side BD]

2BM2 + 2DM2 = 4MN2 + 4 × 1

4BD2

2BM2 + 2DM2 = 4MN2 + BD2 ......(iv)

AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4MN2 [From (iii) and (iv)]

A D

N M

B C


SCHOOL SECTION 65

A D E

M

CB

L

17. Through the midpoint M of the side CD of parallelogram ABCD, the line BMis drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.

Proof : In ELA and BLC,AEL CBL [From (i) and E - L - B]ALE CLB [Vertically oppsoite angles]

ELA ~ BLC [By AA test of similarity]

EL

BL =

EA

BC.....(i) [c.s.s.t.]

ABCD is a parallelogram [Given] seg AD || seg BC [By definition] seg AE || seg BC [ A - D - E]

On transversal BE,AEB CBE .....(ii) [Converse of alternate angles test]In DME and CMB,side DM side CM [Given]DME CMB [Vertically opposite angles]DEM CBM [From (ii) and A - D - E, B - M - E]

DME CMB [By SAA test of congruence] DE = BC .....(iii) [c.s.c.t.]

But, AD = BC .....(iv) [Opposite sides of a parallelogram] DE = BC = AD .....(v) [From (iii) and (iv)]

EL

BL =

ED DA

BC

[From (i) and E - D - A]

EL

BL =

BC BC

BC

[From (v)]

EL

BL =

2BC

BC

EL

BL =

2

1 EL = 2BL

MCQ’s1. In ABC, AB = 3 cm, BC = 2 cm and AC = 2.5 cm. DEF ~ ABC, EF = 4 cm.

What is the perimeter of DEF ?(a) 30 cm (b) 22.5 cm(c) 15 cm (d) 7.5 cm

2. The sides of two similar triangles are 4 : 9. What is the ratio of their area ?(a) 2 : 3 (b) 4 : 9(c) 81 : 16 (d) 16 : 81

3. The areas of two similar triangles are 18 cm2 and 32 cm2 respectively.What is the ratio of their corresponding sides ?(a) 3 : 4 (b) 4 : 3(c) 9 : 16 (d) 16 : 9


SCHOOL SECTION66

4. ABC ~ PQR, AB = 6 cm, BC = 8 cm, CA = 10 cm and QR = 6 cm. What isthe length of side PR ?(a) 8 cm (b) 10 cm(c) 4.5 cm (d) 7.5 cm

5. In XYZ, ray YM is the bisector of XYZ where XY = YZ and X - M - Z, thenwhich of the relation is true ?(a) XM = MZ (b) XM MZ(c) XM > MZ (d) None

6. ABC is an equialteral triangle. seg AD side BC which of the followingrelations is true ?(a) 2AB2 = 3AD2 (b) 3AB2 = 2AD2

(c) 3AB2 = 4AD2 (d) 4AB2 = 3AD2

7. In ABC, AB = 10 cm, BC = 8 cm and AC = 6 cm. What is the type of ABC ?(a) Scalene (b) Isosceles(c) Right angled (d) Equilateral

8. ABC ~ PQR, A = 47º, Q = 83º. What is the measure of C ?(a) 50º (b) 55º(c) 60º (d) 65º

9. The sides of a triangle are 24 cm, 32 cm and 40 cm. What is the length ofthe median on the longest side ?(a) 12 cm (b) 16 cm(c) 18 cm (d) 20 cm

10. In PQR, Q = 90º, P = 60º, PR = 36 cm. Find PQ ?(a) 12 cm (b) 15 cm

(c) 18 cm (d) 18 3 cm

11. In ABC, AB = 6 cm, BC = 8 cm and AC = 10 cm. ABC is enlarged to PQRsuch that the largest side is 12.5 cm. What is the length of the smallestside of PQR ?(a) 7.5 cm (b) 9 cm(c) 8 cm (d) 10 cm

12. In ABC, B - D - C and BD = 6 cm, DC = 4 cm. What is the ratio ofA (ABC) to A (ACD) ?(a) 2 : 3 (b) 5 : 2(c) 3 : 2 (d) 5 : 3

13. In MNP, N = 90º, seg NQ side MP. If MQ = 12 cm and PQ = 27 cm. Whatis the length of side NQ ?(a) 6 cm (b) 13.5 cm(c) 19.5 cm (d) 18 cm

14. In XYZ, PQ || YZ, X - P - Y and X - Q - Z. If XP

PY =

4

13 and XQ = 4.8 cm.

What is XZ ?(a) 15.6 cm (b) 20.4 cm(c) 7.8 cm (d) 10.2 cm


SCHOOL SECTION 67

15. In ABC, P is a point on side BC such that BP = 4 cm and PC = 7 cm.A (APC) : A (ABC) = .................... .(a) 11 : 7 (b) 7 : 11(c) 4 : 7 (d) 7 : 4

16. The length of the altitude of an equilateral triangle whose side is 6 cm is.............. .(a) 3 (b) 5 3(c) 2 3 (d) 3 3

17. Among the following group which of them form the sides of a right angled triangle.(a) 20, 30, 40 (b) 11, 12, 15(c) 9, 40, 41 (d) 10, 15, 17

18. In PQR, seg RS is the bisector, of PRQ, PS = 8, SQ = 6, PR = 20 thenQR = ................. .(a) 10 (b) 15(c) 30 (d) 40

19. A man goes 15 m due east and then 20 m due north. His distance from thestarting point will be ................ .(a) 35 m (b) 5 m(c) 25 m (d) 15 m

20. In ABC, C = 90, AC = 6, BC = 8, then median CD = ............ .(a) 3.5 (b) 5(c) 4 (d) 4.5

21. In ABC, perpendicular AD from A meets BC at D. If BD = 8 cm, DC = 2 cm,AD = 4 cm then .............. .(a) ABC is isosceles (b) AC = 2AB(c) ABC is equilateral (d) ABC is right angled

22. If PQR is an equilateral triangle such that PS QR, then PS2 is ............ .

(a)3

SR2

2(b) 2SR2

(c) 3SR2 (d) 4SR2

23. In ABC, line PQ || side BC, AP = 3, BP = 6, AQ = 5 then the value of CQ is............ .(a) 20 (b) 10(c) 5 (d) 16

24. In the adjoining figure,AB2 + AC2 = 122, BC = 10then the length of AQ is ............... .

(a) 3 (b) 6(c) 12 (d) 36

25. In XYZ, Y = 90º, Z = aº, X = a + 30º then the value of ‘a’ is ............. .(a) 30 (b) 45(c) 60 (d) 90

A

B Q C


SCHOOL SECTION68

: ANSWERS :

1. (c) 15 cm 2. (d) 16 : 81

3. (a) 3 : 4 4. (d) 7.5 cm

5. (a) XM = MZ 6. (c) 3AB2 = 4AD2

7. (c) Right angled 8. (a) 50º

9. (d) 20 cm 10. (c) 18 cm

11. (a) 7.5 cm 12. (b) 5 : 2

13. (d) 18 cm 14. (b) 20.4 cm

15. (b) 7 : 11 16. (d) 3 317. (c) 9, 40, 41 18. (b) 15 m

19. (c) 25 m 20. (b) 5

21. (d) ABC is right angled 22. (c) 3SR2

23. (b) 10 24. (b) 6

25. (a) 30

1 Mark Sums1. In ABC, D is a point on side BC such that BD = 6 cm and DC = 4 cm. Find

A (ABD) : A (ADC).Sol. ABD and ADC have a common vertex A

and their bases BD and DC lie on thesame line BC

their heights are equal.

A ( ABD)

A ( ADC)

=

BD

DC

[Triangles having

equal heights]

A ( ABD)

A ( ADC)

=

6

4

A ( ABD)

A ( ADC)

=

3

2

A (ABC) : A (ADC) = 3 : 2

2. For ABC ~ PQR, state all the corresponding congruent angles.Sol. ABC ~ PQR [Given]

A PB Q [c.a.s.t.]C R

3. ABC ~ APQ, if A ( ABC)A ( APQ)

=

14

, find BCPQ .

Sol. ABC ~ APQ [Given]

A ( ABC)

A ( APQ)

=

BC

PQ

2


1

4 =

BC

PQ

2

2 [Given]

BC

PQ = 1


A

BD

C6 4


SCHOOL SECTION 69

4. In the adjoining figure,ray NQ is the bisectorof MNP. If MN = 25,NP = 40, MQ = 12.5, findthe length of seg PQ.

Sol. In MNP,ray NQ bisects MNP [Given]

MN

NP =

MQ

PQ [Property of angle bisector of a triangle]

25

40 =

12.5

PQ

5

8 =

12.5

PQ

PQ = 8 12.5

5

PQ = 8 × 2.5

PQ = 20 units

5. The ratio of the areas of two triangles with the common base is 6 : 5.Height of the larger triangle is 9 cm. Then find the corresponding heightof the smaller triangle.

Sol. Let the areas of larger and smaller triangle be A1 and A2 respectively.Let their respective heights be h1 and h2A

A1

2 =

6

5 and h1 = 9 cm [Given]

The two triangles have common base [Given]

A

A1

2 =

h

h1

2[Triangles having equal bases]

6

5 =

9

h2

h2 = 5 9

6

h2 = 15

2 h2 = 7.5

Corresponding height of the smaller triangle is 7.5 cm

6. In the adjoining figure,ABCD is a trapezium.seg AD || seg PQ || seg BCAP = 10, PB = 12, DQ = 15.Find the value of QC.

Sol. seg AD || seg PQ || seg BC[Given]On transversals AB and DC,

AP

PB =

DQ

QC

[Property of intercepts made


10

12 =

15

QC

5

6 =

15

QC

M

N

Q

P×

×

A D

P Q

B C

15

?

10

12


SCHOOL SECTION70

QC = 6 15

5

QC = 18 units

7. ABC ~ PQR. State which ratio of sides are equal to ABPQ .

Sol. ABC ~ PQR

AB

PQ = BC

QR = AC

PR[c.s.s.t.]

8. If two triangles are similar, then what is relation between their sidesand corresponding altitudes ?

Sol. If two triangles are similar, then the ratio of corresponding altitudes isequal to the ratio of their corresponding sides.

9. In PQR, m P = 90º. S is the midpoint of side QR. If QR = 10 cm, whatis the length of PS ?

Sol.

In QPR,m QPR = 90º [Given]seg PS is the medianon hypotenuse QR

PS = 1

QR2

[In a right angled triangle, the length of the

median drawn to the hypotenuse is half ofthe hypotenuse]

PS = 1

102

PS = 5 cm

10. In ABC, m A = 60º, m C = 30º and AC = 6 cm. What is the length ofside AB ?

Sol. In ABC,m A = 60ºm C = 30º

[Given]

m B = 90º [Remaining angle] ABC is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem,

AB = 1

AC2 [Side opposite to 30º]

AB = 1

62

AB = 3 cm

11. In ABC, ray AD is the bisector of BAC, B - D - C. If AB = 7.5 cm andAC = 4.5 cm find BD : DC.

Sol. In ABC,ray AD bisects BAC [Given]

AB

AC =

BD

DC[Property of angle bisector of a triangle]

Q

P R

S


SCHOOL SECTION 71

7.5

4.5 =

BD

DC

BD

DC =

75

45

BD

DC =

5

3

BD : DC = 5 : 3

12. A (PQR) = 24 cm2, the height QS is 8 cm. What is the length of side PR ?

Sol. Area of a triangle = 1

base height2

A (PQR) = 1

× PR × QS2

24 = 1

PR 82

24 = PR × 4

PR = 24

4

PR = 6 m

13. In XYZ, m X = 90º, m Y = 60º. If XZ = 5 3 cm, what is the length of YZ ?Sol. In XYZ,

m X = 90º [Given]m Y = 60º

m Z = 30º [Remaining angle] XYZ is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem,

XZ = 3

YZ2

[Side opposite to 60º]

5 3 = 3

YZ2

YZ = 5 3 2

3

YZ = 10 cm

14. If ABC ~ DEF, A (ABC) = 36 cm2, A (DEF) = 64 cm2, what is the ratioof the length of sides BC and EF ?

Sol. ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

BC

EF

2


36

64 =

BC

EF

2

2

BC

EF =

6


BC

EF =

3

4

A

BD

C

4.57.5

Q

P SR

8 cm


SCHOOL SECTION72

15. PQRS is a trapezium. seg PQ || seg SR. A (QSR) = 10 cm2. FindA (PSR).Sol. PQRS is a trapezium [Given]

seg PQ || seg PRQSR and PSR lie between thesame two parallel lines PQ and SR

their heights are equal and theyhave a common base RS

A (QSR) = A (PSR) [Triangles having equal base and equal height]But, A (QSR) = 10 cm2 [Given]

A (PSR) = 10 cm2

16. If DEF ~ MNK, DE = 5, MN = 6, find the value of A ( DEF)A ( MNK)

.

Sol. DEF ~ MNK [Given]

A ( DEF)

A ( MNK)

=

DE

MN

2

2 [Area of similar triangles]

A ( DEF)

A ( MNK)

=

5

6

2

2 [Given]

A ( DEF)

A ( MNK)

=

25

36

17. In ABC, AB = 3 cm, BC = 5 cm and AC = 4 cm. State the vertex of thetriangle which contains the right angle.

Sol. AB2 + AC2 = 32 + 42

AB2 + AC2 = 9 + 16 AB2 + AC2 = 25 ......(i)

BC2 = 52

BC2 = 25 .....(ii) AB2 + AC2 = BC2 [From (i) and (ii)] ABC is a right angled triangle [By converse of Pythagoras theorem] The vertex of ABC containing right angle is A.

18. In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR.Sol. In PQR,

m P = 30ºm R = 60º [Given]m Q = 90º

PQR is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem,

QR = 1

2 PR [Side opposite to 30º]

19. In DEF, m D = 90º, m E = 45º, m F = 45º. If EF = 8 2 cm, find DE.

Sol. In DEF,D = 90º, E = 45º, F = 45º [Given]

DEF is 45º - 45º - 90º triangle

DE = 1

EF2

DE = 1

8 22

DE = 8 cm

P Q

R S


SCHOOL SECTION 73

20. The height and the base of ABC and PQR are equal. Find A ( ABC)A ( PQR)

.

Sol. ABC and PQR have equal base and equal height [Given] A (ABC) = A (PQR) [Triangles with equal base and equal height]

A ( ABC)

A ( PQR)

= 1

21. PQR ~ XYZ. If m Q = 60º then find m Y.Sol. PQR ~ XYZ [Given]

Q Y [c.a.c.t.]But, m Q = 60º [Given]

m Y = 60º

22. If ABC ~ DBA and BC × DB = 16, find AB.Sol. ABC ~ DBA [Given]

BC

AB =

AB

DB[c.s.s.t.]

AB2 = BC × DB AB2 = 16 [Given]

AB = 4 units [Taking square roots]

23. If ABC ~ DEF and DEF ~ XYZ and m A = 70º, then find m X.Sol. ABC ~ DEF [Given]

DEF ~ XYZ ABC ~ XYZ

A = X [c.a.c.t.] But, m A = 70º [Given]

m X = 70º

24. If ABC ~ DAC and AC = 3, then find BC × CD.Sol. ABC ~ DAC

AC

CD =

BC

AC[c.s.c.t.]

AC2 = BC × CD 32 = BC × CD

BC × CD = 9

25. If PQR XYZ, PRXZ

= 23

and PQ = 12, then find XY.

Sol. PQR ~ XYZ [Given]

PQ

XY =

PR

XZ[c.s.c.t.]

12

XY =

2

3

XY = 12 3

2

XY = 6 × 3

XY = 18 units

SCHOOL SECTION 129

Construction of various geometrical figures is a very important part of the study ofgeometry for understanding the concepts learnt in theoretical geometry.

BASIC CONSTRUCTIONS

(i) To draw a perpendicular bisector of a given line segment.

(ii) To draw an angle bisector of a given angle.

(iii) To draw a perpendicular to a line at a given point on it.

T BA

A

D

CB

••

•

•

•

R QP

Geometric Constructions3.


SCHOOL SECTION130

(iv) To draw a perpendicular to a given line from a point outside it.

(v) To draw an angle congruent to a given angle.

(vi) To draw a line parallel to a given line through a point outside it.

L

NM

P

RQ

A

m

P

B

R

m


SCHOOL SECTION 131


1. Draw perpendicular bisector of seg AB of length 8.3 cm. (2 marks)


2. Draw an angle of 125º and bisect it. (2 marks)


3. Construct LMN, such that LM = 6.2 cm, MN = 4.9 cm, LN = 5.6 cm.(2 marks)

P

RQ• •

125º

M

L M6.2 cm

5.6 cm 4.9 cm

N(Rough Figure)

L M

5.6 cm

6.2 cm

4.9 cm

N

A B8.3 cm


SCHOOL SECTION132


4. Construct PQR such that PQ = 5.7 cm, P = Q = 50º. (2 marks)


5. Construct DEF such that, DE = 6.5 cm, E = 50º, F = 30º; and drawEM DF, measure the length EM. (3 marks)

Analysis : In DEF,m D + m E + m F = 180º

m D + 50 + 30 = 180º m D + 80 = 180º m D = 180º – 80º m D = 100º

30º

F

D E

M

100º6.5 cm

50º

(Rough Figure)

50º

30º

F

D E

M

100º

6.5 cm

P Q5.7 cm50º 50º

R

P Q5.7 cm

50º 50º

R(Rough Figure)


SCHOOL SECTION 133

[A] Constructing tangents to a circle from a point on the circle.

Example : Draw a tangent to a circle of radius 2 cm at a point on it.

Steps of construction :1. Draw a circle with radius 2 cm.

Let ‘M’ be the centre of the circle.

2. Take any point ‘P’ on the circle

3. Draw ray MP.

4. Draw the line ‘l’ perpendicular to the ray MP at point ‘P’. Line ‘l’ is the

required tangent to the circle.


1. Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm andcentre ‘O’. (2 marks)

TYPE : 1

PM

2 cm

MO 2.9 cm

MO 2.9 cm

(Rough Figure)


SCHOOL SECTION134


2. Draw a tangent at any point R on the circle of radius 3.4 cm and centre ‘P’.(2 marks)


3. Draw a circle of radius 2.6 cm. Draw tangent to the circle from anypoint on the circle using centre of the circle. (2 marks)

RP 3.4 cm

(Rough Figure)

P 3.4 cm R

PO 2.6 cm

(Rough Figure)

O 2.6 cm P


SCHOOL SECTION 135


12. Draw a tangent to a circle of a radius 3.1 cm and centre O at any point‘R’ on the circle. (2 marks)

[B] Constructing tangents to a circle from a point on the circlewithout using centre.

Example : Given a circle, with a point P on it. Draw a tangent to the circle without using its centre.

Steps of construction :1. Draw the required circle.2. Take any point ‘P’ on it.3. Draw chord PQ.4. Take any point ‘R’ on the alternate

arc of arc PXQ other points than P and Q.5. Join QR and RP.6. Draw a ray PN making an angle congruent to QRP, taking QP as one side

and point P as vertex.

7. The line containing ray PN is the required tangent.

TYPE : 1

R

Q

X

PN M

•

•

RO 3.1 cm

(Rough Figure)

O 3.1 cm R


SCHOOL SECTION136


6. Draw a circle of radius 2.7 cm and draw chord PQ of length 4.5 cm.Draw tangents at P and Q without using centre. (3 marks)


7. Draw a circle having radius 3 cm draw a chord XY = 5 cm. Draw tangentsat point X and Y without using centre. (3 marks)

(Rough Figure)

Q

R

P 4.5 cm

•

••

Q

R

P 4.5 cm• •

•

• •

ST

T S

Y

Z

X 5 cm• •

•

AB

Y

Z

X 5 cm• •

•

(Rough Figure)

••B A


SCHOOL SECTION 137


13. Draw a circle of radius 3.6 cm, take a point M on it. Draw a tangent tothe circle at M without using centre of the circle. (2 marks)


14. Draw a circle of suitable radius and draw a chord XY of length 4.6 cm.Draw tangents at points X and Y without using centre. (3 marks)

N

M•

•L

(Rough Figure)

•A

N M•

•

L

A

X Y•

Z•

• 4.6 cm

P

(Rough Figure)

X Y

Z

4.6 cm

P

•

• •


SCHOOL SECTION138

[A] Constructing tangents to a circle from a point outside the circle.

Example : Draw a tangent to the circle of radius 1.7cm from a point at a distance of 5.2 cm from the centre.

Steps of construction :1. Draw a circle with radius 1.7 cm.

Let O be the centre of the circle.2. Take a point P such that OP = 5.2cm.3. Draw perpendicular bisector of seg OP

and mark the midpoint of seg OP as ‘M’.4. With ‘M’ as a centre and radius MP draw a semicircle .5. Let ‘A’ be the point of intersection of semicircle

and the circle.6. Draw a line joining P and A. Line PA is the

required tangent.


8. Draw a tangent to the circle from the point B, having radius 3.6 cm andcentre ‘C’. Point B is at a distance 7.2 cm from the centre. (3 marks)

TYPE : 2

A

B7.2 cm

3.6

cm

C

(Rough Figure)

A

O M P5.2 cm

1.7

cm

A

B7.2 cm

3.6

cm

C M


SCHOOL SECTION 139


9. Draw a tangent to the circle from the point L with radius 2.8 cm. Point‘L’ is at a distance 5 cm from the centre ‘M’. (3 marks)


10. Draw a tangent to the circle with centre O and radius 3.3 cm from apoint A such that d (O, A) = 7.5 cm. Measure the length of tangentsegment. (3 marks)

The length of tangent segnment AB is 6.7 cm.

A

O M P5.2 cm

1.7

cm

B

A7.5 cm

3.3

cm

C

B

A7.5 cm

3.3

cm

C

(Rough Figure)

A

O P5.2 cm

1.7

cm

(Rough Figure)

M


SCHOOL SECTION140


15. Construct tangents to the circle from point B with radius 3.5 cm andcentre A. Point B is at a distance 7.3 cm from the centre. (3 marks)


16. Draw tangents to the circle with centre P and radius 2.9 cm. From apoint Q which is at a distance 8.8 cm from the centre. (3 marks)

P 8.8 cm

2.9

cm

Q

A

2.9 cm

(Rough Figure)

B

P

8.8 cm

2.9

cm

Q

A

B

2.9

cm

M

(Rough Figure)

7.3 cm B

C

A

D

3.5

cm

3.5 cm

M7.3 cm

C

A

D

B

3.5

cm

3.5 cm


SCHOOL SECTION 141

Constructing circumcircle of triangles

1. A circle passing through the vertices of the triangle is called the circumcircleof a triangle.

2. Circumcentre can be obtained by drawing perpendicular bisectors of anytwo sides of a triangle.

3. The point of intersection of the perpendicular bisectors is calledcircumcentre and it is equidistant from the vertices of the triangle.

The position of circumcentre depends upon the type of a triangle.(i) If the triangle is an obtuse angled triangle, the circumcentre lies outside

the triangle.(ii) If the triangle is an acute angled triangle, the circumcentre lies inside

the triangle.(iii) If the triangle is a right angled triangle, the circumcentre lies on the

midpoint of the hypotenuse.

Example : Draw ABC, with AB = 4.1 cm, BC = 6.5 cm and AC = 5 cm. Constructcircumcircle of ABC. Measure the radius of the circle.

Steps of construction :

1. Construct ABC, with AB = 4.1 cm, BC = 6.5 cm and AC = 5 cm.

2. Draw perpendicular bisectors of any two sides of ABC and let them intersect

at point O.

3. Draw a circle with centre O and radius OA.

4. This circle is the circumcircle of ABC.

TYPE : 3

B C

A

m

6.5 cm

4.1

cm

5 cm

O


SCHOOL SECTION142


1. Draw the circumcircle of PMT such that, PM = 5.4 cm, P = 60º, M = 70º.(3 marks)


3. Construct the circumcircle of KLM in which KM = 7 cm, K = 60º,M = 55º. (3 marks)

P M5.4 cm

60º 70º

T

O

P M5.4 cm60º 70º

T(Rough Figure)

K60º 55º

7 cm M

L

O

K60º 55º

7 cm M

L(Rough Figure)


SCHOOL SECTION 143


4. Construct a right angled triangle PQR where PQ = 6 cm, QPR = 40º,PRQ = 90º. Draw circumcircle of PQR. (3 marks)


6. Construct LEM such that, LE = 6cm, LM = 7.5 cm, LEM = 90º anddraw its circumcircle. (3 marks)

R P6 cm40º

Q

O

R P6 cm40º

Q(Rough Figure)

E 6 cm

7.5 cm

L

M

O

E 6 cm

7.5 cm

L

M

(Rough Figure)


SCHOOL SECTION144


26. In PQR, Q = 90º, seg QM is the median. PQ2 + QR2 = 169. Draw acircumcircle of PQR. (4 marks)

Analysis : PQ2 + QR2 = 169 ......(i) [Given]But,PQ2 + QR2 = PR2 ......(ii) [By Pythagoras theorem]

PR2 = 169 PR = 13

PM = MR =1

PR2

[By definition of median]

=1

×132

PM = MR = 6.5 cmIn PQR,m PQR = 90º

QM =1

PR2

=1

×132

QM = 6.5 cm

Q

P R

Q

13 cmM

[Median drawn to the hypotenuseis half of hypotenuse]

Q

P RM

(Rough Figure)

13 cm


SCHOOL SECTION 145


25. Construct a circumcircle of ABC such that AB = 5 cm, AC = 12 cm,BAC = 90º. (3 marks)

5 c

m

B

A C12 cm

O

5 c

m

B

A C12 cm

(Rough Figure)


SCHOOL SECTION146


2. Construct the circumcircle of SIM in which SI = 6.5 cm, I = 125º,IM = 4.4 cm. (3 marks)

I

4.4 cm

M

125º

6.5 cm S

O

I4.4 cm

M

125º6.5 cm S

(Rough Figure)


SCHOOL SECTION 147


7. Construct DCE, such that, DC = 7.9 cm, C = 135º, D = 20º and drawcircumcircle. (3 marks)

Note : This figure is drawn proportionally and not with given measurements.

C D20º

7.9 cm

135º

E

(Rough Figure)

C D20º

7.9 cm

135º

E

O


SCHOOL SECTION148

Constructing incircle of triangles

1. A circle which touches all the sides of a triangle is called the incircle ofthe triangle. The centre of the incircle is called incentre.

2. Incentre is obtained by drawing angle bisectors of the triangle.3. The angle bisectors are concurrent and their point of intersection is

equidistant from the sides of the triangle.

Example : Construct SRP such that RP = 6 cm, R = 750 and P = 550.

Steps of construction :1. Draw SRP with RP = 6 cm, R = 75º

and P = 55º2. Draw angle bisectors of R and P.3. Let ‘I’ be the point of intersection

of these angle bisectors .3. Draw seg IM side RP.4. Draw a circle with centre I and radius IM.

The circle so obtained is the incircle of SRP.


5. Construct the incircle of RST in which RS = 6 cm, ST = 7 cm andRT = 6.5 cm. (3 marks)

TYPE : 4

S

I

R P

75º 55º

M6 cm

•• ××

S7 cm

6 cm 6.5 cm

R

T××

••

OS

7 cm

6 cm 6.5 cm

R

T

(Rough Figure)

M


SCHOOL SECTION 149


7. Construct the incircle of DEF in which DE = DF = 5.8 cm, EDF = 65º.(3 marks)


8. Construct incircle of SGN such that SG = 6.7 cm, S = 70º,G = 50ºand draw incircle of SGN. (3 marks)

D E5.8 cm

5.8 cm

F

××

••65º

O

S G6.7 cm

N

••70º

××50º

OS G

6.7 cm

N

70º 50º

(Rough Figure)

M

M

D E5.8 cm

5.8 cm

F

65º

(Rough Figure)


SCHOOL SECTION150


8. Construct any right angled triangle and draw incircle of that triangle.(3 marks)

ABC is the required right angled triangle.Such that AB = 5 cm, BC = 7 cm and m ABC = 90º


9. Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm,R = 95º. (3 marks)

B 7 cm C

A

5 cm

×× ••

O

5.9 cm NR

4.9 cm

S

95º ××••

O

5.9 cmNR

4.9 cm

S

95º

(Rough Figure)

M

M

B7 cm

C

A

5 cm

(Rough Figure)


SCHOOL SECTION 151


6. Construct the incircle of STU in which, ST = 7 cm, T = 120º, TU = 5 cm.(3 marks)


10. Construct DAT such that DA = 6.4 cm, D = 120, A = 25 and drawincircle of DAT. (3 marks)

7 cm ST

5 cm

U

120º• ×• ×

O

D A6.4 cm

•120º•

T

×25º ×

O

M

M

7 cmS

T

5 cm

U

120º

(Rough Figure)

(Rough Figure)

D A6.4 cm

120º

T

25º


SCHOOL SECTION152


9. Construct the circumcircle and incircle of an equilateral XYZ withside 6.3 cm. (3 marks)


11. Draw the circumcircle and incircle of an equilateral triangle ABC withside 6.6 cm. (3 marks)

6.3 cmY Z

6.3 cm6.3 cm

×

••

X

×

O

B C6.6 cm

6.6

cm

6.6 cm

××

••

A

O

6.3 cmY Z

6.3 cm 6.3 cm

X(Rough Figure)

B C6.6 cm

6.6

cm

6.6 cm

A(Rough Figure)


SCHOOL SECTION 153

To construct an arc having the given segment as its chord andsubtending a given angle at any point on the arc.

Example : Draw an arc such that seg AB of length 5.4 cm subtends an AQB of 50º on it.

Step of construction :

1. Draw seg AB of length 5.4 cm.

2. Draw rays AO and BO making an angle of 400 with seg AB on the same side.

3. Draw an arc with O as the centre and radius OA.

4. Take any point Q on the arc. Draw AQB.

5. Arc AQB is the required arc.


11. Draw an arc with seg AB = 6.3 cm, inscribing ACB = 65º. (3 marks)

TYPE : 5

Q

A B400 400

1000

500

5.4 cm

O

A25º

6.3 cm B

C

65º

25º

130º

O

A 6.3 cm B

C

65º

(Rough Figure)


SCHOOL SECTION154


13. Draw an arc such that chord ST = 5.6 cm, inscribing SVT = 80º.(3 marks)


17. Construct an arc PQM such that seg PM of length 6.2 cm subtends anangle of 40º on it. (3 marks)

S

80º

V

T5.6 cm10º 10º160º

O

MP 6.2 cm

40º

Q(Rough Figure)

MP 6.2 cm50º 50º

40º

Q

O

80º

S

80º

V

T5.6 cm

(Rough Figure)


SCHOOL SECTION 155


12. Draw an arc with seg MN = 8.9 cm, inscribing MPN = 125º. (2 marks)


18. Construct an arc DCV such that seg DV of length 9.5 cm subtends anangle of 135º on it. (2 marks)

M N35º

125º

P

35º 8.9 cm

110º

O

D

C

V

9.5 cm 45º45º

135º

M N

125º

P

8.9 cm

(Rough Figure)

O

D

C

V9.5 cm

135º

(Rough Figure)


SCHOOL SECTION156

Constructing triangles with a given base, angle oppositeto the base and median.


1. Construct LMN such that LM = 6.6 cm, LNM = 65º and ND is medianND = 5 cm. (4 marks)


2. Construct GHI such that GI = 5.4 cm, GHI = 75º. HR is median.HR = 3.2 cm. (4 marks)

TYPE : 6

DL

N N

M25º25º

65º

5 cm6.6 cm

O

RG

H H

I15º

75º3.2 cm

5.4 cm

15º

O

DL

N

M

65º5 cm

6.6 cm

(Rough Figure)

RG

H

I

75º3.2 cm

5.4 cm

(Rough Figure)

130º

150º


SCHOOL SECTION 157


19. Construct SAB such that SB = 7.6 cm, SAB = 50º seg AD is medianand AD = 5 cm. (4 marks)


21. Construct DEF such that DF = 8.1 cm, DEF = 140º and median EM = 2.5 cm.(4 marks)

D F

E

140º

2.5 cm

(Rough Figure)

8.1 cmM

S B

A

7.6 cm

A

40º 40º

5 cm

D

O100º

D50º 50º

F

E E

140º2.5 cm

M8.1 cm

O

80º

50º

S B

7.6 cm

A

5 cm

(Rough Figure)

D

50º


SCHOOL SECTION158


3. Construct ABC such that BC = 7.8 cm, BAC = 100º and median AM = 3.5 cm.(4 marks)


4. Construct XYZ such that XY = 9.5 cm, XZY = 115º, ZP is median.ZP = 3.3 cm. (4 marks)

X Y

Z

115º

3.3

cm

P9.5 cm

(Rough Figure)

ZZ

Y9.5 cm25ºX P

3.3 cm

115º

25º

O

130º

AA

C7.8 cm 10ºB M

3.5 cm

10º

100º

O160º

B C

A

100º

3.5 cm

(Rough Figure)

7.8 cmM


SCHOOL SECTION 159

Constructing triangles with a given base, angle oppositeto the base and an altitude.


5. Construct DEF such that DF = 6.2 cm, DEF = 60º, EM DF andEM = 4.4 cm. (4 marks)


6. Construct RST such that RT = 5.7 cm, RST = 55º, SD RT, SD = 3.4 cm.(4 marks)

TYPE : 7

MD

E E

F30º

6.2 cm

60º

4.4

cm

30º

4.4

cmO

RD

S S

T35º

5.7 cm

35º

55º

3.4

cm

3.4

cm

O

D F

E

60º

4.4

cm

M6.2 cm

(Rough Figure)

R T

S

55º

3.6

cm

D5.7 cm

(Rough Figure)

120º

B

A

110º

A

B


SCHOOL SECTION160


9. Construct KLM such that KM = 7.2 cm, KLM = 72º, LA KM,KA = 4.8 cm. (4 marks)


22. Construct LAC such that LC = 6.7 cm, LAC = 72º and altitude AB haslength 4 cm. (4 marks)

L C

A

72º4 c

m

B6.7 cm

(Rough Figure)

L

A

C6.7 cm

4 c

m

72º

18º

A

B18º

O

M

N

144º

4 c

m

L

O

K18º

4.8 cm18º

M

144º

7.2 cm

72º

A

K M

L

72º

4.8 cm A7.2 cm

(Rough Figure)


SCHOOL SECTION 161


20. Construct KPM such that KM = 7 cm, KPM = 90º and length of altitudePS is 2.9 cm. (4 marks)


8. Construct CVX such that CX = 9.1 cm, CVX = 130º, VD CX andVD = 1.7 cm. (4 marks)

K M

P

2.9

cm

S7 cm

(Rough Figure)

C X

V

130º

1.7

cm

D9.1 cm

(Rough Figure)

A

B

K S

P

2.9

cm

P

7 cm M

2.9

cm

V V

130º

C X40º D

1.7 cm

9.1 cm 40º

O

100º

A

B

1.7 cm


SCHOOL SECTION162


10. Construct SPQ such that SQ = 8.3 cm, SPQ = 127º, PM SQ,PM = 1.6 cm. (4 marks)


7. Construct PQR such that PQ = 9.2, PRQ = 112º, RK is an attitude,RK = 2.4 cm. (4 marks)

S Q

P

1.6

cm

M8.3 cm

(Rough Figure)127º

P Q

R

2.4

cm

K9.2 cm

(Rough Figure)

112º

A

B

106º

O

B

A

P P

S Q37º

127º

M 8.3 cm 37º

O

1.6 cm1.6 cm

KP

R R

Q22º9.2 cm22º

112º

2.4 cm 2.4 cm

136º


SCHOOL SECTION 163

Constructing similar triangles


1. ABC ~ DEF, In ABC, AB = 5.2 cm, BC = 4.6 cm, B = 45º and BCEF

:23

;

construct DEF. (4 marks)Analysis :ABC ~DEF [Given]

AB

DE =

BC

EF =

AC

DF=

2

3......(i) [c.s.s.t.]

B = E = 45º [c.a.s.t.]

AB

DE =

2

3 [From (i)]

BC

EF=

2

3 [From (i)]

5.2

DE =

2

3

4.6

EF=

2

3

15.6

2 = DE

13.8

2= EF

DE = 7.8 cm EF = 6.9 cm

Information for constructing DEFis complete.

TYPE : 8

D

E F6.9 cm45º

7.8 cm

(Required triangle)

5.2 cm

4.6 cm

45º

B C

A

(Given triangle)


SCHOOL SECTION164


2. LMN ~ XYZ, In LMN, LM = 6 cm, MN = 6.8 cm, LN = 7.6 cm and

LMXY

= 43

; construct XYZ. (4 marks)

Analysis : LMN ~ XYZ [Given]

LM

XY =

MN

YZ =

LN

XZ=

4

3...... (i) [c.s.s.t.]

LM

XY =

4

3 [From (i)]

MN

YZ=

4

3 [From (i)]

LN

XZ =

4

3 [From (i)]

6

XY =

4

3

6.8

YZ=

4

3

7.6

XZ =

4

3

18

4= XY

20.4

4= YZ

22.8

4= XZ

XY = 4.5 cm YZ = 5.1 cm XZ = 5.7 cm

Information for constructing XYZ is complete.

X

Y Z

4.5

cm

5.7 cm

5.1 cm

(Required triangle)

7.6 cm

6 c

m

6.8 cm

L

M N

(Given triangle)


SCHOOL SECTION 165


3. RHP ~ NED, In NED, NE = 7 cm, D = 30º, N = 20º and HPED

= 45

;

construct RHP. (4 marks)Analysis :RHP ~ NED [Given]

RH

NE =

HP

ED =

RP

ND =

4

5......(i) [c.s.s.t.]

R = N = 20º

P = D = 30º [c.a.s.t.]

H = E = 130º

RH

NE =

4

5[From (i)

RH

7 =

4

5

RH = 28

5 = 5.6 cm

Information for constructing RHP is complete.

20º

30º

130º

P

R H5.6 cm

(Required triangle)

7 cm EN

D

130º

30º

20º

(Given triangle)


SCHOOL SECTION166

T

E

A MH

6.3 cm

4.9 cm

120º

A1

A2

A3

A4

A5

A6

A7

× ×

•

•


4. LTR ~ HYD, In HYD, HY = 7.2 cm, YD = 6 cm, Y = 40º andLRHD

= 56

’

construct LTR. (4 marks)

Sol. RHP ~ NED [Given]

LT

HY =

TR

YD =

LR

HD =

5

6.....(i) [c.s.s.t.]

T = Y = 40º [c.a.s.t.]

LT

HY=

5

6 [From (i)]

TR

YD =

5

6 [From (i)]

LT

7.2=

5

6

TR

6 =

5

6

LT =36

6 TR =

30

6 LT = 6 TR = 5

Information for constructing HYD is complete.


5. AMT ~ AHE, In AMT, MA = 6.3 cm, MAT = 120º, AT = 4.9 cm andMAHA

= 75

’ construct AHE. (4 marks)

L

T R5 cm

6 cm

40º

(Given triangle)

6 cm

7.2 c

m

H

Y D

40º

(Required triangle)


SCHOOL SECTION 167


6. SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and

SHSV

= 35

; construct SVU. (4 marks)


23. ABC ~ LMN. In ABC, AB = 5.1 cm, B = 55º, C = 65º and ACLN

= 35

.

then construct LMN. (4 marks)Sol. ABC ~ LMN [Given]

AB

LM=

BC

MN =

AC

LN =

3

5.....(i) [c.s.s.t]

AB

LM=

3

5[From (i)]

5.1

LM=

3

5

25.5

3= LM

LM = 8.5Information for constructing LMN is complete.

U

R

S H V

5.2

cm

4.5 cm

S1

S2

S3

S4

S5

××

5.8

cm

•

•

U

SH

V

R

(Rough Figure)

4.5 cm

5.2 cm

5.8

cm


SCHOOL SECTION168


24. XYZ ~ DEF, in DEF, DE = 5.5 cm, E = 40º, EF = 4 cm and XYDE

= 65

then construct XYZ. (4 marks)Ans. XYZ ~ DEF [Given]

XY

DE =

YZ

EF =

XZ

DF =

6

5 ......(i) [c.s.s.t.]

XY

DE=

6

5 [From (i)]

YZ

EF=

6

5 [From (i)]

XY

5.5=

6

5

YZ

4=

6

5

xy =33

5 YZ =

24

5 XY = 6.6 YZ = 4.8

Information for constructing XYZ is complete.

(Required triangle)

N

L M8.5 cm

65º

55º

(Given triangle)

65º

55º

C

A B5.1 cm60º

Y Z6.6 cm

40º

6.6 c

m

X(Required triangle)

(Given triangle)

D

FE 4 cm

40º

5.5 c

m


SCHOOL SECTION 169


14. To draw seg AB of length 65 without using Pythagoras theorem.

(4 marks)Analysis : In ABC,

ABC = 90ºseg BD hypotenuse AC

ABC ~ ADB [Theorem on similarity of right angled triangle]

AB

AD =

AC

AB[c.s.s.t.]

AB2 = AD × AC AB2 = 5 × 13 AB2 = 65

AB = 65

A C

B

5 cm

13 cm

D

A C

B

5 cm D13 cm

(Analytical Figure)


SCHOOL SECTION170

OR

Analysis : In CAD,m CAD = 90ºseg AB hypotenuse CD

AB2 = CB × BD [Property of Geometric mean] AB2 = 5 × 13 AB2 = 65

AB = 65 [Taking square roots]


A

C DB 5 cm

18 cm

C D

A

5 cm B 13 cm

(Analytical Figure)


SCHOOL SECTION 171

16. Draw segment AB of any length. Take point D on AB such that AD2 = 3BD2.(4 marks)

Analysis : In CDB,

tan CBD =CD

BD[By definition]

tan 60 =AD

BD[ CD = AD]

3 =AD

BD

3 =AD

BD

2

2 [Squaring both sides]

AD2 = 3BD2

22. Draw a triangle ABC with side BC = 6 cm, B = 45º and A = 100º, then

construct a triangle whose sides are 47

times the corresponding sides

of ABC. (4 marks)

Analysis : In ABC,m A = 100º [Given]m B = 45º

m C = 35º [Remaining angle]

C

BDAl

60º

C

BDA

60º

(Analytical Figure)

•

A

B Q C

100º

45º 35º

P

(Rough Figure)

6 cm


SCHOOL SECTION172

PBQ is the required triangle

whose sides are 4

7times the corresponding sides of ABC

23. Construct a triangle ABC, in which BC = 3.5 cm, B = 60º and altitudeAD = 2.5 cm and draw its incircle and measure its radius. (4 marks)


A

B CD

3.5 cm

2.5

cm

60º

(Rough Figure)

A

B C3.5 cmD

2.5

cm

60º

2.5

cm

••

××

M

I

A

B C

100º

B1

P

45º 35º

B2

B3

B4

B5

B6

B7

6 cm Q

×

×

• •


SCHOOL SECTION 173

24. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm.Draw its circumcircle and measure its radius. (4 marks)

Analysis : ABC is an isosceles triangle with AB = ACseg AD side BC

BD = DC = 1

2BC [Perpendicular drawn to the base, bisects the base]

BD = DC = 1

2 × 8

BD = DC = 4 AD = BD = DC = 4 cm

25. In PQR, QR = 7.5 cm, QPR = 110º and PQ + PR = 8.3 cm then constructPQR and measure PQR. Construct its circumcircle. (5 marks)

Analysis : line l is perpendicular bisector of side TR PT = PR .......(i) [Perpendicular bisector theorem]

QT = 8.3 cm PQ + PT = 8.3 [Q - P - T] PQ + PR = 8.3 [From (i)]

In PTR,side PT side PR [From (i)]

PTR PRT [Isosceles triangle theorem]Let, PTR = PRT = xNow, QPR is an exterior angle of PTR,

QPR = PTR + PRT [Remote interior angles theorem] 110 = x + x 110 = 2x x = 55 PTR = PRT = 55º Information to draw RQT is complete.

(Rough Figure)

RQ

P

110º

7.5 cm

A

B C

4 cm

8 cmD

A

B C8 cm

4 c

m

(Rough Figure)

D


SCHOOL SECTION174

26. Construct LMN, such that LN = 8 cm and LMN = 80º and LM – MN = 3cm.Construct its circumcircle. (5 marks)

Analysis : Line l is a perpendicular bisector of side TN TM = MN .......(i) [Perpendicular bisector theorem]

LM = LT + TM [L - T - M] LM = 3 + MN [From (i)] LM – MN = 3 cm

In MTN,side MT side MN [From (i)]

MTN MNT [Isosceles triangle theorem]Let, MTN = MNT = x

x + x + M = 180 x + x + 80 = 180 2x = 180 – 80 2x = 100 x = 50 MTN = MNT = 50º

LTN + MTN = 180 LTN + 50 = 180 LTN = 180– 50 LTN = 130º

Information for drawing LTN is complete.

R

Q P S

110º

8.3 cm

7.5

cm

55º

O

l

(Analytical Figure)

NL

T 80º

130º

8 cm

M

3 cm

55º

RQ

P

T

110º8.

3 cm

7.5 cm

55º

(Analytical Figure)

L N

M

80º

8 cm

(Rough Figure)


SCHOOL SECTION 175

27. ConstructXYZ such that, YZ = 6.2 cm, Z = 65º and XY – XZ = 2.4 cmand draw incircle of it. (4 marks)

Analysis : Line l is a perpendicular bisector of side YW XY = XW .......(i) [Perpendicular bisector theorem]

XW = XZ + ZW [X - Z - W] XY = XZ + 2.4 [From (i)] XY – XZ = 2.4

X

65ºY Z6.2 cm

(Rough Figure)

N

L T

80º130º

8 cm

M3 cm

O

2.4

cm

65ºY

Z6.2 cm

X• •

××

W

l

Y Z

X

W

(Analytical Figure) l

6.2 cm65º

2.4

cm


SCHOOL SECTION176

28. In RST, RS = 5 cm, RT = 6.8 cm and median RM = 5.3 cm construct acircumcircle of RST. (4 marks)

Analysis : In RST extend median RX to point P such that R - X - P and RX = XPalso SX = XT

PSRT is a parallelogramInformation to constructing parallelogram PSRTis complete and RST can be obtained.Hence draw its circumcircle.

29. In ABC, BC = 6 cm and median AM = 5.1 cm. G is the centroid of ABCand BGC = 130º. Construct ABC. (4 marks)

Analysis : In ABC, G is the centroid on median AM

GM = 1

3AM [Centroid bisects each median]

GM = 1

3 × 5.1 = 1.7 cm

Also, BGC = 130º and BC = 6 cmInformation for constructing BGC is complete.Position of A can be obtained an line GM.Hence draw ABC.

A A

G

B CM

6 cm

G

40º 40º

5.1 cm

1.7 cm

B C

A

130º

5.1 cm

(Rough Figure)

6 cmM

G

R

S

MP10.6 cm

6.8 cm

5 cm

T

6.8 cm

6.8 cm

R

S M

6.8 cm

5 cm

T

(Rough Figure)

5.3 cm

P

R

S M T

(Analytical Figure)


SCHOOL SECTION 177

30. Draw a triangle ABC, right angled at B such that, AB = 3 cm andBC = 4 cm. Now construct a triangle similar to ABC, each of whose

sides is 75

times the corresponding side of ABC. (4 marks)

MCQ’s1. What is the point of concurrence of the medians of a triangle called ?

(a) Circumcentre (b) Incentre(c) Orthocentre (d) Centroid

2. What is the point of concurrence of the altitudes of a triangle called ?(a) circumcentre (b) incentre(c) orthocentre (d) centroid

3. What is the point of concurrence of the angle bisectors of a triangle called ?(a) circumcentre (b) incentre(c) orthocentre (d) centroid

4. An arc of a circle containing an angle of 70º is to be drawn on the upper side ofseg AB. What are the measures of the angles to be drawn at points A and B ?(a) 20º on the upper side of seg AB (b) 70º on the upper side of seg AB(c) 20º on the lower side of seg AB (d) 70º on the lower side of seg AB

5. An arc of a circle containing an angle of 140º is to be drawn on the upper side ofseg AB. What are the measures of the angles to be drawn at points A and B.(a) 70º on the upper side of seg AB (b) 50º on the upper side of seg AB(c) 50º on the lower side of seg AB (d) 70º on the lower side of seg AB

6. To find the circumcentre of ABC, we bisect .............. of ABC.(a) side AB (b) all sides(c) any two sides (d) any two angles

P

A

B

B1

C R

B2

B3

B4

B5

B6

B7

4 cm

3 c

m 3 cm

P

A

B CR

4 cm

(Rough Figure)


SCHOOL SECTION178

7. To find incentre of a given triangle, we bisect ..............(a) any two angles (b) all sides(c) all angles (d) one side and one angle

8. From a point outside a circle, .................. tangents can be drawn(a) one (b) two(c) at the most two (d) none of these

9. The circumcentre of an acute angled triangle is ................. of the triangle.(a) on one side (b) in the interior(c) in the exterior (d) none of these

10. If the circumcentre lies in the exterior of the triangle, then it is ..........triangle.(a) a right angled (b) an acute angled(c) an isosceles (d) an obtuse angled

11. Tangent drawn from a point M on the circle is perpendicular to the ............. .(a) chord MP (b) diameter MN(c) chord AB (d) radius OP

12. To draw arc of measure 120º on seg AB, we first construct isosceles trianglewith base angle of .............. .(a) 30º (b) 60º(c) 90º (d) 120º

13. Three sides of ABC are given. To construct similar PQR, at least.................. of PQR must be given.(a) one angle (b) any two angles(c) any one side (d) all sides

14. The circumcentre and incentre of ............... triangle are at the same point.(a) a scalene (b) an isosceles(c) an equilateral (d) an acute angled

15. To construct ABC of base AB = 5 cm and height CP = 6 cm, we drawparallel line at a distance of ................ cm.(a) 1 (b) 5(c) 6 (d) 11

16. The sides of ABC are 6 cm, 8 cm, 10 cm. A circumcentre of ABC isdrawn. What is the radius of the circumcircle ?(a) 5 cm (b) 10 cm(c) 4 cm (d) 24 cm

17. ABC ~ XYZ .......... .............. .(a) AB, XY (b) BC, YZ(c) AC, AZ (d) B, Y

18. To draw a tangent at point be on arc ABC .............. must be given.(a) centre (b) none(c) diameter (d) length of chord AC

19. ABC ~ XYZ and AB 2

XY 1

m ABC

m XYZ

= ............. .


SCHOOL SECTION 179

(a)1

2(b) 2

(c) 1 (d)1

3

20. O is the centre of a circle with radius 5 cm, the length of the tangentsegment drawn from the point 13 cm from centre O is .......... cm.(a) 5 (b) 13(c) 12 (d) 18

: ANSWERS :

1. (d) Centroid 2. (c) orthocentre

3. (b) incentre 4. (a) 20º on the upper side of seg AB

5. (b) 50º on the upper side of seg AB 6. (c) any two sides

7. (a) any two angles 8. (b) two

9. (b) in the interior 10. (d) an obtuse angled

11. (b) diameter MN 12. (a) 30º

13. (c) any one side 14. (c) an equilateral

15. (c) 6 16. (a) 5 cm

17. (d) B, Y 18. (a) none

19. (c) 1 20. (c) 12

SCHOOL SECTION180

A

B C

TRIGONOMETRIC RATIOS OF AN ACUTE IN A RIGHT ANGLED TRIANGLE

A

B C

Trigonometry4.` Introduction :

The word TRIGONOMETRY is derived from Greek words Tri meaningthree, gona meaning sides and metron meaning measure.

Thus, Trigonometry deals with measurements of sides and angles of atriangle.In our syllabus, we restrict our learning to right angled triangles.In a right angled triangle,

(i) The side opposite to right angle is called the hypotenuse.(ii) For any acute angle, the side opposite to it is called the opposite side.(iii) For any acute angle, the side adjacent to it other than the hypotenuse is

called the adjacent side.

In ABC,m ABC = 90º

(i) seg AC is the hypotenuse.(ii) For ACB, seg AB is the opposite side.(iii) For ACB, seg BC is the adjacent side.

For any acute angle in a right angled triangle, the three above mentionedsides, can be arranged two at a time, in six different ratios. These ratiosare called Trigonometric ratios.

In ABC,m ABC = 90ºm ACB =

Sine ratio of = sin =Opposite side

Hypotenuse =AB

AC

Cosine ratio of = cos =Adjacent side

Hypotenuse =BC

AC

Tangent ratio of = tan =Opposite side

Adjacent side =AB

BC

Cosecant ratio of = cosec =Hypotenuse

Opposite side =AC

AB

Secant ratio of = sec =Hypotenuse

Adjacent side =AC

BC

Cotangent ratio of = cot =Adjacent side

Opposite side =BC

AB


SCHOOL SECTION 181

RELATIONS BETWEEN TRIGONOMETRIC RATIOS

TRIGONOMETRIC IDENTITIES

Termin

al arm

Initial arm

B

OA

Vertex

Initial arm

B

O

A

Terminal arm

Terminal arm

A

O

B

Initial arm

1. cosec =1

sin 2. sec =1

cos 3. cot = 1

tan

4. tan =sin

cos

5. cot =

cos

sin

(i) sin2 + cos2 = 1 (ii) 1 + tan2 = sec2 (iii) 1 + cot2 = cosec2

` Angle in Standard position or standard angle : Directed angle :

Consider the ray OA and rotate it in anti-clockwisedirection about O, the final position of the ray OA isray OB. In the adjoining figure, the rotation fromthe ray OA to ray OB defines an AOB. It givesthe direction from ray OA to ray OB.It is called as directed angle.

• Initial arm : The initial position of the ray is called the initial arm. In theabove figure OA is the initial arm.

• Terminal arm : The final position of the ray after rotation is called terminalarm. In the above figure OB is the terminal arm.

• Vertex : The point of rotation is called the vertex. In the above figure O isthe vertex.In directed angle we take the following into consideration :

(i) Initial arm(ii) Terminal arm(iii) The amount and the sense

of rotation of the initial ray.(iv) The rotation may be in

anticlockwise or clockwise direction.

NOTE 1. If the rotation of the initial ray is anti-clockwise then the directed angle

is positive.2. If the rotation of the initial ray is clockwise then the directed angle is

negative.

• Standard angle : In rectangular co-ordinate system a directed angle withits vertex at the origin O and the initial ray along positive X-axis is calledStandard Angle or Angle in Standard Position.

• Measure of standard angle : The amount of rotation of the ray from theinitial position to terminal position is the measure of standard angle.


SCHOOL SECTION182

Angle in Quadrant : A directed angle in standard position is said to be inparticular quadrant if its terminal arm lies in that quadrant.

Quadrantal Angle : A directed angle in standard position whose terminalarm lies along co-ordinate axes is called a quadrantal angle.


1. Draw the figure and write the answers :(i) For the angle in standard position if the initial arm rotates 220º in clockwise

direction then terminal arm is in which quadrant ? (1 mark)Sol. Since the initial arm rotates in

clockwise direction and the angle ismore than – 180º but less than – 270º,the terminal arm lies in II quadrant.

(ii) For the angle in standard position if the initial arm rotates 25º inanticlockwise direction then terminal arm is in which quadrant ? (1 mark)

Sol. Since the initial arm rotates inanticlockwise direction and the angleis more than 0º but less than 90º,terminal arm lies in I quadrant.

(iii) For the angle in standard position if the initial arm rotates 305º in anticlockwisedirection then terminal arm is in which quadrant ? (1 mark)

Sol. Since the initial arm rotates inanticlockwise direction and the angleis more than 270º but less than 360º,terminal arm lies in IV quadrant.


2. The terminal arm is in II quadrant, what are the possible angles ?(1 mark)Sol. The terminal arm is in II quadrant,

the angle is in between 90º and 180ºif the initial arm rotates anticlockwisedirection or the angle is between–270º and – 180º if the initial armrotates clockwise.

Y

XXO

Y

220º

Y

Y

XX O

25º

Y

Y

XX305º

O

Y

Y

XX O


SCHOOL SECTION 183


3. The terminal arm is on negative Y-axis, what are the possible angles ? What can you say about this angle ? (1 mark)

Sol. The terminal arm is on negative y-axis,the possible angles are 270º and – 90º.These angles are called quadrantalangles.

We define the trigonometric ratios in terms of co-ordinates of apoint P (x, y) as follows :

(i) sin = y

r(ii) cos =

x

r

(iii) tan = y

x, where x 0 (iv) cosec =

r

y , where y 0

(v) sec = r

x, where x 0 (v) cot =

x

y , where y 0

Signs of trigonometric ratios in different quadrants :If x is positive, cosine is positive. If x is negative, cosine is negative.If y is positive, sine is positive. If y is negative, sine is negative.


1. Find the trigonometric ratios in standard position whose terminal armpasses through the points :

(i) (4, 3) (2 marks)Sol. The terminal arm passes through P (4, 3)

x = 4 and y = 3

r = x y+2 2

= (4) (3)+2 2

= 16 9+

= 25

r = 5 units

Let the angle be

sin =y

r=

3

5cosec =

r

y =5

3

cos =x

r=

4

5sec =

r

x=

5

4

tan =y

x=

3

4cot =

x

y =4

3

X X

Y

Y

O


SCHOOL SECTION184

(iii) (– 24, – 7) (2 marks)Sol. The terminal arm passes through P (– 24, – 7)

x = – 24 and y = – 7

r = x y+2 2

= (–24) (–7)+2 2

= 576 49+

= 625 r = 25 units

Let the angle be

sin =y

r=

– 7

25cosec =

r

y =–25

7

cos =x

r=

–24

25sec =

r

x=

–25

24

tan =y

x=

–7

–24 = 7

24cot =

x

y =–24

–7 = 24

7

(iv) ( )–1 , 3 (2 marks)

Sol. The terminal arm passes through P –1, 3

x = – 1 and y = 3

r = x y+2 2

= 1 3 22

= 1 3

= 4 r = 2 units

Let the angle be

sin =y

r=

3

2cosec =

r

y =2

3

cos =x

r=

–1

2sec =

r

x=

– 2

1

tan =y

x=

3

–1 = – 3 cot =x

y =–1

3

(ii) (5, – 12) (2 marks)Sol. The terminal arm passes through P (5, – 12)

x = 5 and y = – 12

r = x y+2 2

= (5) (–12)+2 2

= 25 144+

= 169 r = 13 units


SCHOOL SECTION 185

Let the angle be

sin =y

r=

–12

13cosec =

r

y =–13

12

cos =x

r=

5

13sec =

r

x=

13

5

tan =y

x=

–12

5cot =

x

y =–5

12

(v) (1, – 1) (2 marks)Sol. The terminal arm passes through P (1, – 1)

x = 1 and y = – 1

r = x y+2 2

= (1) + (–1)2 2

= 1 1

= 2

r = 2 unitsLet the angle be

sin =y

r=

–1

2cosec =

r

y =2

–1 = – 2

cos =x

r=

1

2sec =

r

x=

2

1 = 2

tan =y

x=

–1

1 = – 1 cot =

x

y =1

–1 = – 1

(vi) (– 2, – 3) (2 marks)Sol. The terminal arm passes through P (– 2, – 3)

x = – 2 and y = – 3

r = x y+2 2

= (–2) (–3)2 2

= 4 9

= 13

r = 13 units

Let the angle be

sin =y

r=

– 3

13cosec =

r

y =13

– 3 = – 13

3

cos =x

r=

– 2

13sec =

r

x=

13

– 2 = – 13

2

tan =y

x=

– 3

– 2 = 3

2cot =

x

y =– 2

– 3 = 2

3


1. If the terminal arm passes through the point (1, –1) making an anglefind the value of sec . (2 marks)

Sol. The terminal arm passes through point (1, – 1) x = 1 and y = – 1


SCHOOL SECTION186

r = x y2 2

r = (1) (–1)2 2

r = 1 1

r = 2 units

Let the angle be

sec =r

x

sec =2

1

sec = 2


3. Find where the angle lies if the terminal arm passes through :

(i) (5, – 7) (1 mark)Sol. (5, – 7)

x is positive and y is negative The terminal arm is in

IV quadrant.

(ii) (– 8, 1) (1 mark)Sol. (– 8, 1)

x is negative and y is positive The terminal arm is in

II quadrant.

(iii) (– 3, – 3) (1 mark)Sol. (– 3, – 3)

x is negative and y is negative The terminal arm is in

III quadrant.

(iv) (0, 2) (1 mark)Sol. (0, 2)

x is zero and y = 2 The terminal arm is on

positive Y-axis.


2. If the angle = – 60º, find the value of sin , cos , sec and tan .(2 marks)Sol. = – 60º

sin (– ) = – sin sin (– 60) = – sin 60

sin (– 60) =3

–2

sec (– ) = sec sec (– 60)= sec 60

sec (– 60)= 2

cos (– ) = cos cos (– 60)= cos 60

cos (– 60)=1

2

tan (– ) = – tan tan (– 60)= – tan 60

tan (– 60)= – 3


SCHOOL SECTION 187


3. If the angle = – 60º, find cos and cosec . (1 mark)Sol. = – 60º

cos (– ) = cos cos (– 60) = cos 60

cos (– 60) =1

2cosec (– ) = – cosec

cosec (– 60) = – cosec 60

cosec (– 60) =2

–3


1. If sin = 5

13. where is an acute angle, find the value of other

trigonometric ratios using identities. (3 marks)

Sol. sin =5

13[Given]

cosec =1

sin

=1

513

cosec =13

5

sin2 + cos2 = 1

5

13

2

+ cos2 = 1

25

169 + cos2 = 1

cos2 =25

1 –169

cos2 =169 – 25

169

cos2 =144

169

cos =12


sec =1

cos

=1

1213

sec =13

12


SCHOOL SECTION188

tan =sin

cos

=

513

1213

tan =5

12

cot =1

tan

=1

512

cot =12

5


4. If cos = 725

and is in fourth quadrant, find the other five trigonometric

ratios. (3 marks)

Sol. cos = 7

25

sec = 25

7

1sec

cos

sin2 + cos2 = 1

sin2 = 1 – cos2

=7

1 –25

2

=49

1 –625

=625 – 49

625

=576

625

sin =24


cosec =25

24

1cosec

sin

sin

cos

= tan

2425

725

= tan

tan =24

7

cot =7

24

1cot

tan

is in the fourth quadrant

sin , cosec , tan and cot are negative

sin = –24

25, cosec =

–25

24, tan =

–24

7 and cot =

– 7

24


SCHOOL SECTION 189


2. If cot = 7

–24

, then find the values of sin and sec , is in IV quadrant.

(3 marks)

Sol. cot =7

–24

tan =24

–7

1tan

cot

1 + tan2 = sec2

1 + 24

–7

2

= sec2

1 + 576

49= sec2

49 576

49

= sec2

625

49= sec2

sec =25

7

cos =7

25

1cos

sec

sin

cos

= tan

sin = tan × cos

sin =–24 7

×7 25

sin =–24

25


3. 3 sin – 4 cos = 0, then find the values of tan , sec and cosec ,where is an acute angle. (3 marks)

Sol. 3 sin – 4 cos = 0 3 sin = 4 cos

sin

cos

=

4

3

tan =4

31 + tan2 = sec2

1 + 4

3

2

= sec2

1 + 16

9= sec2

9 16

9

= sec2

25

9= sec2

sec =5



SCHOOL SECTION190

cot =1

tan

=1

43

cot =3

41 + cot2 = cosec2

1 + 3

4

2

= cosec2

1 + 9

16= cosec2

16 9

16

= cosec2

25

16= cosec2

cosec =5



4. If tan = 1, then find the value of sin + cos

sec + cosec , where is an acute

angle. (3 marks)Sol. tan = 1

sin

cos

= 1

sin = cos .......(i)1 + tan2 = sec2

1 + (1)2 = sec2 1 + 1 = sec2 2 = sec2 sec = 2 [Taking square roots]

cos =1

sec

cos =1

2

sin =1

2[From (i)]

cosec =1

sin

=1

12

cosec = 2

sin cos

sec cosec

=

1 1

2 22 2

=

22

2 2


SCHOOL SECTION 191

=2

2 2 2

=2

2 2

=2

4

sin cos

sec cosec

=

1

2


5. If sec = 2

3, then find the value of

1 – cosec1 + cosec

, where is in IV quadrant.

(3 marks)

Sol. sec =2

3

cos =1

sec

=

12

3

cos =3

2sin2 + cos2 = 1

sin2 + 3

2

2

= 1

sin2 + 3

4= 1

sin2 =3

1 –4

sin2 =4 – 3

4

sin2 =1

4

sin =1


cosec =1

sin

=112

cosec = 2 is in IV quadrant

cosec = – 21 – cosec

1 cosec =

1 – (–2)

1 (– 2)

1 – cosec

1 cosec =

1 2

1 – 2

1 – cosec

1 cosec = –3


SCHOOL SECTION192


14. If 3 tan = 3 sin , find the value of sin2 – cos2 , where 0.

(3 marks)

Sol. 3 tan = 3 sin

sin3

cos

= 3 sin

3

cos = 3

cos =3

3

cos2 =3

9

cos2 =1

3.......(i)

sin2 + cos2 = 1 sin2 = 1 – cos2

=1

1 –3

[From (i)]

=3 – 1

3

sin2 =2

3......(ii)

sin2 – cos2 =2 1

–3 3

[From (i) and (ii)]

=2 – 1

3

sin2 – cos2 =1

3


8. Eliminate , if(i) x = a sec , y = b tan (2 marks)Sol. x = a sec

sec =x

a......(i)

y = b tan

tan =y

b......(ii)

1 + tan2 = sec2

1 + y

b

2

=x

a

2

[From (i) and (ii)]

1 + y

b

2

2 =x

a

2

2

2

2

xa

– 2

2

yb

= 1


SCHOOL SECTION 193

(ii) x = 2 cos – 3 sin , y = cos + 2 sin (3 marks)Sol. x = 2 cos – 3 sin ......(i)

y = cos + 2 sin ......(ii)Multiplying (ii) by 2,

2y = 2 cos + 4 sin .....(iii)Subtracting (iii) from (i),x – 2y = 2 cos – 3 sin – (2 cos + 4 sin )

x – 2y = 2 cos – 3 sin – 2 cos – 4 sin x – 2y = – 7 sin

sin =–(x – 2y)

7......(iv)

Substituting sin – x 2y

7

in equation (ii)

y = cos + 2 x 2y

7

y = cos 2 x 2y

7

y + 2 x 2y

7

= cos

7y 2 x 2y

7

= cos

7y 2x 4y

7

= cos

cos =2x 3y

7

We know,sin2 + cos2 = 1

– (x – 2y) 2x 3y

7 7

2 2

= 1

(x – 2y) (2x 3y)

49 49

2 2

= 1

Multiplying throughout by 49,(x – 2y)2 + (2x + 3y) = 49

(iii) x = 3 cosec + 4 cot , y = 4 cosec – 3 cot (3 marks)Sol. x = 3 cosec + 4 cot ......(i)

y = 4 cosec – 3 cot .....(ii)Multiplying (i) by 4,

4x = 12 cosec + 16 cot .....(iii)Multiplying (ii) by 3,

3y = 12 cosec – 9 cot .....(iv)Subtracting (iv) from (iii),4x – 3y = 12 cosec + 16 cot – (12 cosec – 9 cot )

4x – 3y = 12 cosec + 16 cot – 12 cosec + 9 cot 4x – 3y = 25 cot

cot = 4x – 3y

25

Substituting cot = 4x – 3y

25 in equation (i)


SCHOOL SECTION194

x = 3cosec + 4 4x – 3y

25

x = 3cosec + 16x – 12y

25

x – 16x – 12y

25= 3cosec

25x – 16x 12y

25

= 3cosec

9x 12y

25

= 3cosec

3 (3x 4y)

3 25

= cosec

cosec = 3x 4y

25

We know, cosec2 – cot2 = 1

3x 4y 4x – 3y

–25 25

2 2

= 1

(3x 4y) (4x – 3y)

–625 625

2 2

= 1

Multiplying throughout by 625,(3x + 4y)2 – (4x – 3y)2 = 625


6. Find the possible values of sin x if 8 sin x – cos x = 4. (3 marks)Sol. 8 sin x – cos x = 4

8 sin x – 4 = cos x ..... (i) sin2 x + cos2 x = 1 sin2 x + (8 sin x – 4)2 = 1 [from (i) sin2 x + 64 sin2 x – 64 sin x + 16 = 1 sin2 x + 64 sin2 x – 64 sin x + 16 – 1 = 0 65 sin2 x – 64 sin x + 15 = 0 64 sin2 x + sin2 x – 64 sin x + 16 – 1 = 0 65 sin2 x – 64 sin x + 15 = 0 65 sin2 x – 39 sin x – 25 sin x + 15 = 0 13 sin x (5 sin x – 3) – 5 (5 sin x – 3) = 0 (5 sin x – 3) (13 sin x – 5) = 0 5 sin x – 3 = 0 or 13 sin x – 5 = 0 5 sin x = 3 or 13 sin x = 5

sin x =3

5or sin x =

5

13


8. Find the possible value of cos x if cot x + cosec x = 5. (3 marks)Sol. cot x + cosec x = 5

cos x 1

sin x sin x = 5

cos x 1

sin x

= 5

cos x 1

5

= sin x ....... (i)

sin2 x + cos2 x = 1


SCHOOL SECTION 195

cos x 1

5

2

+ cos2 x = 1 [from (i)]

cos x 1

25

2

+ cos2 x = 1

(cos x + 1)2 + 25cos2 x = 25 [Multiplying throughout by 25] cos2 x + 2 cos x + 1 + 25cos2 x – 25 = 0 26cos2 x + 2 cos x – 24 = 0 2 (13 cos2 x + cos x – 12) = 0 13 cos2 x + cos x – 12 = 0 13 cos2 x + 13 cos x – 12 cos x – 12 = 0 13 cos x (cos x + 1) – 12 (cos x + 1) = 0 (cos x + 1) (13 cos x – 12) = 0 cos x + 1 = 0 or 13 cos x – 12 = 0 cos x = – 1 or 13 cos x = 12

cos x = – 1 or cos x =12

13


11. If x = a sin , y = b tan then prove that 2 2

2 2

a b–

x y = 1. (2 marks)

Proof : x = a sin

1

sin =a

x

cosec =a

x......(i)

1cosec

sin

y = b tan

1

tan =b

y

cot =b

y .....(ii)1

cotta n

We know,

1 + cot2 = cosec2 cosec2 – cot2 = 1

a b

–x y

22

= 1 [From (i) and (ii)]

2 2

2 2

a b–

x y = 1


12. If a cos + b sin = m and a sin – b cos = n, prove that a2 + b2 = m2 + n2.(3 marks)

Proof : a cos + b sin = m (a cos + b sin )2 = m2 .......(i) [Squaring both sides]

(a sin – b cos ) = n (a sin – b cos )2 = n2 ......(ii) [Squaring both sides]

Adding (i) and (ii),(a cos + b sin )2 + (a sin – b cos )2 = m2 + n2

a2 cos2 + 2ab cos . sin + b2 sin2 + a2 sin2 – 2ab sin . cos + b2

cos2 = m2 + n2

a2 cos2 + b2 cos2 + a2 sin2 + b2 sin2 = m2 + n2

cos2 (a2 + b2) + sin2 (a2 + b2) = m2 + n2

(a2 + b2) (cos2 + sin2 ) = m2 + n2

a2 + b2 = m2 + n2 [ sin2 + cos2 = 1]


SCHOOL SECTION196


9. If tan + sin = m and tan – sin = n, show that m2 – n2 = 4 mn .(4 marks)

Proof : tan + sin = mtan – sin = nm2 – n2 = (tan + sin )2 – (tan – sin )2

= tan2 + 2 tan .sin + sin2 – [tan2 – 2 tan sin + sin2 ]= tan2 + 2 tan .sin + sin2 – tan2 + 2 tan .sin – sin2 = 4 tan .sin .......(i)

4 mn = 4 tan sin tan – sin

= 4 tan – sin2 2

=sin

4 – sincos

22

2

=1

4 sin – 1cos

22

= 4 sin sec – 1 2 2

= 4 ta n . sin 2 21 tan sec

tan sec – 1

2 2

2 2

= 4 × sin × tan .....(ii)From (i) and (ii),m2 – n2 = 4 mn


10. If sec + tan = p, show that 2

2

p – 1p + 1 = sin . (3 marks)

Proof : sec + tan = p

1 sin

cos cos

= p

1 sin

cos

= p

(1 sin )

cos

2

2 = p2

1 sin

1 – sin

2

2 = p2sin cos 1

cos 1 – sin

2 2

2 2

1 sin

1 sin 1 – sin

2

= p2

1 sin

1 – sin

= p2

1 sin + 1 – sin

1 sin – 1 + sin

=

p 1

p – 1

2

2 [By Componendo-Dividendo]

2

2 sin =p 1

p – 1

2

2

1

sin =p 1

p – 1

2

2

2

2

p – 1p + 1 = sin [By Invertendo]


SCHOOL SECTION 197


13. If sin + sin2 = 1, prove that cos2 + cos4 = 1. (2 marks)Proof : sin + sin² = 1 [Given]

sin = 1 – sin²

sin = cos2 sin + cos = 1

1 – sin = cos

2 2

2 2

sin2 = cos4 [Squaring both sides]

1 – cos2 = cos4 sin + cos = 1

1 – cos = sin

2 2

2 2

cos² + cos4 = 1


7. Prove the following :

(x) If tan A + 1

tan A = 2, show that tan2 A + 2

1tan A = 2. (2 marks)

Proof : tan A + 1

tan A = 2

1

tan Atan A

2

= 4 [Squaring both sides]

tan2 A + 2 tan A . 1 1

tan A tan A 2 = 4

tan2 A + 2 + 1

tan A2 = 4

tan2 A + 1

tan A2 = 4 – 2

tan2 A + 2

1tan A = 2

(iii) sec2 + cosec2 = sec2 . cosec2 (3 marks)Proof : L.H.S. = sec2 + cosec2

=1 1

cos sin2 2

1 1sec , cosec

cos sin

=sin + cos

cos . sin

2 2

2 2

=1

cos . sin2 2 [ sin2 + cos2 = 1]

= sec2 . cosec2 = R.H.S.

sec2 + cosec2 = sec2 . cosec2

(viii) 2 2sec + cosec = tan + cot (3 marks)

Proof : L.H.S. = sec cosec2 2

= 1 tan 1 cot 2 2

= tan 2 cot 2 2

= (tan cot ) 2

1tan

cot

tan cot 1


SCHOOL SECTION198

= tan + cot = R.H.S.

2 2sec + cosec = tan + cot

(v)tan sec + 1

+sec + 1 tan

= 2 cosec (3 marks)

Proof : L.H.S. =tan sec 1

sec 1 tan

=tan (sec + 1)

(sec 1) tan

2 2

=tan sec 2sec 1

(sec 1) tan

2 2

=sec sec 2sec

(sec 1) tan

2 2 [ 1 + tan2 = sec2 ]

=2sec 2sec

(sec 1) tan

2

=2sec (sec + 1)

(sec 1) tan

=2sec

tan

=1 cos

2cos sin

sin 1

tan , seccos cos

=2

sin = 2 cosec = R.H.S.

tan sec + 1

+sec + 1 tan

= 2 cosec

(i)1 – cos A1 + cos A = cosec A – cot A (3 marks)

Proof : L.H.S. =1 – cos A

1 cos A

= 1 – cos A 1 – cos A

1 cos A 1 – cos A

= 1 – cos A

1 – cos A

2

2

= 1 – cos A

sin A

2

2

sin A cos A 1

sin A 1 – cos A

2 2

2 2

=1 – cos A

sin A

=1 cos A

–sin A sin A


SCHOOL SECTION 199

= cosec A – cot A 1 cos

cosec = , cotsin sin

= R.H.S.

1 – cos A1 + cos A = cosec A – cot A

(ii) cosec x – 1 1cosec x + 1 sec x + tan x (4 marks)

Proof : L.H.S. =cosec x – 1

cosec x 1

= cosec x – 1 cosec x – 1

cosec x 1 cosec x – 1

= cosec x – 1

cosec x – 1

2

2

= cosec x – 1

cot x

2

2

1 cot x cosec x

cot x cosec x – 1

2 2

2 2

=cosec x – 1

cot x

=cosec x 1

–cot x cot x

=

1sin x

– tan xcos xsin x

=1

– tan xcos x

= sec x – tan x

=(sec x – tan x) (sec x tan x)

(sec x tan x)

=sec x – tan x

sec x tan x

2 2

=1

sec x tan x1 tan x sec x

sec x – tan x 1

2 2

2 2

= R.H.S.

cosec x – 1cosec x + 1 =

1sec x + tan x

(iv) sec6 x – tan6 x = 1 + 3 sec2 x.tan2 x (3 marks)Proof : L.H.S. = sec6 x – tan6 x

= (sec2 x)3 – (tan2 x)3

= (sec2 x – tan2 x)[(sec2 x)2 + sec2 x . tan2 x + (tan2 x)2]= 1 × [(sec2 x)2 + (tan2 x)2 + sec2 x . tan2 x]= (sec2 x – tan2 x)2 + 2 sec2 x . tan2 x + sec2 x . tan2 x

[(a – b)2 = a2 – 2ab + b2, a2 + b2 = (a – b)2 + 2ab]= 1 + 3 sec2x.tan2x.= R.H.S.

sec6 x – tan6 x = 1 + 3 sec2 x . tan2 x


SCHOOL SECTION200

(ix)1 1 1 1

– = –cosec A – cot A sin A sin A cosec A + cot A

i.e. 1 1 1 1

+ = +cosec A – cot A cosec A + cot A sin A sin A (3 marks)

Proof : L.H.S. =1 1

+cosec A – cot A cosec A + cot A

=cosec A + cot A + cosec A – cot A

(cosec A – cot A) (cosec A + cot A)

=2cosec A

cosec A – cot A2 2

=2 cosec A

1

1 cot A cosec A

cosec A – cot A 1

2 2

2 2

= cosec A + cosec A

=1 1

sin A sin A

= R.H.S.

1 1 1 1

+ = +cosec A – cot A cosec A + cot A sin A sin A

1 1 1 1

– = –cosec A – cot A sin A sin A cosec A + cot A


6. Show that : 2 3cos sin

+ 1 – tan sin – cos

= 1 + sin . cos (3 marks)

Proof : L.H.S. =cos sin

1 – tan sin – cos

2 3

=

cos sinsin sin – cos1 –cos

2 3

=

cos sincos – sin sin – cos

cos

2 3

=cos sin

cos – sin sin – cos

3 3

=cos sin

–cos – sin cos – sin

3 3

=cos – sin

cos – sin

3 3

=(cos – sin ) (cos + cos . sin + sin )

(cos – sin )

2 2

= cos2 + sin2 + sin . cos = 1 + sin . cos [ sin2 + cos2 = 1]= R.H.S.

2 3cos sin

+1 – tan sin – cos

= 1 + sin . cos


SCHOOL SECTION 201


4. Show that : tan cot

+ 1 – cot 1 – tan

= 1 + sec . cosec . (4 marks)

Proof : L.H.S. =tan cot

1 – cot 1 – tan

=

sin coscos sin

cos sin1 – 1 –sin cos

=

sin coscos sin

sin – cos cos – sinsin cos

=sin cos

cos (sin – cos ) sin (cos – sin )

2 2

=sin cos

–cos (sin – cos ) sin (sin – cos )

2 2

=1 sin cos

–sin – cos cos sin

2 2

=1 sin – cos

sin – cos cos sin

3 3

= sin – cos sin sin cos cos1

sin – cos cos sin

2 2

=sin sin . cos cos

cos . sin

2 2

=1 sin . cos

cos . sin

[ sin2 + cos2 = 1]

=1 sin . cos

cos . sin cos . sin

=1 1

× 1cos sin

= sec . cosec + 1 1 1

sec , coseccos sin

= R.H.S.

tan cot

+1 – cot 1 – tan

= 1 + sec . cosec .


7. Prove the following :

(vi)1 + sin A 1 + sin A + cos A

=cos A 1 + cos A – sin A (3 marks)

Proof : 1 – sin2 A = cos2 A (1 – sin A) (1 + sin A) = cos A . cos A


SCHOOL SECTION202

1 sin A

cos A

=

cos A

1 – sin A

1 sin A

cos A

=

cos A

1 – sin A =1 sin A cos A

cos A 1 – sin A


1 + sin A

cos A =1 + sin A + cos A1 + cos A – sin A

(vii)tan A tan A + sec A + 1

=sec A – 1 tan A + sec A – 1 (3 marks)

Proof : 1 + tan2 A = sec2 A tan2 A = sec2 A – 1 tan A . tan A = (sec A – 1) (sec A + 1)

tan A

sec A – 1 =sec A 1

tan A

tan A

sec A – 1 =sec A 1

tan A

=

tan A sec A 1

sec A – 1 tan A


tan A

sec A – 1 = tan A + sec A + 1tan A + sec A – 1


5. Show that : sin – cos + 1sin + cos – 1

=

1sec – tan (4 marks)

Proof : sin2 + cos2 = 1 cos2 = 1 – sin2 cos . cos = (1 – sin ) (1 + sin )

cos

1 – sin

=

1 sin

cos

By theorem on equal ratios,

1 + sin – cos

cos – (1 – sin )

=

cos

1 – sin

=

1 sin

cos

1 sin – cos

cos – (1 – sin )

=

cos

1 – sin

Dividing the numerator and denominator of R.H.S. by cos

1 + sin – cos

cos – 1 + sin

=

coscos

(1 – sin )cos

1 + sin – cos

cos – 1 + sin

=

11 sin

–cos cos

sin – cos + 1sin + cos – 1

=

1sec – tan


7. If 3 tan2 – 4 3 tan + 3 = 0, find the value of . (3 marks)

Sol. 3 tan2 – 4 3 tan + 3 = 0

3 3 tan – 4tan 32 = 0

3 tan – 4tan 32 = 0

3 tan – 3tan – tan + 32 = 0


SCHOOL SECTION 203

3 tan tan – 3 – 1 tan – 3 = 0

tan – 3 3 tan – 1 = 0

tan – 3 = 0 OR 3 tan – 1 = 0

tan = 3 3 tan = 1

But, tan 60 = 3 tan =1

3

tan = tan 60 But, tan 30 = 1

3 = 60 tan = tan 30

= 30

` Heights and Distances:Many times , we require to find the height of a tower, building, tree ordistance of a ship from the lighthouse or width of the river etc. We cannotmeasure them actually, we can find the heights and distances with thehelp of trigonometric ratios.

(i) Line of vision : The line connecting the eye of the observer and the objectis called as the Line of vision.

(ii) Angle of Elevation : If A, B are twopoints such that B is at higher levelthan A and AM is horizontal linethrough A, then MAB is the angle ofelevation of B with respect to A.

(iii) Angle of Depression : If A, B are twopoints such that B is at lower level thanA and AM is the horizontal line throughA, then MAB is the angle of depressionof B with respect to A.


1. For a person standing at a distance of 80 m from a church, the angle ofelevation of its top is of measure 45º. Find the height of the church.

(3 marks)Sol. seg AB represents the church

C represents the position of observer.BC = 80 mACB is the angle of elevationm ACB = 45ºIn right angled ABC,

tan 45º =AB

BC[By definition]

1 =AB

80 AB = 80 m

The height of the church is 80 m.

Line of V

ision

Angle of Elevation

Horizontal Line

B

AM

Angle of DepressionLine of Vision

Horizontal LineA

M

B

A

B C80 m

45º


SCHOOL SECTION204


6. A kite is flying at a height of 60 m above the ground. The string attachedto the kite is temporarily tied to the ground. The inclination of thestring with the ground is 60º. Find the length of the string, assumingthat there is no slack in the string. ( 3 = 1.73) (3 marks)

Sol. seg AB represents the distance of a kite from ground.

AB = 60 m

seg AC represents the length of the string

m ACB = 60º

In right angled ABC,

sin 60º =AB

AC[By definition]

3

2=

60

AC

AC =120

3

AC =120 3

3

AC = 40 3 m

AC = 40 × 1.73

AC = 69.2 m

The length of the string, assuming that there is no slack in the

string is 69.2 m.


19. A circus artist is climbing from the ground along a rope stretched fromthe top of a vertical pole and tied to a peg at the ground. The height ofthe pole is 12 m and the angle made by the rope with the ground level is30º. Calculate the distance covered by the artist in climbing to the topof the pole. (3 marks)

Sol. seg AB represents the height of the pole.seg BC represents the distance of the polefrom where the rope is tied to the ground.ACB is angle made by rope with groundseg AC represents the length of ropeIn right angle ABC,

sin 30º =AB

AC[By definition]

1

2=

12

AC AC = 24 m

The distance covered by the artist in climbing the top of the pole is24 m.

A

B 60ºC

60 m

A

B C

12 m

30º


SCHOOL SECTION 205


3. Two buildings are in front of each other on either side of a road of width10 metres. From the top of the first building, which is 30 metres high,the angle of elevation of the top of the second is 45º. What is the heightof the second building ? (4 marks)

Sol. seg AB and seg CD represents the two buildingsAB = 30 mseg BD represents the width of the roadBD = 10 mA represents the position of observer.CAE is the angle of elevationm CAE = 45ºABDE is a rectangleAB = DE = 30 m [Opposite sides of rectangle]BD = AE = 10 mIn right angled CEA,

tan 45º =CE

AE[By definition]

1 =CE

10 CE = 10 m

CD = CE + ED [ C - E - D] CD = 10 + 30 CD = 40 m

The height of second building is 40 m.


5. A tree is broken by the wind. The top struck the ground at an angle of30º and at a distance of 30 m from the root. Find the whole height of

the tree. ( 3 = 1.73) (5 marks)

Sol. seg AB represents the height of the treeThe tree breaks at point Dseg AD is the broken part of tree whichthen takes the position of DC

AD = DCmDCB = 30ºBC = 30 mIn right angled DBC,

tan 30º =DB

BC[By definition]

1

3=

DB

30

DB =30

3

DB =30 3

3

DB = 10 3 m

A

D

B C30 m30º

A

C

E

B D

45º

10 m

30 m


SCHOOL SECTION206

cos 30º =BC

DC[By definition]

3

2=

30

DC

DC =30 2

3

DC =30 3 2

3

DC = 20 3 m

AD = DC = 20 3 m

AB = AD + DB [ A - D - B]

AB = 20 3 + 10 3

AB = 30 3 m

AB = 30 × 1.73 AB = 51.9 m

The height of tree is 51.9 m.


15. A tree 12m high, is broken by the wind in such a way that its top touchesthe ground and makes an angle 60º with the ground. At what height from

the bottom, the tree is broken by the wind ? ( 3 = 1.73) (5 marks)

Sol. seg AB represents the treeAB = 12 mThe tree breaks at point Dseg AD is the broken part of treewhich then takes the position of DC

AD = DCm DCB = 60ºLet DB = x m

AD + DB = AB [ A - D - B] AD + x = 12 AD = (12 – x) m DC = (12 – x) m

In right angled DBC,

sin 60º =DB

DC[By definition]

3

2=

x

12 – x

3 12 – x = 2x

12 3 – 3 x = 2x

12 3 = 2x 3 x

x 2 3 = 12 3

x =12 3

2 3

A

B C60º

D12 m


SCHOOL SECTION 207

DB =12 3

2 3 m

DB =

12 3 2 – 3

2 3 2 – 3

DB = 24 3 – 12 (3)

(2) – 322

DB =24 3 – 36

4 – 3

DB =24 (1.73) – 36

1 DB = 41.52 – 36 DB = 5.52 m

The height at which the tree is broken from the bottom by the windis 5.52 m.


16. A person standing on the bank of a river observes that the angle ofelevation of the top of a tree standing on the opposite bank is 60º. Whenhe moves 40 m away from the bank, he finds the angle of elevation to

be 30º. Find the height of the tree and the width of the river. ( 3 = 1.73)

(5 marks)Sol. Let seg AB represents the tree

seg BC represents width of riverLet BC = x mC and D represents the initial andfinal positions of the observerDC = 40 mACB and ADB are theangles of elevationm ACB = 60º and m ADB = 30ºIn right angled ACB,

tan 60º =AB

BC[By definition]

3 =AB

x

AB = 3 x m .....(i)

In right angled ADB,

tan 30º =AB

DB[By definition]

1

3=

AB

40 x

AB =40 x

3

m .....(ii)

From (i) and (ii) we get,

A

BC

D40 m

30º 60º


SCHOOL SECTION208

3 x =40 x

3

3x = 40 + x 3x – x = 40 2x = 40 x = 20 BC = 20 m

AB = 20 3 m [From (i)]

AB = 20 × 1.73 AB = 34.6 m

Height of tree is 34.6 m and width of river is 20 m.


4. Two poles of height 18 metres and 7 metres are erected on the ground.A wire of length 22 metres tied to the top of the poles. Find the anglemade by the wire with the horizontal. (4 marks)

Sol. seg AB and CD represents two poles. AB = 18 m, CD = 7 mseg AC represent the length of the wire. AC = 22 mEBDC is a rectangle

EB = CD = 7 m [Opposite sides of rectangle]AB = AE + EB [ A - E - B]

18 = AE + 7 18 – 7 = AE AE = 11 m

In right angled AEC,

sin C =AE

AC[By definition]

sin C =11

22

sin C =1

2......(i)

But,

sin 30º =1

2......(ii)

sin C = sin 30º C = 30º

The angle made by the wire with horizontal is 30º.


2. From the top of a lighthouse, an observer looks at a ship and find theangle of depression to be 60º. If the height of the lighthouse is 90metres then find how far is that ship from the lighthouse ? ( 3 = 1.73)

(3 marks)Sol. seg AB represents the lighthouse

C represents the position of ship.A represents the position of observer.EAC is the angle of depression.AB = 90 mm EAC = 60º

A

E

BD

C

7 m

22 m

18 m

A

B C

E

60º

90 m

60º


SCHOOL SECTION 209

EAC ACB [Converse of alternate angle test] m ACB = 60º


tan 60º =AB

BC[By definition]

3 =90

BC

BC =90

3

BC =90 3

3

BC = 30 3 m

BC = 30 × 1.73 BC = 51.9 m

The ship is 51.9 m far from the lighthouse.


17. The angle of elevation of a cloud from a point 60 m above a lake is 30ºand the angle of depression of the reflection of cloud in the lake is 60º.Find the height of the cloud. (5 marks)

Sol.

Let E be the position of the cloudand let BC represent the surfaceof the lake.Let A be the point of observer andlet F be the reflection of the cloud

EC = CFLet EC = CF = x mABCD is a rectangle [By definition]

AB = CD = 60 m [Opposite sidesof rectangle]

EC = ED + DC [E - D - C] x = ED + 60 ED = (x – 60)m

Also,DF = DC + CF [D - C - F]

DF = (60 + x) DF = (x + 60) m

In right angled ADE,

tan 30º =ED

AD[By definition]

1

3=

x – 60

AD

AD = 3 x – 60 m

In right angled ADF,

tan 60º =DF

AD[By definition]

3 =x + 60

3 (x – 60)

A

B

E

D

C

60º30º

60 m

F

60 m

x

x


SCHOOL SECTION210

3 (x – 60) = x + 60 3x – 180 = x + 60 3x – x = 60 + 180 2x = 240 x = 120

The height of the cloud above the lake is 120 m.


18. A man on cliff observes a boat at an angle of depression 30º, which issailing towards the point of the shore immediately beneath him. Threeminutes later the angle of depression of the boat is found to be 60º.Assuming that the boat sails at a uniform speed, determine how muchmore time it will take to reach the shore. (5 marks)

Sol. D and C are the initial andfinal positions of the ship.A represents the position of observer.EAD and EAC are the anglesof depression.m EAD = 30º, m EAC = 60ºm EAD = m ADB = 30º [By converse of alternate angles test]m EAC = m ACB = 60ºThe ship took 3 mins to travel from D to CLet the speed of the boat be x units/minute

Distance = Speed × Time CD = x × 3 CD = 3x units ......(i)


tan 60º =AB

BC[By definition]

3 =AB

BC

3 BC = AB ......(ii)

In right angled ABD,

tan 30º =AB

BD[By definition]

1

3=

AB

BD

BD

3= AB ......(iii)

From (ii) and (iii) we get,

3 BC =BD

3 3BC = BD 3BC = BC + CD [ B - C - D] 3BC – BC = CD 2BC = CD

BC =CD

2

BC =3x

2 BC = 1.5 x units

A

BC

E

D

30º60º

30º60º


SCHOOL SECTION 211

Time =Dis tance

Speed

=BC

x

=1.5x

x= 1.5 minutes= 1.5 × 60 seconds= 90 seconds

The time taken by the ship to reach the shore is 90 seconds.


20. A bird was flying in a line parallel to the ground from north to south ata height of 2000 metres. Tom, standing in the middle of the field, firsthe observed the bird in the north at an angle of 30º. After 3 minutes, heagain observed it in the south at an angle of 45º. Find the speed of thebird in kilometers per hour. ( 3 = 1.73) (5 marks)

Sol. A and C represents the first and thesecond position of the bird at northand south respectively.seg BD is the distance covered bybird in 3 mins.seg AB and seg CD represents the height at which the bird is flying.AB = CD = 2000 mAEB and CED are the angles of elevation m AEB = 30º and mCED = 45ºIn right angled ABE,

tan 30º =AB

BE[By definition]

1

3=

2000

BE

BE = 2000 3 m

In right angled CDE,

tan 45º =CD

ED[By definition]

1 =2000

ED

ED = 2000 mBD = BE + ED [ B - E - D]

BD = 2000 3 2000

BD = 2000 3 1 m

BD = 2000 3 1

1000

km [ 1 km = 1000 m]

BD = 2 3 1 km

Speed =Distance

Time

Speed = 2 3 1

360

[ 1 hour = 60 minutes]

30º

A

BE D

C

45º

200m


SCHOOL SECTION212

Speed = 2 3 1 60

3

Speed = 40 3 1 Speed = 40 (1.73 + 1) Speed = 40 (2.73) Speed = 109.2 km/hr.

Speed of the bird is 109.2 km/hr.


18. P is the circumcentre of an acute angled triangle ABC with cirumradius R. Midpointof BC is D. Show that the perimeter of ABC is 2R (sin A + sin B + sin C).

(5 marks)Proof :

Let, m BAC = x ......(i) m BPC = 2x .....(ii) [ An angle subtended by an arc at the

centre is double the angle subtended byit at any point on the remaining part ofthe circle]

In BDP and CDP,seg BD seg CD [ D is midpoint of side BC]seg DP seg DP [Common side]seg PB seg DC [Radii of the same circle]

BDP CDP [By SSS test of congruence] BPD CPD [c.s.c.t.]

m BPD = m CPD = 1

2m BPC

m BPD = m CPD = 1

2 × 2x [From (ii)]

m BPD = m CPD = x .....(iii)In right angled BDP,

sin BPD = BD

BP

sin x = BD

R......(iv) [From (i)]

sin x = 2BD

2R [Multiplying and dividing the R.H.S. by 2]

sin x = BC

2R [ D is midpoint of seg BC]

sin A = BC

2R 2R . sin A = BC .......(v)

Similarly,2R . sin B = AC ......(vi)2R . sin C = AB .....(vii)

A

B CD

R RP


SCHOOL SECTION 213

Adding (v), (vi) and (vii),2R . sin A + 2R . sin B + 2R . sin C = BC + AC + AB

2R . (sin A + sin B + sin C) = Perimeter of ABC Perimeter of ABC = 2R (sin A + sin B + sin C)

20. Find the area of regular polygon having 15 sides which is inscribed in acircle of radius 4 cm. (sin 24 = 0.407) (4 marks)

Sol. A regular polygon having 15 sides is inscribed in a circle

Measure of arc () =360

15 = 24º

Radius (r) = 4 cm

Area of OAB =1

r sin2

2

=1

× 4 × 4 × sin 242

= 8 × 0.407 Area of OAB = 3.256 cm2

Area of a regular polygon = 15 × 3.256= 48.84 cm2

Area of a regular polygon is 48.84 cm2.

21. Prove that (1 + tan )2 + (1 + cot )2 = (sec + cosec )2. (4 marks)Sol. L.H.S. = (1 + tan )2 + (1 + cot )2

= 1 + 2 tan + tan2 + 1 + 2 cot + cot2 = 1 + tan2 + 1 + cot2 + 2 tan + 2 cot = sec2 + cosec2 + 2 (tan + cot )

[ 1 + tan2 = sec2 , 1 + cot2 = cosec2 ]

= sec2 + cosec2 + sin cos

2cos sin

= sec2 + cosec2 + sin + cos

2cos × sin

2 2

= sec2 + cosec2 + 2 × 1

cos × sin [ sin2 + cos2 = 1]

= sec2 + cosec2 + 2 × sec × cosec = (sec + cosec )2

= R.H.S. (1 + tan )2 + (1 + cot )2 = (sec + cosec )2

31. If sec – tan = P. Obtain the values of tan, sec and sin in terms of P.(4 marks)

Sol. sec – tan = p [Given] sec = (p + tan ) ......(i)

1 + tan2 = sec2 1 + tan2 = (p + tan )2 [From (i)] 1 + tan2 = p2 + 2p . tan + tan2 1 = p2 + 2p . tan 1 – p2 = 2p . tan

1 – p

2p

2

= tan

tan =1 – p

2p

2

Substituting tan = 1 – p

2p

2

in (i)

24º

O

4 4

A B


SCHOOL SECTION214

sec =1 – p

p2p

2

sec =2p 1– p

2p

2 2

sec =p 1

2p

2

sec =1 p

2p

2

cos =1

sec

cos =2p

1 p 2

sin

cos

= tan

sin = tan × cos

sin =1 – p 2p

2p 1 p

2

2

sin =1 – p

1 p

2

2

32. Prove that 1 + sin x – cos x 1 + sin x + cos x

+ = 2 cosec x1 + sin x + cos x 1 + sin x – cos x . (4 marks)

Proof : L.H.S. =1 + sin x – cos x 1 + sin x + cos x

+1 + sin x + cos x 1 + sin x – cos x

=

1 + sin x – cos x + 1 + sin x + cos x

1 + sin x + cos x 1 + sin x – cos x

2 2

=

1 + sin x – cos x + 1 + sin x + cos x

1 + sin x + cos x 1 + sin x – cos x

2 2

=

1 sin x – 2 1 sin x . cos x cos x 1 sin x 2 1 sin x . cos x cos x

1 sin x – cos x

2 22 2

2 2

= 2 1 sin x 2 cos x

1 2 sin x sin x – cos x

2 2

2 2

= 2 1 + 2 sin x + sin x + 2 cos x

1 – cos x + sin x + 2 sin x

2 2

2 2

=2 + 4 sin x + 2 sin + 2cos x

sin x + sin x + 2 sin x

2 2

2 2

sin x + cos x 1

sin x 1 – cos x

2 2

2 2

= 2 + 4 sin x + 2 sin x +cos x

2 sin x + 2 sin x

2 2

2

=2 + 4sin x + 2 (1)

2 sin x (sin x + 1) [ sin2 x + cos2 x = 1]

=4 + 4sin x

2 sin x (sin x + 1)

=4 (1 + sin x)

2 sin x (sin x + 1)

=2

sin x


SCHOOL SECTION 215

= 2 cosec x= R.H.S.

1 + sin x – cos x 1 + sin x + cos x

+1 + sin x + cos x 1 + sin x – cos x = 2 cosec x

33. The angle of elevation of a jet plane from a point on the ground is 60º.After a flight of 30 secs., the angle of elevation changes to 30º. If the jet

plane is flying at a constant height of 3600 3 m, find the speed of the

jet plane. (5 marks)Sol. Let A and B be the first and

second position of a jet plane.P be the point of observerAPD and BPC are the anglesof elevationm APD = 60º, mBPC = 30º

AD and BC represent constant height at which the jet plane is flying.

AD = BC = 3600 3 m

In right angled ADP,

tan 60º =AD

PD[By definition]

3 =3600 3

PD PD = 3600 m

In right angled BCP,

tan 30º =BC

PC[By definition]

1

3=

3600 3

PC PC = 3600 × 3 PC = 10800 m

PD + DC = PC [ P - D - C] 3600 + DC = 10800 DC = 10800 – 3600 DC = 7200 m

ABCD is a rectangle [By definition] AB = DC = 7200 m [Opposite sides of rectangle] Distance covered in 30 seconds = 7200 m

Speed =Dis tance

Time

=AB

30 secs

=7200m

30 secs

= 7200 30

1000 3600 [1 km = 1000 m and 1 hr = 3600 seconds]

= 7200 3600

1000 30

= 864 km /hr.

Speed of the jet plane is 864 km/hr.

B

D CP

60º30º

3600 3

A


SCHOOL SECTION216

34. A straight highway leads to the foot of tower. A man standing at the topof tower observes a car at an angle of depression of 30º, which isapproaching the foot of the tower with uniform speed. Six seconds later,the angle of depression of the car is found to be 60º. Find the time takenby the car to reach the foot of the tower from this point. (5 marks)

Sol. Seg AB represents a tower. Let A be a position of an observer.D and C are the initial and final position of the car.EAD and EAC are angles of depressionm EAD = m ADB = 30º [Converse of alternate anglesm EAC = m ACB = 60º test]The car took 6 sec. to travel from D to CLet the speed of car be x units/secondsDistance = speed × time

CD = x × 6 CD = 6x units ......(i)

In right angled ABC

tan 60 =AB

BC[By definition]

3 =AB

BC

AB = 3 BC .....(ii)

In right angled ABD,

tan 30 =AB

BD

1

3=

AB

BD

AB =BD

3.....(iii)

3 BC =BD


3BC = BD 3BC = BC + CD [ B - C - D] 3BC – BC = CD 2BC = CD 2BC = 6x [From (i)]

BC =6x

2 BC = 3x

Time =Dis tance

Speed

=BC

x

=3x

x= 3 seconds

The time taken by the car to reach the foot of the tower is 3 seconds.

AE

B C D

30º 60º

60º 30º


SCHOOL SECTION 217

35. A pilot in an aeroplane observes that Vashi bridge is on one side of theplane and Worli sea-link is just on the opposite side. The angle ofdepression of Vashi bridge and Worli sea-link are 60º and 30ºrespectively. If the aeroplane is at a height of 5500 3 m at that time,what is the distance between Vashi bridge and Worli sea-link ? (5 marks)

Sol. Let A be the point of observerLet B and C represent the positions of Vashi bridge and Worli sea-link respectively.AD represents the height of a plane from the groundAD = 5500 3 mEAB and FAC are angles of depressionm EAB = m ABD = 60º

[Converse of alternate angle test]m FAC = m ACD = 30ºIn right angled ADB,

tan 60º =AD

BD[By definition]

3 =5500 3

BD BD = 5500 m

In right angled ADC,

tan 30 =AD

DC[By definition]

1

3=

5000 3

DC DC = 5500 × 3 DC = 16500 m

BC = BD + DC [B - D - C] BC = 5500 + 16500 BC = 22000 m

Distance between Vashi bridge and Worli sea-link is 22 km.

46. In a right angled ABC, A = 90º and 2 2

2

5 sin B + 7 cos C + 4 7273 + 8 tan 60 . If

AC = 3 find the perimeter of ABC. (5 marks)Sol. In ABC,

m A = 90º [Given]

sin B =AC

BC......(i)

cos C =AC

BC.....(ii)

5 sin B 7 cos C 4

3 8 tan 60

2 2

2 = 7

27

5 sin B 7 cos C 4

3 8 3

2 2

2 = 7

27

5 sin B 7 cos C 4

3 8 3

2 2

= 7

27

5 sin B 7 cos C 4

27

2 2

= 7

27

5 sin B 7 cos C 4 2 2 = 7 [Multiplying throughout by 27]

5 sin B 7 cos C2 2 = 3

A

B D C

30ºE F

60º

60º 30º

550 3 m

C

A B

3


SCHOOL SECTION218

AC AC

5 × + 7 ×BC BC

2 2

= 3 [From (i) and (ii)]

12 AC

BC

2

= 3

12 × 3

BC

2

= 3

12 × 9

BC2 = 3

BC2 =12 × 9

3 BC2 = 36

BC = 6 units [Taking square roots]

In ABC,

m A = 90º [Given)

BC2 = AC2 + AB2 [By Pythagoras theorem]

(6)2 = (3)2 + AB2

36 = 9 + AB2

AB2 = 36 – 9

AB2 = 27

AB = 9 × 3

AB = 3 3 unitsPerimeter of ABC = AB + BC +AC

= 3 3 + 6 + 3

= 3 3 + 9

Perimeter of ABC = 3 3 3 units

60. Prove that 2 2 2 4

1 1 11 + 1 + =

tan A cot A sin A – sin A . (4 marks)

Proof : L.H.S. =1 1

1 1tan A cot A

2 2

= (1 + cot2 A) (1 + tan2 A)= cosec2 A × sec2 A [ 1 + cot2 A = cosec2 A

1 + tan2 A = sec2 A]

=1 1

×sin A cos A2 2

= 1

sin A 1 – sin A2 2 [sin2 A + cos2 A = 1

cos2 A = 1 – sin2 A]

=1

sin A – sin A2 4

= R.H.S.

2 2 2 4

1 1 11 + 1 + =

tan A cot A sin A – sin A


SCHOOL SECTION 219

MCQ’s1. For the angle in standard position if the initial arm rotates 300º in

anticlockwise direction, then in which quadrant will the terminal arm be(a) I (b) II(c) III (d) IV

2. For the angle in standard position, if the terminal arm passes through thepoint (8, – 15) then what is the value of sin :

(a)15

17(b)

–15

17

(c)8

17(d)

–8

17

3. If 5 sin = 12 cos and ‘’ is an acute angle. What is the value of cosec ?

(a)12

13(b)

13

12

(c)5

13(d)

13

5

4. If tan = 8

7 and is in III quadrant then what is the value of

1 sin 1 – sin

1 cos 1 – cos

(a)7

8(b)

8

7

(c)64

49(d)

49

64

5. If tan = 3

4, what is the value of cos2 – sin2 ?

(a)– 4

25(b)

425

(c)– 7

25(d)

7

25

6. What is the value of (sin2 17º – cos2 73º) ?(a) 1 (b) – 1

(c) 0 (d)1

3

7. If x sin (90º – ) cot (90º – ) = cos (90º – ) then what is the value of ‘x’ ?(a) – 1 (b) 1(c) – 2 (d) 2

8. If and (2 – 45º) are acute angles such that sin = cos (2 – 45º) thenwhat is the value of tan ?

(a) 0 (b)1

3

(c) 1 (d) 3


SCHOOL SECTION220

9. The ratio of the length of a rod and its shadow is 1 : 3 , what is the angleof elevation of the sun ?(a) 30º (b) 45º(c) 60º (d) 98º

10. If the angle of elevation of the top of a tower from a distance of 100 m fromits foot is 60º, then what is height of the tower ?(a) 50 3 m (b) 100 3 m

(c)100

3m (d)

50

3m

11. The tops of two poles of height 20 m and 16 m are connected by a wire. Ifthe wire makes an angle of 30º with the horizontal then what is the lengthof the wire ?(a) 6 m (b) 8 m(c) 10 m (d) 12 m

12. If ‘5’ and ‘4’ are acute angles such that sin 4 = cos 5 then what is thevalue of (2 sin 3 – 3 tan 3) ?(a) 1 – 3 (b) 1(c) 0 (d) – 1

13. sec2 + cosec2 = ............... .(a) sin2 . cos2 (b) sin + cos (c) sec . cosec (d) sec2 . cosec2

14. The value of 1 – cos

1 cos is ................. .

(a)sin

1 – cos

(b)

sin

1 cos

2

(c)1 – cos

sin

(d)

1 cos

sin

15. Which of the following is not a measure of quadrantal angle ?(a) 180º (b) 270º(c) 450º (d) 420º

16. If cot a = 2 and ‘a’ is in IIIrd quadrant, then cosec a is .............. .

(a) 5 (b) – 5

(c) 5 (d)1

5

17. (sec + tan ) (sec – tan ) = ............. .(a) 1 (b) – 1(c) 2 sec (d) 2 tan

18. If the terminal arm of an angle passes through the point (0, 2) then theterminal arm lies ........... .(a) on X axis (b) on Y axis(c) in 3rd quadrant (d) in 4th quadrant


SCHOOL SECTION 221

19. If the possible angles are 90 and – 270, the terminal arm is on the ......... .(a) negative X axis (b) positive X axis(c) negative Y axis (d) positive Y axis

20. If and are complementary angles, then which of following is true.(a) sin = sin (b) cos = cos (c) tan = tan (d) sec = cosec

21. If 3 sec – 4 cosec = 0 then tan = ............. .

(a)3

4(b)

5

4

(c)3

5(d)

4

3

22. The slope of a hill makes an angle of 60º with the horizontal. If one has towalk 500 m to reach the top of the hill then the height of the hill is ........... .

(a) 500 3 m (b) 250 3 m

(c) 500 m (d) 250 m

23. In the IInd quadrant ................. is positive.(a) sin (b) cos (c) tan (d) cot

24. If tan = 1

3 what is the value of sec ?

(a)2

3(b)

3

2

(c) 3 (d)4

3

25. The terminal arm passes through the point –1, 3 then cos = .......... .

(a)3

2(b)

–1

2

(c) 2 (d)1

2

: ANSWERS :

1. (d) IV 2. (b) –15

17

3. (b)13

124. (d)

49

64

5. (d)7

256. (c) 0

7. (b) 1 8. (c) 1

9. (a) 30º 10. (b) 100 3 m

11. (b) 8 m 12. (c) 0


SCHOOL SECTION222

13. (d) sec2 . cosec2 14. (c) 1 – cos

sin

15. (d) 420º 16. (b) – 517. (a) 1 18. (b) on Y axis19. (d) positive Y axis 20. (d) sec = cosec

21. (d) 4

322. (b) 250 3 m

23. (a) sin 24. (a) 2

3

25. (b)–1

2

1 Mark Sums1. For the angle in standard position, if the initial arm rotates 260º in clockwise

direction, then state the quadrant in which the terminal arm lies.Sol. For an angle in standard position, if the initial arm rotates 260º in

clockwise direction, then the terminal arm lies in the first quadrant.

2. If the angle is in quadrant I and initial arm rotates in clockwise direction,write the possible values of angles.

Sol. If the angle is in I quadrant and initial arm rotates in clockwise direction,then its lies between – 360º and – 270º.

3. Find the value of 3 sin2 + 3 cos2 .Sol. 3 sin2 + 3 cos2 = 3 (sin2 + cos2 )

= 3 (1) [ sin2 + cos2 = 1]= 3

4. If x--coordinate of point A is negative and y-coordinate is positive, thenin which quadrant point A lie ?

Sol. If x-co-ordinate of point A is negative and y-co-ordinate is positive.Then, point A lies in the II quadrant.

5. If = – 30º, find the value of sin .Sol. = – 30º [Given]

sin = sin (– 30)= – sin 30

=1

–2

sin (– 30) =1

–2

6. What is the directed angle, whose terminal arm lies along the coordinateaxes, called ?

Sol. If the terminal arm of a directed angle lies along the co-ordinate axes,then it is called a quadrantal angle.

7. If the initial arm rotates 70º in clockwise direction, then in which quadrantwill the terminal arm lie ?

Sol. If the initial arm rotates 70º in clockwise direction then the terminalarm lies in the IV quadrant.


SCHOOL SECTION 223

8. If sin = 1, what is the value of ?Sol. sin = 1 [Given]

But, sin 90 = 1 sin = sin 90

= 90º

9. If cos = 13

, what is value of cosec2 ?

Sol. sin2 + cos2 = 1

sin2 + 1

3

2

= 1

sin2 =1

1 –9

sin2 =9 – 1

9

sin2 =8

9

cosec2 =1

sin2

cosec2 =1

89

cosec2 =9

8

10. If the terminal arm lies on the positive y-axis, what are the possible angles ?Sol. If the terminal arm lies on the positive Y-axis, then angle made is 90º

or – 270º.

11. If + = 90º and tan = 34

then what is the value of cot ?

Sol. + = 90º [Given]

tan = 3

4[Given]

cot = tan [ cot = tan (90 – )]

cot = 3

4

12. If sec = 2, what is the value of tan2 ?Sol. sec = 2 [Given]

But, sec 60 = 2 sec = sec 60

= 60º

13. What is the value of (1 + cot2 ) (1 + cos ) (1 – cos ) ?Sol. (1 + cot2 ) (1 + cos ) (1 – cos )

= cosec2 (1 – cos2 ) [ 1 + cot2 = cosec2 ]= cosec2 × sin2 [sin2 + cos2 = 1, sin2 = 1 – cos2 ]

=1

sinsin

22

= 1


SCHOOL SECTION224

14. If sin = 45

, what is the value of cot2 ?

Sol. sin =4

5[Given]

cosec =1

sin

=1

45

cosec =5

41 + cot2 = cosec2

cot2 = cosec2 – 1

cot2 =5

– 14

2

cot2 =25

– 116

cot2 =25 – 16

16

cot2 =9

16

15. What is the value of cot2 – 2

1sin ?

Sol. cot2 – 1

sin2 = cot2 – cosec2

= – 1 [ 1 + cot2 = cosec2 cot2 – cosec2 = – 1]

16. What is the value of (1 + tan2 ) (1 – sin ) (1 + sin ) ?Sol. (1 + tan2 ) (1 – sin ) (1 + sin )

= sec2 (1 – sin2 ) [ 1 + tan2 = sec2 ]= sec2 × cos2 [sin2 + cos2 = 1

cos2 = 1 – sin2 ]

=1

coscos

22

= 1

17. If tan = 3 , then = ?

Sol. tan = 3 [Given]

But, tan 60 = 3 tan = tan 60

= 60


SCHOOL SECTION 225

18. If sin2 – cos2 = 12

, what is the value of sin4 – cos4 ?

Sol. sin2 – cos2 = 1

2[Given]

sin4 – cos4 = (sin2 )2 – (cos2 )2

= (sin2 + cos2 ) (sin2 – cos2 )= 1 (sin2 – cos2 ) [ sin2 + cos2 = 1]= sin2 – cos2

=1

2

19. If 3 sin – 4 cos = 0, what is the value of tan ?Sol. 3 sin – 4 cos = 0

3 sin = 4 cos

sin

cos

=

4

3

tan =4

3

20. What are the possible angles, if the terminal arm is in the I quadrant ?Sol. If the terminal arm is in the I quadrant and it in the anticlockwise are

0º and 90º. If it moves in the clockwise direction, the possible anglesare – 270º and – 360º.

21. If tan = 15 , what is the value sec ?

Sol. tan = 15 1 + tan2 = sec2

1 + 152

= sec2

1 + 15 = sec2 sec2 = 16

sec = 4 [Taking square roots]

22. If + = 90º and cosec = 2 , then find the value of sec ?

Sol. + = 90º [Given]cosec = 2 [Given]

sec = cosec [ sec = cosec (90 – )]

sec = 2

23. Find tan , for the angle , whose terminal arm passes through (3, 4).Sol. The terminal arm passes through (3, 4)

x = 3y = 4

tan = y

x

tan = 4

3


SCHOOL SECTION226

24. For the angle in standard position, if the initial arm rotates 340º in theanticlockwise direction, state the quadrant in which the terminal armlies.

Sol. For the standard angle, if the initial arm rotates 340º in the anticlockwisedirection then the terminal arm lies in the IV quadrant.

25. State the value of tan (– 60).Sol. tan (– 60)

= – tan 60

= – 3

tan (– 60) = – 3

26. If the terminal arm passes through the point (1, 1) making an angle ,find the value of sec .

Sol. The terminal arm passes through point (1, 1) x = 1 and y = 1

r = x y2 2

r = (1) (1)2 2

r = 1 1

r = 2

sec =r

x

sec =2

1

sec = 2

27. If sec = 2

3, find the value of ?

Sol. sec = 2

3[Given]

But, sec 30º = 2

3 sec = sec 30º

= 30º

SCHOOL SECTION 227

` Co-ordinate Geometry :Co-ordinate geometry is a branch of mathematics where we follow aalgebraic approach to geometry.Revision of concepts and formulae studied in Std. IX

1. Distance formula : AB = (x – x ) + (y – y )2 22 1 2 1

2. Section formula for internal division : P mx + nx my + ny

,m + n m + n

2 1 2 1

3. Section formula for external division : P mx – nx my – ny

,m– n m– n

2 1 2 1

4. Midpoint formula : P 1 2 1 2x + x y + y

,2 2

5. Area of a triangle : A (ABC) = 1

2 [x1 (y2 – y3) + x

2 (y3 – y1) + x3 (y1 – y2)]

` Inclination of a line :In the adjoining figure,Let l be the line in the XY- plane.Let be the angle made by theline l with positive direction of X-axis.The inclination of a line l is thesmallest positive angle made byit with positive direction ofX-axis (anticlockwise).Thus is called the inclination of aline where 0 < < 180.

Inclination of X-axis is zero degree and inclination of Y-axis is 90º.

` Slope of a line :In mathematics, slope or gradient of a line describesits steepness or incline. A higher slope value indicatesa steeper incline.The slope of inclined plane is the ratio of vertical (rise)height and horizontal distance (run).

i.e. slope = Vertical height

Horizontal dis tance

If the inclination of the line is , then tangent of is called the slope ofthe line. It is denoted by m.

B

Inclination

A

Y

Y

X XO

Co-ordinate Geometry5.


SCHOOL SECTION228

Slope of a line = tan = mSlope of X-axis is 0.Slope of Y-axis is not defined.Slopes of parallel lines are equal.


1. Find the slope of line with inclinations :(i) 45º (1 mark)Sol. Inclination of the line = 45º

Slope of the line = tan = tan 45º= 1

Slope of the line is 1

(ii) 30º (1 mark)Sol. Inclination of the line = 30º

Slope of the line = tan = tan 30º

=1

3


3

(iii) 90º (1 mark)Sol. Inclination of the line = 90º

Slope of the line = tan = tan 90º= Not defined

Slope of the line is not defined


1. Find the slope of the inclination of the line of the following :(i) = 30º (1 mark)Sol. Inclination of the line = 30º

Slope of the line = tan = tan 30º

=1

3


3

(ii) = 60º (1 mark)Sol. Inclination of the line = 60º



A

O

Y

Y

X X

B


SCHOOL SECTION 229

` Slope of a line passing through two points :Slope of a line passing through two points A (x1, y1) and B (x2, y2) is

m = 2 1

2 1

y – yx – x


2. Find the slope of the line passing through the following points :(i) A (– 2, 1) and B (0, 3) (1 mark)Sol. A (– 2, 1) (x1, y1)

B (0, 3) (x2, y2)

Slope of line AB =y – y

x – x2 1

2 1

=3 – 1

0 – (–2)

=2

2= 1

Slope of line AB is 1

(ii) P (1, – 1) and Q (– 2, 5) (1 mark)Sol. P (1, – 1) (x1, y1)

Q (– 2, 5) (x2, y2)

Slope of line PQ =y – y

x – x2 1

2 1

=5 – (–1)

– 2 – 1

=5 1

– 2 – 1

=6

– 3= – 2

Slope of line PQ is – 2.

(iii) C (3, 5) and D (– 2, – 3) (1 mark)Sol. C (3, 5) (x1, y1)

D (– 2, – 3) (x2, y2)

Slope of line CD =y – y

x – x2 1

2 1

=– 3 – 5

– 2 – 3

=– 8

–5

=8

5

Slope of line CD is 8

5

(iv) G (– 4, – 5) and H (– 1, – 2) (1 mark)Sol. G (– 4, – 5) (x1, y1)

H (– 1, – 2) (x2, y2)

Slope of line GH =y – y

x – x2 1

2 1

=(–2) – (–5)

(–1) – (–4)

=–2 5

–1 4

=3

3= 1

Slope of line GH is 1.

(v) M (4, 0) and N (– 3, – 2) (1 mark)Sol. M (4, 0) (x1, y1)

N (– 3, – 2) (x2, y2)

Slope of line MN =y – y

x – x2 1

2 1

=–2 – 0

–3 – 4

=– 2

– 7

=2

7

Slope of line MN is 2

7.

(vi) B (0, – 5) and D (1, 2) (1 mark)Sol. B (0, – 5) (x1, y1)

D (1, 2) (x2, y2)

Slope of line BD =y – y

x – x2 1

2 1

=2 – (–5)

1 – 0

=2 5

1

=7

1= 7

Slope of line BD is 7.


SCHOOL SECTION230


2. Find the slope of the line passing through the following pairs :(i) (– 1, 3) and (3, 5) (2 marks)Sol. Let, A (– 1, 3) (x1, y1)

B (3, 5) (x2, y2)


x – x2 1

2 1

=5 – 3

3 – (–1)

= 2

3 1

=2

4

=1

2

Slope of line passing through points (– 1, 3) and (3, 5) is 1

2.

(ii) (– 4, 5) and (2, 3) (2 marks)Sol. Let, A (– 4, 5) (x1, y1)

B (2, 3) (x2, y2)


x – x2 1

2 1

=3 – 5

2 – (–4)

= 3 – 5

2 4

=– 2

6

=–1

3

Slope of line passing through points (– 4, 5) and (2, 3) is –1

3

(iii) (7, 8) and (3, 4) (2 marks)Sol. Let, A (7, 8) (x1, y1)

B (3, 4) (x2, y2)


x – x2 1

2 1

=4 – 8

3 – 7

=– 4

– 4= 1

Slope of line passing through points (7, 8) and (3, 4) is 1.


SCHOOL SECTION 231

(iv) (3, 6) and (– 6, – 7) (2 marks)Sol. Let, A (3, 6) (x1, y1)

B (– 6, – 7) (x2, y2)


x – x2 1

2 1

=–7 – 6

– 6 – 3

=–13

–9

=13

9

Slope of line passing through points (3, 6) and (– 6, – 7) is 13

9.

(v) –1, 2 3 and – 2, 3 (2 marks)

Sol. Let, A –1, 2 3 (x1, y1)

B – 2, 3 (x2, y2)


x – x2 1

2 1

=3 – 2 3

– 2 – (–1)

= – 3

– 2 1

=– 3

–1

= 3

Slope of line passing through points –1, 2 3 and – 2, 3 is 3 .

(vi) –0, 3 and (3, 0) (2 marks)

Sol. Let, A 0, – 3 (x1, y1)B (3, 0) (x2, y2)


x – x2 1

2 1

= 0 – – 3

3 – 0

=3

3

Slope of line passing through points 0, – 3 and (3, 0) is 3

3.


6. A (0, 0), B (7, 2), C (7, 7) and D (2, 7) are the vertices of a quadrilateral.Find the slope of each diagonal. (2 marks)

Sol. For ABCD,A (0, 0), B (7, 2), C (7, 7), D (2, 7)seg AC and seg BD are the diagonals of ABCD


SCHOOL SECTION232

Slope of a line =y – y

x – x2 1

2 1

Slope of diagonal AC =7 – 0

7 – 0

=7

7= 1

Slope of diagonal AC is 1

Slope of diagonal BD =7 – 2

2 – 7

=5

–5= – 1

Slope of diagonal BD is –1


7. The vertices of a triangle are A (3, – 4), B (5, 7) and C (– 4, 5). Find theslope of each side of the triangle ABC. (3 marks)

Sol. For ABC,A (3, – 4), B (5, 7), C (– 4, 5)


x – x2 1

2 1

Slope of side AB =7 – (–4)

5 – 3

=7 4

2

=11

2

Slope of side AB =11

2

Slope of side BC =5 – 7

– 4 – 5

=– 2

– 9

=2

9

Slope of side BC is 2

9

Slope of side AC =5 – (– 4)

– 4 – 3

=5 4

–7

=9

– 7

=– 9

7

Slope of side AC is – 9

7


SCHOOL SECTION 233


3. Using slope concept. Check whether the following points are collinear :(i) A (7, 8), B (– 5, 2) and C (3, 6) (3 marks)Sol. A (7, 8) (x1, y1)

B (– 5, 2) (x2, y2)C (3, 6) (x

3, y3)


x – x2 1

2 1

=2 – 8

–5 – 7

=–6

–12

=1

2

Slope of line BC =y – y

x – x3 2

3 2

=6 – 2

3 – (–5)

=4

3 5

=4

8

=1

2 Slope of line AB and slope of line BC are equal and point B is a

common point for both the lines Points A, B and C are collinear.

(ii) P (– 2, 3), Q (7, – 4) and R (2, 1) (3 marks)Sol. P (– 2, 3) (x1, y1)

Q (7, – 4) (x2, y2)R (2, 1) (x

3, y3)

Slope of line PQ =y – y

x – x2 1

2 1

=– 4 – 3

7 – (–2)

=–7

7 2

=– 7

9

Slope of line QR =y – y

x – x3 2

3 2

=1 – (–4)

2 – 7

=1 4

– 5

=5

– 5= – 1

Slope of line PQ and slope of line QR are not equal. Points P, Q and R are not collinear.


SCHOOL SECTION234

(iii) X (– 1, 3), Y (8, – 3) and Z (2, 1) (3 marks)Sol. X (– 1, 3) (x1, y1)

Y (8, – 3) (x2, y2)Z (2, 1) (x

3, y3)

Slope of line XY =y – y

x – x2 1

2 1

=–3 – 3

8 – (–1)

=–6

8 1

=– 6

9

=– 2

3

Slope of line YZ =y – y

x – x3 2

3 2

=1 – (–3)

2 – 8

=1 3

– 6

=4

– 6

=– 2

3 Slope of line XY and slope of line YZ are equal and point Y is a

common point for both the lines. Points X, Y and Z are collinear.

(iv) M (1, – 2), N (2, – 1) and T (3, 0) (3 marks)Sol. M (1, – 2) (x1, y1)

N (2, – 1) (x2, y2)T (3, 0) (x

3, y3)

Slope of line MN =y – y

x – x2 1

2 1

=–1 – (–2)

2 – 1

=–1 2

1

= 1

Slope of line NT =y – y

x – x3 2

3 2

=0 – (–1)

3 – 2

=0 1

1

= 1 Slope of line MN and slope of line NT are equal and point N is a

common point for both the lines. Points M, N and T are collinear.


SCHOOL SECTION 235

(v) A (– 2, – 2), B (1, 1) and D (3, 3) (3 marks)Sol. A (– 2, – 2) (x1, y1)

B (1, 1) (x2, y2)D (3, 3) (x

3, y3)


x – x2 1

2 1

=1 – (–2)

1 – (–2)

=1 2

1 2

=3

3= 1

Slope of line BD =y – y

x – x3 2

3 2

=3 – 1

3 – 1

=2

2= 1

Slope of line AB and slope of line BD are equal and point B is acommon point for both the lines.

Points A, B and D are collinear.

(vi) V (– 7, 8), W (– 5, 2) and U (3, 6) (3 marks)Sol. V (– 7, 8) (x1, y1)

W (– 5, 2) (x2, y2)U (3, 6) (x

3, y3)

Slope of line VW =y – y

x – x2 1

2 1

=2 – 8

–5 – (–7)

=–6

–5 7

=– 6

2= – 3

Slope of line WU =y – y

x – x3 2

3 2

=6 – 2

3 – (–5)

=4

3 5

=4

8

=1

2 Slope of line VW and slope of line WU are not equal. Points V, W and U are not collinear.


SCHOOL SECTION236


3. Check whether points are collinear or not :(i) (– 1, 2), (3, 4), (5, – 6) (3 marks)Sol. Let, A (– 1, 2) (x1, y1)

B (3, 4) (x2, y2)C (5, – 6) (x3, y3)


x – x2 1

2 1

=4 – 2

3 – (–1)

= 2

3 1

=2

4

=1

2

Slope of line BC = y – y

x – x3 2

3 2

=–6 – 4

5 – 3

=–10

2= – 5

Slope of line AB and slope of line BC are not equal. The points (1, – 2), (3, 4) and (5, – 6) are not collinear.

(ii) (4, – 5), (7, 8), (– 2, – 3) (3 marks)Sol. Let, A (4, – 5) (x1, y1)

B (7, 8) (x2, y2)C (– 2, – 3) (x3, y3)


x – x2 1

2 1

=8 – (–5)

7 – 4

=8 5

3

=13

3

Slope of line BC = y – y

x – x3 2

3 2

=–3 – 8

–2 – 7

=–11

–9

=11

9 Slope of line AB and slope of line BC are not equal. The points (4, – 5), (7, 8) and (– 2, – 3) are not collinear.


SCHOOL SECTION 237

(iii) (7, – 1), (0, 3), (4, 0) (3 marks)Sol. Let, A (7, – 1) (x1, y1)

B (0, 3) (x2, y2)C (4, 0) (x3, y3)


x – x2 1

2 1

=3 – (–1)

0 – 7

=3 1

–7

=4

– 7

=– 4

7


x – x3 2

3 2

=0 – 3

4 – 0

=– 3

4 Slope of line AB and slope of line BC are not equal. The points (7, – 1), (0, 3) and (4, 0) are not collinear.

(iv) (– 1, 8), (9, – 2), (3, 4) (3 marks)Sol. Let, A (– 1, 8) (x1, y1)

B (9, – 2) (x2, y2)C (3, 4) (x3, y3)


x – x2 1

2 1

=–2 – 8

9 – (–1)

= –10

9 1

=–10

10= – 1


x – x3 2

3 2

=4 – (–2)

3 – 9

=4 2

– 6

=6

– 6= – 1

Slope of line AB and slope of line BC are equal and point B is acommon point for both the lines..

The points (– 1, 8), (9, – 2) and (3, 4) are collinear.


SCHOOL SECTION238

(v) (0, 6), (3, 0), (+ 2, + 4) (3 marks)Sol. Let, A (0, 6) (x1, y1)

B (3, 0) (x2, y2)C (2, 4) (x3, y3)


x – x2 1

2 1

=0 – 6

3 – 0

=– 6

3= – 2


x – x3 2

3 2

=4 – 0

2 – 3

=4

–1

= – 4 Slope of line AB and slope of line BC are not equal. The points (0, 6), (3, 0) and (2, 4) are not collinear.

(vi)

–10 ,

4 , 1

, 13 ,

2 1

,3 4 (3 marks)

Sol. Let, A

–10,

4 (x1, y1)

B 1

, 13 (x2, y2)

C 2 1

,3 4 (x3, y3)


x – x2 1

2 1

=

–1

1 –4

1– 0

3

=

11

413

=

5413

=5 3

×4 1

=15

4


SCHOOL SECTION 239


x – x3 2

3 2

=

1– 1

42 1

–3 3

=

–3413

=–3 3

×4 1

=– 9

4 Slope of line AB and slope of line BC are not equal.

The points

–10,

4 , 1

, 13 and

2 1

,3 4 are not collinear.


9. If the slope of the line joining points (k, – 3) and (4, 5) is 12

then find the

value of k. (3 marks)Sol. Let A (k, – 3) (x1, y1)

B (4, 5) (x2, y2)

Slope of line AB =1

2[Given]


x – x2 1

2 1

1

2=

5 – (–3)

4 – k

1

2=

5 3

4 – k

1

2=

8

4 – k 4 – k = 16 – k = 16 – 4 – k = 12 k = – 12

The value of k is – 12.


8. Find x if the slope of line joining (x, – 2) and (8, – 11) is – 34

. (3 marks)

Sol. Let, A (x, – 2) (x1, y1)B (8, – 11) (x2, y2)

Slope of line AB =– 3

4[Given]


SCHOOL SECTION240


x – x2 1

2 1

– 3

4=

–11 – (–2)

8 – x

– 3

4=

–11 2

8 – x

– 3

4=

–9

8 – x 3 (8 – x) = 9 × 4 24 – 3x = 36 3x = 24 – 36 3x = – 12

x =–12

3 x = – 4

The value of x is – 4.


4. Find the value of k if (– 3, 11), (6, 2) and (k, 4) are collinear points.(3 marks)Sol. Let, A (– 3, 11) (x1, y1)

B (6, 2) (x2, y2)C (k, 4) (x

3, y3)

Points A, B and C are collinearSlope of line AB = Slope of line BC

y – y

x – x2 1

2 1=

y – y

x – x3 2

3 2

2 – 11

6 – (–3) =4 – 2

k – 6

–9

6 3 =2

k – 6

– 9

9=

2

k – 6

– 1 =2

k – 6 – (k – 6) = 2 – k + 6 = 2 – k = 2 – 6 – k = – 4 k = 4

The value of k is 4


4. If (1, 2), 1

, 32 and (0, k) are collinear points find the value of k.(3 marks)

Sol. Let, A (1, 2) (x1, y1)

B 1

, 32 (x2, y2)

C (0, k) (x3, y3)


SCHOOL SECTION 241


y – y

x – x2 1

2 1=

y – y

x – x3 2

3 2

3 – 21

– 12

=k – 3

10 –

2

11

–2

=k – 3

–12

1 = k – 3 k = 1 + 3 k = 4

The value of k is 4


5. If the points 2 1

,5 3 ,

1

, k2 and

4

, 05 are collinear then find the value

of k. (3 marks)

Sol. Let, A 2 1

,5 3 (x1, y1)

B 1

, k2 (x2, y2)

C 4

, 05 (x

3, y3)

Points A, B and C are collinear

Slope of line AB = Slope of line BC

y – y

x – x2 1

2 1=

y – y

x – x3 2

3 2

1k –

31 2

–2 5

=0 – k4 1

–5 2

3k – 131

10

=– k3

10

3k – 1

×103

=10

– k ×3

3k – 1 = – k 3k + k = 1 4k = 1

k =1

4

The value of k is 1

4


SCHOOL SECTION242


9. The points (k, 3), (2, – 4) and (– k + 1, – 2) are collinear, find k. (3 marks)Sol. Let, A (k, 3), B (2, – 4), C (– k + 1, – 2)


– 4 – 3

2 – k = –2 – (–4)

(– k 1) – 2

–7

2 – k =

–2 4

– k 1 – 2

–7

2 – k =2

– k – 1

– 7 (– k – 1) = 2 (2 – k) 7k + 7 = 4 – 2k 7k + 2k = 4 – 7 9k = – 3

k =– 3

9

k =–1

3

The value of k is –1

3.


8. Show that the line joining (– 1, 1) and (– 9, 6) is parallel to the linejoining (– 2, 14) and (6, 9). (3 marks)

Sol. Let, A (– 1, 1), B (– 9, 6), C (– 2, 14), D (6, 9)


x – x2 1

2 1

Slope of side AB =6 – 1

–9 – (–1)

=5

–9 1

=5

– 8

Slope of line AB =–5

8

Slope of line CD =9 – 14

6 – (–2)

=–5

6 2

Slope of line CD =–5

8


SCHOOL SECTION 243

Slope of line AB and slope of line CD are equal. line AB || line CD The line joining (– 1, 1) and (– 9, 6) is parallel to the line joining

(– 2, 14) and (6, 9).


5. Show that (– 2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of a parallelogram.(4 marks)

Sol. Let, P (– 2, 1), Q (0, 3), R (2, 1), S (0, – 1)


x – x2 1

2 1

Slope of line PQ =3 – 1

0 – (–2)

2

0 2

2

2 Slope of line PQ = 1

Slope of line RS =–1 – 1

0 – 2

=– 2

– 2 Slope of line RS = 1 Slope of line PQ = Slope of line RS line PQ || line RS .......(i)

Slope of line QR =1 – 3

2 – 0

=– 2

2 Slope of line QR = – 1

Slope of line PS =–1 – 1

0 – (–2)

=– 2

0 2

=– 2

2 Slope of line PS = – 1

Slope of line QR = Slope of line PS

line QR || line PS ......(ii)

In PQRS,

side PQ || side RS [From (i)]

side QR || side PS [From (ii)]

PQRS is a parallelogram [By definition]

The points (–2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of

parallelogram.

P (– 2, 1) S (0, –1)

R (2, 1)Q (0, 3)


SCHOOL SECTION244


6. If P (– 2, 4), Q (4, 8), R (10, 5) and S (4, 1) are the vertices of a quadrilateralshow that it is a parallelogram. (4 marks)

Sol. P (– 2, 4), Q (4, 8), R (10, 5), S (4, 1)


x – x2 1

2 1

Slope of line PQ =8 – 4

4 – (–2)

= 4

4 2

=4

6

Slope of line PQ =2

3

Slope of line RS =1 – 5

4 – 10

=– 4

– 6

Slope of line RS =2

3 Slope of line PQ = Slope of line RS line PQ || line RS ........(i)

Slope of line QR =5 – 8

10 – 4

=– 3

6

Slope of line QR =–1

2

Slope of line PS =1 – 4

4 – (–2)

= –3

4 2

=– 3

6

Slope of line PS =–1

2 Slope of line QR = Slope of line PS line QR || line PS ........(ii)

In PQRS,side PQ || side RS [From (i)]side QR || side PS [From (ii)]

PQRS is a parallelogram [By definition]


7. Find the value of k if line PQ will be parallel to line RS where P (2, 4),Q (3, 6), R (8, 1) and S (10, k). (4 marks)

Sol. P (2, 4), Q (3, 6), R (8, 1), S (10, k)line PQ || line RS [Given]

P (– 2, 4) S (4, 1)

R (10, 5)Q (4, 8)


SCHOOL SECTION 245

Slope of line PQ = Slope of line RS

6 – 4

3 – 2 =k – 1

10 – 8

2

1=

k – 1

2 k – 1 = 4 k = 4 + 1 k = 5

The value of k is 5

` Equation of a line :An equation of a line essentially defines the conditions which must besatisfied by every point on the line.


4. If (4, – 3) is a point on the line 5x + 8y = c, then find c. (2 marks)Sol. The equation of the line is 5x + 8y = c

Let P (4, – 3)Point P lies on the line 5x + 8y = c

Co-ordinates of P satisfies the equation of the line 5 (4) + 8 (– 3) = c 20 – 24 = x c = – 4

The value of c is – 4


5. If (– 2, – 3) is a point on the line 2y = mx + 5, find m. (2 marks)Sol. The equation of the line is 2y = mx + 5

Let P (– 2, – 3)Point P lies on the line 2y = mx + 5

Co-ordinates of point P satisfies the equation of the line 2 (– 3) = m (– 2) + 5 – 6 = – 2m + 5 – 2m = – 6 – 5 – 2m = – 11

m =11

2

The value of m is 11

2

` Intercept of a line :A line can have two intercepts viz., x intercept and y intercept.

• The x intercept of a line is the point at which the line intersects the X-axis x intercept (x, 0)Hence, the x intercept of a line is the value of the x-co-ordinate of thepoint at which the line intersects the X-axis.

• The y intercept of a line is the point at which the line intersects the Y-axis. y intercept (0, y)Hence, the y intercept of a line is the value of the y-co-ordinate of thepoint at which the line intersects the Y-axis.


SCHOOL SECTION246

In the adjoining figure,Line l intersects the X-axis at point (– 3, 0) The x intercept of line l is – 3Line l intersects the Y-axis at point (0, 4) The y intercept of line l is 4.

(I) Equation of a line in slope intercept form :If the slope of a line is m and its y intercept is c then the equation of the lineis y = mx + c.The equation of a line passing through the origin and having slope ‘m’ isy = mx.


2. Write the equation of the lines if m and c are given as follows :(i) m = 5, c = – 1 (1 mark)Sol. Slope of the line (m) = 5

y intercept of the line (c) = – 1By slope intercept form,The equation of the line is y = mx + c

y = 5 (x) + (–1) y = 5x – 1

The equation of the given line is y = 5x – 1

(ii) m = – 4, c = – 3 (1 mark)Sol. Slope of the line (m) = – 4


y = (– 4)x + (– 3) y = – 4x – 3

The equation of the given line is y = – 4x – 3

(iii) m = – 2, c = 3 (1 mark)Sol. Slope of the line (m) = – 2

y intercept of the line (c) = 3By slope intercept form,The equation of the line is y = mx + c

y = (– 2)x + 3 y = – 2x + 3

The equation of the given line is y = – 2x + 3

(iv) m = 4, c = 0 (1 mark)Sol. Slope of the line (m) = 4


y = (4)x + (0) y = 4x

The equation of the given line is y = 4x

(0, 4)

Y

X

l

(–3, 0)O


SCHOOL SECTION 247

(v) m = 0, c = 2 (1 mark)Sol. Slope of the line (m) = 0


y = (0)x + 2 y = 2

The equation of the given line is y = 2

(vi) m = 0, c = – 3 (1 mark)Sol. Slope of the line (m) = 0


y = (0)x + (– 3) y = 0 – 3 y = – 3

The equation of the given line is y = – 3


1. Slope of line is 3 and y intercept is – 4. Write the equation of that line.(1 mark)

Sol. Slope of the line (m) = 3y intercept of the line (c) = – 4By slope intercept form,The equation of the line is y = mx + c

y = (3)x + (– 4) y = 3x – 4

The equation of the given line is y = 3x – 4.


2. Line PQ intersects Y-axis in (0, 3) with slope 1

–2

write the equation of

line PQ. (2 marks)

Sol. Slope of the line PQ (m) = 1

–2

line PQ intersects Y-axis in (0, 3) Its y intercept (c) is 3

By slope intercept form,The equation of the line y = mx + c

y = 1

–2

x + 3

The equation of the given line is y = 1

–2

x + 3


3. Write the equation of line in slope intercept form :(i) 2y – 3x + 5 = 0 (1 mark)Sol. 2y – 3x + 5 = 0

2y = 3x – 5

y =3 5

x –2 2

[Dividing throughout by 2]

Where m = 3

2 and c =

–5

2


SCHOOL SECTION248

(ii) 3y – x = 1 (1 mark)Sol. 3y – x = 1

3x = x + 1

y =1 1

x3 3


Where m = 1

3 and c =

1

3

(iii) x + 2y – 4 = 0 (1 mark)Sol. x + 2y – 4 = 0

2y = – x + 4

y =–1 4

x2 2


y =–1

x 22

Where m = –1

2 and c = 2

(iv) 3x – 2y = 5 (1 mark)Sol. 3x – 2y = 5

– 2y = – 3x + 5

y =–3 5

x–2 –2

[Dividing throughout by – 2]

y =3 5

x –2 2

Where m = 3

2 and c =

–5

2


1. Find the slope and y-intercept of the lines given below :(i) y = 3x – 5 (1 mark)Sol. y = 3x – 5

y = (3) x + (– 5)Comparing with the equation of a line in slope intercept form y = mx + cm = 3 and c = – 5

Slope of the line is 3The y intercept of the line is – 5.

(ii) y = 23

x + 4 (1 mark)

Sol. y = 2

3x + 4

Comparing with the equation of a line in slope intercept form y = mx + c

m = 2

3 and c = 4


The y intercept of the line is 4.

(iii) y = – 2x + 3 (1 mark)Sol. y = – 2x + 3

Comparing with the equation of a line in slope intercept form y = mx + cm = – 2 and c = 3

Slope of the line is – 2The y intercept of the line is 3.


SCHOOL SECTION 249

(iv) y = – 3x – 5 (1 mark)Sol. y = – 3x – 5

y = (– 3) x + (– 5)Comparing with the equation of a line in slope intercept form y = mx + cm = – 3 and c = – 5

Slope of the line is – 3The y intercept of the line is – 5.


3. Write the slope of each of the line stated below :(i) y – 5 = 2 (x – 7) (1 mark)Sol. y – 5 = 2 (x – 7)

Comparing with the equation of a line in slope point form,y – y1 = m (x – x1)

m = 2

Slope of the line y – 5 = 2 (x – 7) is 2

(ii) y – 2 = 5 (x – 2) (1 mark)Sol. y – 2 = 5 (x – 2)


m = 5

Slope of the line y – 2 = 5 (x – 2) is 5

(iii) y + 3 = 12

(x – 5) (1 mark)

Sol. y + 3 =1

2 (x – 5)


m = 1

2

Slope of the line y + 3 = 1

2 (x – 5) is

1

2

(iv) y = – 5 (x + 3) (1 mark)Sol. y = – 5 (x + 3)


m = – 5

Slope of the line y = – 5 (x + 3) is – 5

(v) 3 (x + 3) = y – 1 (1 mark)Sol. 3 (x + 3) = y – 1

y – 1 = 3 (x + 3)Comparing with the equation of a line in slope point form,y – y1 = m (x – x1)

m = 3

Slope of the line 3 (x + 3) = y – 1 is 3


SCHOOL SECTION250

(II) Equation of a line in slope point form :If a line l having slope m and passes through point A (x1, y1) then,the equation of line l is y – y1 = m (x – x1).


4. Write the equation of the line passing through the point P and havingslope m :

(i) P (3, 5) and m = 2 (2 marks)Sol. P (3, 5) (x1, y1)

m = 2The equation of the line passing through P (3, 5) and having slopem = 2 is given by slope point form

(y – y1) = m (x – x1) (y – 5) = 2 (x – 3) y – 5 = 2x – 6 2x – y – 6 + 5 = 0 2x – y – 1 = 0

The required equation of the line is 2x – y – 1 = 0

(ii) P (– 3, 7) and m = 12

(2 marks)

Sol. P (– 3, 7) (x1, y1)

m = 1

2The equation of the line passing through P (– 3, 7) and having slope

m = 1

2 is given by slope point form

(y – y1) = m (x – x1)

(y – 7) =1

2 [x – (– 3)]

y – 7 =1

2 (x + 3)

2 (y – 7) = x + 3 2y – 14 = x + 3 x – 2y + 3 + 14 = 0 x – 2y + 17 = 0

The required equation of the line is x – 2y + 17 = 0

(iii) P (– 2, – 3) and m = 35

(2 marks)

Sol. P (– 2, – 3) (x1, y1)

m = 3

5The equation of the line passing through P (– 2, – 3) and having slope

m = 3


(y – y1) = m (x – x1)

y – (– 3) =3

5 [x – (– 2)]

y + 3 =3

5 (x + 2)

5 (y + 3) = 3 (x + 2)


SCHOOL SECTION 251

5y + 15 = 3x + 6 3x – 5y + 6 – 15 = 0 3x – 5y – 9 = 0

The required equation of the line is 3x – 5y – 9 = 0

(iv) P (0, 6) and m = 67

(2 marks)

Sol. P (0, 6) (x1, y1)

m = 6

7The equation of the line passing through P (0, 6) and having slope

m = 6


(y – y1) = m (x – x1)

y – 6 =6

7 (x – 0)

7 (y – 6) = 6x 7y – 42 = 6x 6x – 7y + 42 = 0

The required equation of the line is 6x – 7y + 42 = 0


12. Write down the equation of a the line whose slope is 32

and which

passes through P where P divides the line segment joining A (– 2, 6) andB (3, – 4) in the ratio 2 : 3. (3 marks)

Sol. A (– 2, 6), B (3, – 4)Point P divides seg AB internally in the ratio 2 : 3Let, P (x, y)By section formula for internal division,

x =mx nx

m + n

2 1y =

my + ny

m + n2 1

=2 (3) + 3 (–2)

2 + 3 =y – 7

7 – 3

=6 – 6

5=

–8 + 18

5

=0

5=

10

5= 0 = 2

P (0, 2)

The line having slope 3

2 passes through the point P (0, 2)

The equation of the line by slope point form is,(y – y1) = m (x – x1)

(y – 2) =3

2 (x – 0)

y – 2 = 3x 2y – 4 = 3x 3x – 2y + 4 = 0

The equation of the required line is 3x – 2y + 4 = 0


SCHOOL SECTION252


10. In the adjoining figure,two lines are intersecting at point (3, 4).Find the equation of line PA and line PB. (3 marks)

Sol. Inclination of line PA is 45ºSlope of line PA = tan = tan 45º = 1Line PA passes through point P (3, 4)

The equation of the line by slope point form is,(y – y1) = m (x – x1)

(y – 4) = 1 (x – 3) y – 4 = x – 3 x – y – 3 + 4 = 0 x – y + 1 = 0

The equation of line PA is x – y + 1 = 0

Inclination of line PB is 60º Slope of line PB = tan = tan 60º = 3

Line PB passes through point P (3, 4) The equation of the line by slope point form is, (y – y1) = m (x – x1)

(y – 4) = 3 (x – 3)

y – 4 = 3 x – 3 3

3x – y 4 – 3 3 = 0

The equation of PB is 3x – y 4 – 3 3 = 0

(III) Equation of a line in two point form :If a line l passes through point A (x1, y1) and B (x2, y2) then,

the equation of line l is 1 1

1 2 1 2

x – x y – y=

x – x y – y


7. Two points of each line are given below write the equation of these lines :(i) A (– 3, 4), B (4, 5) (2 marks)Sol. A (– 3, 4) (x1, y1)

B (4, 5) (x2, y2)The equation of line AB by two point form is,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–3)

(–3) – 4 =y – 4

4 – 5

x 3

–7

=

y – 4

– 1 x + 3 = 7 (y – 4) x + 3 = 7y – 28 x – 7y + 3 + 28 = 0 x – 7y + 31 = 0

The equation of line AB is x – 7y + 31 = 0

P (3, 4)

A

BXX

45º 60º

Y

Y


SCHOOL SECTION 253

(ii) C (4, – 5), D (– 1, – 2) (2 marks)Sol. C (4, – 5) (x1, y1)

D (– 1, – 2) (x2, y2)The equation of line CD by two point form is,

x – x

x – x1

1 2 =

y – y

y – y1

1 2

x – 4

4 – (–1) =y – (–5)

–5 – (–2)

x – 4

4 1 =y 5

–5 2

x – 4

5=

y 5

– 3

– 3 (x – 4) = 5 (y + 5) – 3x + 12 = 5y + 25 – 3x – 5y + 12 – 25 = 0 – 3x – 5y – 13 = 0 3x + 5y + 13 = 0

The equation of line CD is 3x + 5y + 13 = 0

(iii) D (5, 6), E (– 5, – 6) (2 marks)Sol. D (5, 6) (x1, y1)

E (– 5, – 6) (x2, y2)The equation of line CD by two point form is,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 5

5 – (–5) =y – 6

6 – (–6)

x – 5

5 5 = y – 6

6 6

x – 5

10=

y – 6

12 12 (x – 5) = 10 (y – 6) 12x – 60 = 10y – 60 12x – 10y – 60 + 60 = 0 12x – 10y = 0 6x – 5y = 0 [Dividing throughout by 2]

The equation of line DE is 6x – 5y = 0

(iv) E (– 2, – 3), F (– 4, 7) (2 marks)Sol. E (– 2, – 3) (x1, y1)

F (– 4, 7) (x2, y2)The equation of line EF by two point form is,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–2)

–2 – (– 4) =y – (–3)

–3 – 7


SCHOOL SECTION254

x 2

–2 4 =y 3

–10

x 2

2=

y 3

–10 – 10 (x + 2) = 2 (y + 3)

– 10x – 20 = 2y + 6

– 10x – 2y – 20 – 6 = 0

– 10x – 2y – 26 = 0

5x + y + 13 = 0 [Dividing throughout by – 2]

The equation of line EF the 5x + y + 13 = 0

(v) G (0, 2), H (– 3, – 1) (2 marks)Sol. G (0, 2) (x1, y1)

H (– 3, – 1) (x2, y2)The equation of line GH by two point form is,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 0

0 – (–3) =y – 2

2 – (–1)

x

3=

y – 2

2 1

x

3=

y – 2

3 x = y – 2 x – y + 2 = 0

The equation of line GH is x – y + 2 = 0

(vi) R (4, – 5), T (– 3, 0) (2 marks)Sol. R (4, – 5) (x1, y1)

T (– 3, 0) (x2, y2)The equation of line RT by two point form is,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 4

4 – (–3) =y – (–5)

–5 – 0

x – 4

4 3 =y 5

–5

x – 4

7=

y 5

–5

– 5 (x – 4) = 7 (y + 5) – 5x + 20 = 7y + 35 – 5x – 7y + 20 – 35= 0 – 5x – 7y – 15 = 0 5x + 7y + 15 = 0

The equation of line RT is 5x + 7y + 15 = 0


SCHOOL SECTION 255


5. Write the equation of the line passing through each of the pair of pointsgiven below in the form y = mx + c

(i) (2, 3) and (4, 7) (2 marks)Sol. Let, A (2, 3) (x1, y1)

B (4, 7) (x2, y2)The line passes through points A and B

The equation of the line by two point form isx – x

x – x1

1 2=

y – y

y – y1

1 2

x – 2

2 – 4 =y – 3

3 – 7

x – 2

–2 =y – 3

– 4 4 (x – 2) = 2 (y – 3) 4x – 8 = 2y – 6 2y = 4x – 8 + 6 2y = 4x – 2 y = 2x – 1 [Dividing throughout by 2]

y = 2x – 1 is the equation of the line passing through (2, 3) and (4,7)

(ii) (0, 5) and (5, 6) (2 marks)Sol. Let, A (0, 5) (x1, y1)

B (5, 6) (x2, y2)The line passes through points A and B


x – x1

1 2=

y – y

y – y1

1 2

x – 0

0 – 5 =y – 5

5 – 6

x

–5 =y – 5

–1 x = 5 (y – 5) x = 5y – 25 5y = x + 25

y =1

5x + 5 [Dividing throughout by 5]

y = 1

5x + 5 is the equation of the line passing through (0, 5) and (5, 6)

(iii) (– 3, 5) and (4, – 7) (2 marks)Sol. Let, A (– 3, 5) (x1, y1)

B (4, – 7) (x2, y2)The line passes through points A and B

The equation of the line by two point form is

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–3)

–3 – 4 =y – 5

5 – (– 7)


SCHOOL SECTION256

x 3

– 7 =y – 5

12 12 (x + 3) = – 7 (y – 5) 12x + 36 = – 7y + 35 7y = – 12x + 35 – 36 7y = – 12x – 1

y =–12 1

x –7 7


y = –12 1

x –7 7

is the equation of the line passing through (– 3, 5) and (4, – 7)

(iv) (– 2, – 5) and (– 4, – 3) (2 marks)Sol. Let, A (– 2, – 5) (x1, y1)

B (– 4, – 3) (x2, y2)The line passes through points A and B


x – x1

1 2=

y – y

y – y1

1 2

x – (–2)

–2 – (– 4) =y – (–5)

–5 – (–3)

x 2

–2 4 =

y 5

– 5 3

x 2

2=

y 5

–2 – 2 (x + 2) = 2 (y + 5) – 2x – 4 = 2y + 10 2y = – 2x – 4 – 10 2y = – 2x – 14 y = – x – 7 [Dividing throughout by 2]

y = – x – 7 is the equation of the line passing through (– 2, – 5)and (– 4, – 3).


8. A (3, 7), B (5, 11), C (– 2, 8) are the vertices of ABC. AD is one of themedians of the triangle. Find the equation of the median AD. (3 marks)

Sol. A (3, 7), B (5, 11), C (– 2, 8)Seg AD is median of ABC

D is the midpoint of seg BC By midpoint theorem,

D x + x y + y

,2 2

1 2 1 2

5 (–2) 11 8

,2 2

5 – 2 9

,2 2

3 19

,2 2

The equation of median AD by two point form is

A (3, 7)

B (5, 11) C (– 2, 8)D


SCHOOL SECTION 257

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 3

33 –

2

=y – 7

197 –

2

x – 3

(6 – 3)2

=y – 7

(14 – 19)2

x – 3–3

2

=y – 7–5

2

2 (x – 3)

3=

2 (y – 7)

5 5 (x – 3) = 3 (y – 7) 5x – 15 = 3y – 21 5x – 3y – 15 + 21 = 0 5x – 3y + 6 = 0

The equation of median AD is 5x – 3y + 6 = 0.


9. P (– 3, 4), Q (2, 3), R (– 2, – 5) are the vertices of the PQR. Find theequations of all the medians of PQR. (5 marks)

Sol. P (– 3, 4), Q (2, 3), R (– 2, – 5)Let seg PA, seg QB and seg RC be the medians on the sides QR, PRand PQ respectively.

A, B and C are the midpoints of sides QR,PR and PQ respectively

By midpoint formula,

A x + x y + y

,2 2

1 2 1 2

2 (–2) 3 (– 5)

,2 2

2 – 2 3 – 5

,2 2

0 –2

,2 2

(0, – 1)

B –3 (–2) 4 (–5)

,2 2

–3 – 2 4 – 5

,2 2

–5 –1

,2 2

C –3 2 4 3

,2 2

–1 7

,2 2

A

P (– 3, 4)

C B

Q (– 3, 4) R (– 2, – 5)


SCHOOL SECTION258

By two point form,Equation of line PA is

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–3)

(–3) – 0 =y – 4

4 – (–1)

x 3

–3

=

y – 4

5 5 (x + 3) = – 3 (y – 4) 5x + 15 = – 3y + 12 5x + 3y + 15 – 12 = 0 5x + 3y + 3 = 0

Equation of line QB is

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 2

–52 –

2

=y – 3

–13 –

2

x – 2

52

2

=y – 3

13

2

x – 2

(4 5)2

=y – 3

(6 1)2

2 (x – 2)

9=

2 (y – 3)

7 7 (x – 2) = 9 (y – 3) 7x – 14 = 9y – 27 7x – 9y – 14 + 27 = 0 7x – 9y + 13 = 0

Equation of line RC,x – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–2)

–1(–2) – 2=

y – (–5)7(–5) – 2

x 2

1– 2 2

=y 5

7–5 – 2

x 2

(– 4 1)2

=

y 5(–10 – 7)

2

2 (x 2)

–3

=

2 (y 5)

–17

17 (x + 2) = 3 (y + 5) 17x + 34 = 3y + 15 17x – 3y + 34 – 15 = 0 17x – 3y + 19 = 0

The equation of the median of PQR are 5x + 3y + 3 = 0, 7x – 9y + 13 = 0and 17x – 3y + 19 = 0.


SCHOOL SECTION 259


13. In triangle ABC the co-ordinates of vertices A, B and C are (4, 7),(– 2, 3) and (0, 1) respectively. Find the equation of medians passingthrough vertices A, B and C. (3 marks)

Sol. A (4, 7), B (– 2, 3), C (0, 1)Let, seg AD, seg BE and seg CF be the medians on sides BC, ACand AB respectively.

D, E and F are the midpoints of sides BC, AC and AB respectively.By midpoint formula,

D x + x y + y

,2 2

1 2 1 2

–2 + 0 3 + 1

,2 2

–2 4

,2 2

(– 1, 2)

E x + x y + y

,2 2

1 2 1 2

4 0 7 1

,2 2

4 8

,2 2

(2, 4)

F x + x y + y

,2 2

1 2 1 2

4 (–2) 7 3

,2 2

4 – 2 10

,2 2

2

, 52

(1, 5)By two point form,The equation of median AD,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 4

4 – (– 1) =y – 7

7 – 2

x – 4

4 1+ =y – 7

5

x – 4

5=

y – 7

5 x – 4 = y – 7 x – y – 4 + 7 = 0 x – y + 3 = 0

D

A (4, 7)

F E

B (– 2, 3) C (0, 1)


SCHOOL SECTION260

The equation of the median BEx – x

x – x1

1 2=

y – y

y – y1

1 2

x – (–2)

–2 – 2 =y – 3

3 – 4

x 2

– 4

=

y – 3

–1 x + 2 = 4 (y – 3) x + 2 = 4y – 12 x – 4y + 2 + 12 = 0 x – 4y + 14 = 0

The equation of the median CFx – x

x – x1

1 2=

y – y

y – y1

1 2

x – 0

0 – 1 =y – 1

1 – 5

x

–1 =y – 1

– 4 4x = y – 1 4x – y + 1 = 0

The equation of the medians of ABC are x – y + 3 = 0, x – 4y + 14 = 0and 4x – y + 1 = 0


14. A (5, 4), B (– 3, – 2) and C (1, – 8) are the vertices of a triangle ABC. Findthe equation of median AD and line parallel to AC passing through point B.

(3 marks)Sol. A (5, 4), B (– 3, – 2), C (1, – 8)

seg AD is the median of seg BC D is midpoint of seg BC

D x + x y + y

,2 2

1 2 1 2

–3 + 1 –2 + (–8)

,2 2

–2 –2 – 8

,2 2

–10

–1 ,2

(– 1, – 5)By two point form,The equation of median AD

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 5

5 – (–1) =y – 4

4 – (–5)

x – 5

5 + 1 =y – 4

4 + 5

x – 5

6=

y – 4

9


SCHOOL SECTION 261

9 (x – 5) = 6 (y – 4) 9x – 45 = 6y – 24 9x – 6y – 45 + 24 = 0 9x – 6y – 21 = 0 3x – 2y – 7 = 0 [Dividing throughout by 3]

The equation of median AD is 3x – 2y – 7 = 0

Slope of line AC =y – y

x – x2 1

2 1

=–8 – 4

1 – 5

=–12

– 4= 3

Slope of parallel lines are equalSlope of the line parallel to line AC is 3The line passes through B (– 3, – 2)

The equation of the line parallel to line AC passing through point B bythe slope point form is

y – y1 = m (x – x1) y – (– 2) = 3 [x – (– 3)] y + 2 = 3 (x + 3) y + 2 = 3x + 9 3x – y + 9 – 2 = 0 3x – y + 7 = 0

The equation of the line parallel to AC passing through point B is3x – y + 7 = 0


11. Find the equation of the line which passes through (2, 7) and whosey-intercept is 3. (2 marks)

Sol. Let A (2, 7)The y intercept of the line is 3

The line intersects the y-axis at point (0, 3)Let B (0, 3)The line passes through point A and B

The equation of the line ABBy two point form

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 2

2 – 0 =y – 7

7 – 3

x – 2

2=

y – 7

4 4 (x – 2) = 2 (y – 7) 4x – 8 = 2y – 14 4x – 2y – 8 + 14 = 0 4x – 2y + 6 = 0 2x – y + 3 = 0 [Dividing throughout by 2]

The required equation of the line is 2x – y + 3 = 0


SCHOOL SECTION262

(III) Equation of a line in double intercept form :If a line l has x intercept a and y intercept b then,

the equation of line l is x y

+ = 1a b


6. Convert the equation in y = mx + c form :

(i)x y

+ = 12 3

(1 mark)

Sol. x y

2 3 = 1

Multiplying throughout by 3,

3x

y2

= 3

y = –3x

32

(ii)x y

– = 13 4

(1 mark)

Sol.x y

–3 4

= 1

Multiplying throughout by 4,4x

– y3

= 4

y = 4x

– 43

(iii)x y

+ = 1– 4 –1 (1 mark)

Sol. x y

– 4 –1 = 1

– x

– y4

= 1

y = –1

x – 14

(iv)x y

+ = 24 3

(1 mark)

Sol. x y

4 3 = 2


3x

y4

= 6

y = –3x

64


3. Write the equation of each of the following line in double interceptform and write x intercept and y intercept.

(i) x + y = 2 (2 marks)Sol. x + y = 2

Dividing throughout by 2,

x y

2 2 = 1

x intercept of line x + y = 2 is 2y intercept of line x + y = 2 is 2

(ii) 2x – y = 3 (2 marks)Sol. 2x – y = 3


2x y

–3 3

= 1

x y3 –32

= 1

x intercept of line 2x – y = 3 is 3

2y intercept of line 2x – y = 3 is – 3


SCHOOL SECTION 263

(iii) 3x + y = 4 (2 marks)Sol. 3x + y = 4


3x y

4 4 =

4

4

x y4 43

= 1

x intercept of line 3x + y = 4 is 4

3y intercept of line 3x + y = 4 is 4

(iv) 4x – y – 7 = 0 (2 marks)Sol. 4x – y – 7 = 0

4x – y = 7Dividing throughout by 7,

4x y

–7 7

= 7

7

x y7 –74

= 1

x intercept of line 4x – y – 7 = 0 is 7

4y intercept of line 4x – y – 7 = 0 is – 7

(v) 2x + 3y – 7 = 0 (2 marks)Sol. 2x + 3y – 7 = 0

2x + 3y = 7Dividing throughout by 7,

2x 3y

7 7 =

7

7

x y7 72 3

= 1

x intercept of line 2x + 3y – 7 = 0 is 7

2

y intercept of line 2x + 3y – 7 = 0 is 7

3

(vi) 2x – y = 11 (2 marks)Sol. 2x – y = 11


2x y

–11 11

= 11

11

x y

11 –112

= 1

x intercept of line 2x – y = 11 is 11

2y intercept of line 2x – y = 11 is – 11


SCHOOL SECTION264


1. Find x and y intercepts of each of the line :(i) y = 2x – 3 (2 marks)Sol. y = 2x – 3 ......(i)

Substituting y = 0 in equation (i),0 = 2x – 3

2x = 3

x =3

2

The x intercept of the line y = 2x – 3 is 3

2Substituting x = 0 in equation (i),

y = 2 (0) – 3 y = 0 – 3 y = – 3

The y intercept of the line y = 2x – 3 is – 3

(ii) y – x = – 5 (2 marks)Sol. y – x = – 5 ......(i)

Substituting y = 0 in equation (i),0 – x = – 5

– x = – 5 x = 5

The x intercept of the line y – x = – 5 is 5

Substituting x = 0 in equation (i),y – 0 = – 5

y = – 5

The y intercept of the line y – x = – 5 is – 5

(iii) y = 23

x – 4 (2 marks)

Sol. y =2

3x – 4 ......(i)

Substituting y = 0 in equation (i),

0 =2

3x – 4

2

3x = 4

x = 4 × 3

2 x = 2 × 3 x = 6

The x intercept of the line y = 2

3x – 4 is 6

Substituting x = 0 in equation (i),

y =2

3 (0) – 4

y = – 4

The y intercept of the line y = 2

3x – 4 is – 4


SCHOOL SECTION 265

(iv) 5x – y – 27

= 0 (2 marks)

Sol. 5x – y – 2

7= 0 ......(i)


5x – 0 – 2

7= 0

5x – 2

7= 0

5x =2

7

x =2

7 ×

1

5

x =2

35

The x intercept of the line 5x – y – 2

7 = 0 is

2


5 (0) – y – 2

7= 0

– y – 2

7= 0

y = – 2

7

The y intercept of the line 5x – y – 2

7 = 0 is

– 2

7

(v) x + 2y – 6 = 0 (2 marks)Sol. x + 2y – 6 = 0 .......(i)

Substituting y = 0 in equation (i),x + 2 (0) – 6 = 0

x – 6 = 0 x = 6

The x intercept of the line x + 2y – 6 = 0 is 6

Substituting x = 0 in equation (i),0 + 2y – 6 = 0

2y = 6 y = 3

The y intercept of the line x + 2y – 6 = 0 is 3

(vi) 3x + y – 53

= 0 (2 marks)

Sol. 3x + y – 5

3= 0 .......(i)


3x + 0 – 5

3= 0

3x – 5

3= 0


SCHOOL SECTION266

3x =5

3

x =5

3 ×

1

3

x =5

9

The x intercept of the line 3x + y – 5

3 = 0 is

5

9

Substituting x = 0 in equation (i),

3 (0) + y – 5

3= 0

0 + y – 5

3= 0

y =5

3

The y intercept of the line 3x + y – 5

3 = 0 is

5

3

(vii) 3y – 2x + 5 = 0 (2 marks)Sol. 3y – 2x + 5 = 0 .......(i)

Substituting y = 0 in equation (i),3 (0) – 2x + 5 = 0

0 – 2x + 5 = 0 – 2x + 5 = 0 – 2x = – 5

x =5

2

The x intercept of the line 3y – 2x + 5 = 0 is 5

2Substituting x = 0 in equation (i),3y – 2(0) + 5 = 0

3y + 5 = 0 3y = – 5

y =–5

3

The y intercept of the line 3y – 2x + 5 = 0 is –5

3

(viii) 7x + 6y – 1 = 0 (2 marks)Sol. 7x + 6y – 1 = 0 .......(i)

Substituting y = 0 in equation (i),7x + 6 (0) – 1 = 0

7x – 1 = 0 7x = 1

x =1

7

The x intercept of the line 7x + 6y – 1 = 0 is 1



SCHOOL SECTION 267

7 (0) + 6y – 1 = 0 6y – 1 = 0 6y = 1

y =1

6

The y intercept of the line 7x + 6y – 1 = 0 is 1

6

(ix) y – 3x = 0 (2 marks)Sol. y – 3x = 0 .......(i)

Substituting y = 0 in equation (i),0 – 3x = 0

– 3x = 0 x = 0

The x intercept of the line y – 3x = 0 is 0

Substituting x = 0 in equation (i),y – 3 (0) = 0

y – 0 = 0 y = 0

The y intercept of the line y – 3x = 0 is 0

(x) 4x – 7y = 0 (2 marks)Sol. 4x – 7y = 0 .......(i)

Substituting y = 0 in equation (i),4x – 7(0) = 0

4x = 0 x = 0

The x intercept of the line 4x – 7y = 0 is 0

Substituting x = 0 in equation (i),4 (0) – 7y = 0

– 7y = 0 y = 0

The y intercept of the line 4x – 7y = 0 is 0

` General equation of a line :The general equation of line is ax + by + c = 0

Slope of the line = m = – coefficient of x

coefficient of y

x intercept of the line = – constant term

coefficient of x

y intercept of the line = – cons tant term

coefficient of y


2. Find the slope and y intercept for each of the following line :(i) 2x – 3y = 7 (2 marks)Sol. 2x – 3y = 7

2x – 3y – 7 = 0

Slope of line 2x – 3y = 7 is – coefficient of x

coefficient of y = – 2

– 3 = 2

3


SCHOOL SECTION268

Slope of line 2x – 3y = 7 is 2

3

y intercept of line 2x – 3y = 7 is – constant term

coefficient of y = – (–7)

–3 = – 7

3

y intercept of line 2x – 3y = 7 is – 7

3

(ii) y – 3x – 6 = 0 (2 marks)Sol. y – 3x – 6 = 0

– 3x + y – 6 = 0

Slope of line y – 3x – 6 = 0 is – coefficient of x


1 = 3

Slope of line y – 3x – 6 = 0 is 3.

y intercept of line y – 3x – 6 = 0 is – constant term


1 = 6

y intercept of line y – 3x – 6 = 0 is 6.

(iii) 2y + 2x – 5 = 0 (2 marks)Sol. 2y + 2x – 5 = 0

2x + 2y – 5 = 0

Slope of line 2x + 2y – 5 = 0 is – coefficient of x

coefficient of y =– 2

2= – 1

Slope of line 2x + 2y – 5 = 0 is – 1.

y intercept of line 2x + 2y – 5 = 0 is – constant term


2 =

5

2

y intercept of line 2x + 2y – 5 = 0 is 5

2.

(iv) 7x – y + 3 = 0 (2 marks)Sol. 7x – y + 3 = 0

Slope of line 7x – y + 3 = 0 is – coefficient of x


–1 = 7

Slope of line 7x – y + 3 = 0 is 7.

y intercept of line 7x – y + 3 = 0 is – constant term


–1 = 3

y intercept of line7x – y + 3 = 0 is 3.

(v) 4x – y = 0 (2 marks)Sol. 4x – y = 0

Slope of line 4x – y = 0 is – coefficient of x


–1 = 4

Slope of line 4x – y = 0 is 4.

y intercept of line 4x – y = 0 is – constant term


–1 = 0

y intercept of line 4x – y = 0 is 0.

(vi) 8x – 4y – 1 = 0 (2 marks)Sol. 8x – 4y – 1 = 0

Slope of line 8x – 4y – 1 = 0 is – coefficient of x


– 4 = 2

Slope of line 8x – 4y – 1 = 0 is 2.


SCHOOL SECTION 269

y intercept of line 8x – 4y – 1 = 0 is – constant term


– 4 = 1

– 4 = –1

4

y intercept of line 8x – 4y – 1 = 0 is –1

4.

(vii) 5x – 2y = 3 (2 marks)Sol. 5x – 2y = 3

5x – 2y – 3 = 0

Slope of line 5x – 2y = 3 is – coefficient of x

coefficient of y = –5

– 2 = 5

2

Slope of line 5x – 2y = 3 is 5

2.

y intercept of line 5x – 2y = 3 is – constant term


–2 = 3

– 2 = – 3

2

y intercept of line 5x – 2y = 3 is – 3

2

(viii) 5x – 8y = – 2 (2 marks)Sol. 5x – 8y = – 2

5x – 8y + 2 = 0

Slope of line 5x – 8y = – 2 is – coefficient of x


– 8 = 5

8

Slope of line 5x – 8y = – 2 is 5

8.

y intercept of line 5x – 8y = – 2 is – constant term


– 8 = 1

4

y intercept of line 5x – 8y = – 2 is 1

4.


10. The equation of a line is 3x – 4y + 12 = 0. It intersects X-axis in point Aand y-axis in point B, find the co-ordinates of points A and B, find thelength of AB. (3 marks)

Sol. The equation of line is 3x – 4y + 12 = 0

y intercept of the line = – cons tant term


– 4 = 3

x intercept of the line = – constant term

coefficient of x = –12

3 = – 4

The line intersects the X-axis at point A

Its y-co-ordinate is 0

A (– 4, 0)

The line intersects the Y-axis at point B

Its x-co-ordinate is 0

B (0, 3)


SCHOOL SECTION270

By Distance formula,

AB = (x – x ) (y – y )2 22 1 2 1

= [0 – (–4)] (3 – 0)2 2

= (0 4) 3 2 2

= 4 32 2

= 16 9

= 25= 5 units

l (AB) = 5 units


4. Find the equation of the line passing through (2, –1) and parallel to3x + 4y = 10. (3 marks)

Sol. The equation of the given line is 3x + 4y = 10i.e., 3x + 4y – 10 = 0

Slope of the given line is – coefficient of x


4The required line is parallel to the given line

Slope of the required line is – 3

4

The required line whose slope is – 3

4 passes through point (2, – 1)

The equation of the line by slope point form is(y – y1) = m (x – x1)

y – (– 1) =– 3

4 (x – 2)

y + 1 =– 3

4 (x – 2)

4 (y + 1) = – 3 (x – 2) 4y + 4 = – 3x + 6 3x + 4y + 4 – 6 = 0 3x + 4y – 2 = 0

The equation of the line passing through (2, –1) and parallel to3x + 4y = 10 is 3x + 4y – 2 = 0.


5. Find the equation of the line passing through (– 3, – 5) and parallel tox – 2y – 7 = 0. (3 marks)

Sol. The equation of the given line is x – 2y – 7 = 0

Slope of the given line is – coefficient of x


– 2 = 1


Slope of the required line is 1

2

The required line whose slope is 1

2 passes through (– 3, – 5)


SCHOOL SECTION 271

The equation of the line by slope point form is,(y – y1) = (x – x1)

y – (– 5) =1

2 [ x – (– 3)]

y + 5 =1

2 (x + 3)

2 (y + 5) = x + 3 2y + 10 = x + 3 x – 2y + 3 – 10 = 0 x – 2y – 7 = 0

The equation of the line passing through (– 3, – 5) and parallel tox – 2y – 7 = 0 is x – 2y – 7 = 0.


6. Find the equation of the line parallel to 4x + 3y = 5 and havingx-intercept (– 3). (3 marks)

Sol. The equation of the given line is 4x + 3y = 5i.e., 4x + 3y – 5 = 0

Slope of the given line 4x + 3y = 5 is – coefficient of x



Slope of the required line is – 4

3The required line has x intercept – 3

The line intersects the X-axis at point (– 3, 0)Let A (– 3, 0)

The required line has slope – 4

3 and passes through point A (– 3, 0)

The equation of the line by slope point form is,y – y1 = m (x – x1)

y – 0 = – 4x – (–3)

3 3y = – 4 (x + 3) 3y = – 4x – 12 4x + 3y + 12 = 0

The equation of required line is 4x + 3y + 12 = 0


7. Write the equation of each of the following line :(i) The X-axis and Y-axis (1 mark)Sol. The equation of X-axis is y = 0

The equation of Y-axis is x = 0

(ii) The line passing through the origin and the point (– 3, 5) (2 marks)Sol. Let O (0, 0) (x1, y1)

A (– 3, 5) (x2, y2) The equation of line OA by two point form is,

x – x

x – x1

1 2 =

y – y

y – y1

1 2

x – 0

0 – (– 3)=

y – 0

0 – 5


SCHOOL SECTION272

x

3=

y

–5 – 5x = 3y 5x + 3y = 0

The equation of the line passing through the origin and the point(– 3, 5) is 5x + 3y = 0.

(iii) The line passing through the points (2, 3) and (4, 5). (2 marks)Sol. Let A (2, 3) (x1, y1), B (4, 5) (x2, y2)

The equation of line AB by two point form is,x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 2

2 – 4 =y – 3

3 – 5

x – 2

–2 =y – 3

–2 x – 2 = y – 3 x – y – 2 + 3 = 0 x – y + 1 = 0

The equation of the line passing through the points (2, 3) and (4, 5)is x – y + 1 = 0

(iv) The line passing through the points (3, 4) and having slope 5.(2 marks)Sol. Let A (3, 4) (x1, y1) and m = 5

The equation of the line passing through A and having slope 5 byslope point form is,

y – y1 = m (x – x1) y – 4 = 5 (x – 3) y – 4 = 5x – 15 5x – y – 15 + 4 = 0 5x – y – 11 = 0

The equation of the line passing through the points (3, 4) andhaving slope 5 is 5x – y – 11 = 0.

(v) The line parallel to X-axis and passing through the point (– 3, 4). (1 mark)Sol. The equation of the line parallel to X-axis and passing through the point

(– 3, 4) is y = 4.

(vi) The line parallel to Y-axis and passing through the point (– 3, 5). (1 mark)Sol. The equation of the line parallel to Y-axis and passing through the point

(– 3, 5) is x = – 3.


36. Point (m, 2m – 1) lies on the line 3x 2y

– = –15 3

find m. (2 marks)

Sol. Let, A (m, 2m – 1)

Point A lies on the line 3x 2y

–5 3

= – 1

Co-ordinates of point A satisfies the equation,

3m 2 (2m – 1)

–5 3

= – 1


SCHOOL SECTION 273


3m 2 (2m – 1)15 – 15

5 3

= 15 (– 1)

3 (3m) – 5 (4m – 2) = – 15 9m – 20m + 10 = – 15 – 11m = – 15 – 10 – 11m = – 25

m =25

11

The value of m is 25

11

37. The line x – 6y + 11 = 0 bisects the segment joining (8, – 1) and (0, k)find the value of k. (3 marks)

Sol. Let, A (8, – 1), B (0, k)Let, line x – 6y + 11 = 0 bisect seg AB at point P

Point P is the midpoint of seg ABBy midpoint formula,

P x x y y

,2 2

1 2 1 2

P 8 0 –1 k

,2 2

P –1 k

4 ,2

Point P –1 k

4 ,2

lies on line x – 6y + 11 = 0

Co-ordinates of point P satisfies the equation,

4 – 6 –1 k

2

+ 11 = 0

4 – 3 (– 1 + k) + 11 = 0 4 + 3 – 3k + 11 = 0 7 – 3k + 11 = 0 – 3k + 18 = 0 – 3k = – 18 k = 6

The value of k is 6.

38. The point P divides the segment AB joining points A (2, 1) andB (– 3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0. Justify.

(4 marks)Proof : Let A (2, 1), B (– 3, 6)

Point P divides seg AB internally in the ratio 2 : 3 By section formula for internal division,

P mx nx my ny

,m n m n

2 1 2 1

P 2 (–3) 3 (2) 2 (6) 3 (1)

,2 3 2 3

P 6 6 12 3

,5 5

P 0 15

,5 5

P (0, 3)


SCHOOL SECTION274

Given equation of the line is x – 5y + 15 = 0Substituting x = 0 and y = 3 on the L.H.S. of the equationL.H.S. = x – 5y + 15

= 0 – 5 (3) + 15= 0 – 15 + 15= 0= R.H.S.

Point P (0, 3) lies on the x – 5y + 15 = 0

39. The side AB of an equilateraltriangle ABC is parallel to X-axis.Find the slopes of all sides.

(4 marks)

Construction : Extend ray CA and ray CB intersecting the X-axis at points D andE respectively. Extend ray AB upto point F such that A - B - F. Takepoint M as shown in the figure.

Sol. ABC is an equilateral triangleside AB || X axis [Given]Slope of X axis is 0

Slope of side AB is 0 [Slopes of parallel lines are equal]

line AB || X axis [Given] On transversal CD,

CAB CDE [Converse of corresponding angles test]But, m CAB = 60º [Angle of equialteral triangle]

m CDE = 60º The inclination of line AC is 60º

Slope of side AC = tan 60º = 3m CBF + m CBA = 180 [Linear pair axiom]

m CBF + 60 = 180 m CBF = 180 – 60 m CBF = 120º ......(i)

line AB || X axis On transversal CE,

CBF CEM [Converse of corresponding angles test] m CEM = 120º Inclination of line CB is 120º Slope of line CB = tan 120º ......(ii) Slope of line CB = cot (90 – 120) [tan = cot (90 – )] Slope of line CB = cot – 30 Slope of line CB = – cot 30

Slope of line CB = – 3

40. Find the value of k so that PQ will be parallel to RS where P (2, 4), Q (3, 6),R (8, 1) and S (10, k). (2 marks)

Sol. P (2, 4), Q (3, 6), R (8, 1), S (10, k)Line PQ is parallel to line RS

Slope of line PQ = Slope of line RS

X

Y

Y

60º 60º 120º

F•

•M X

O

A

D E

B

C


SCHOOL SECTION 275

6 – 4

3 – 2 =k – 1

10 – 8

2

1=

k – 1

2 4 = k – 1 k = 4 + 1 k = 5

Value of k is 5.

41. Find the equation of the straight line passing through the origin and thepoint of intersection of the lines x + 2y = 7 and x – y = 4. (3 marks)

Sol. Let line x + 2y = 7 and x – y = 4 intersect at point Ax + 2y = 7 .......(i)x – y = 4 ......(ii)Subtracting (ii) from (i),x + 2y = 7x – y = 4

(–) (+) (–)3y = 3

y = 1Substituting y = 1 in equation (ii),x – 1 = 4

x = 4 + 1 x = 5 A (5, 1)

The straight line passes through A (5, 1) and O (0, 0) The equation of the line by two point form,

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 5

5 – 0 =y – 1

1 – 0

x – 5

5=

y – 1

1 x – 5 = 5 (y – 1) x – 5 = 5y – 5 x – 5y – 5 + 5 = 0 x – 5y = 0

The equation of the line passing through the origin and the point ofintersection of the lines x + 2y = 7 and x – y = 4 is x – 5y = 0.

42. Lines x = 5 and y = 4 form a rectangle with coordinate axes. Find theequation of the diagonals. (4 marks)

Sol. Let line y = 4 intersect the Y-axis at point A A (0, 4)

Let line x = 5 intersect the X-axis at point C C (5, 0)

Let line y = 4 and x = 5 intersect at point B B (5, 4)

The origin O (0, 0)ABCO is a rectangleseg AC and seg BO are the diagonalsEquation of line AC by two point form,

X

A B

O C

Y

Y

X

y = 4

x =

5


SCHOOL SECTION276

x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 0

0 – 5 = y – 4

4 – 0

x

–5 =y – 4

4 4x = – 5 (y – 4) – 5y + 20 = 4x 4x + 5y – 20 = 0 Equation of diagonal AC is 4x + 5y – 20 = 0.

Equation of line BD by two point from ix,x – x

x – x1

1 2=

y – y

y – y1

1 2

x – 5

5 – 0 =y – 4

4 – 0

x – 5

5=

y – 4

4 4 (x – 5) = 5 (y – 4) 5y – 20 = 4x – 20 4x – 5y = 0

Equation of diagonal BD is 4x – 5y = 0.

43. If the points A (1, 2), B (4, 0), C (3, 5) are the vertices of a triangle ABC.Find the equation of the line passing through the mid points of AB andAC. (4 marks)

Sol. A (1, 2), B (4, 6), C (3, 5)Let D and E be the midpoints of sides AB and AC respectivelyPoint D is the midpoint of side AB

D x x y y

,2 2

1 2 1 2

1 4 2 6

,2 2

5 8

,2 2

5

, 42

Point E is the midpoint of side AC

E x x y y

,2 2

1 2 1 2

1 + 3 2 5

,2 2

4 7

,2 2

7

2 ,2

The required line passes through points D and E The equation of the line by two point form,

x – x

x – x1

1 2=

y – y

y – y1

1 2

5x – 25 – 22

=y – 4

74 – 2


SCHOOL SECTION 277

2x – 52

5 – 42

= y – 4

8 – 72

2x – 5

5 – 4=

(y – 4)12

2x – 5 = 2 (y – 4) 2x – 5 = 2y – 8 2x – 2y – 5 + 8 = 0 2x – 2y + 3 = 0

The equation of the required line is 2x – 2y + 3 = 0

44. Find the equation of the line making an angle of 45º with positive

X-axis and at a distance 2 2 from the origin. (3 marks)

Sol. Let line l be the line making an angle of 45º with positive X axis atpoint ASlope of line l = tan 45 = 1Let seg OP line l as shown in the figure,

OP = 2 2 units [Given]

OAP = 45 [Vertically opposite angles]In OPA,m OPA = 90ºm OAP = 45ºm PAO = 45º

OPA is a 45º - 45º - 90º triangle By 45º - 45º - 90º triangle theorem,

OP = 1

2OA

2 2 = 2 2 OA

OA = 2 2 2 OA = 4 units A is at distance of 4 from the origin A (4, 0)

line l has slope 1 and passes through point A (4, 0)Equation of l by slope point

(y – y1) = m (x – x1) y – 0 = 1 (x – 4) y = x – 4 x – y – 4 = 0

45. Find the equation of the line passing through the point of intersectionof 4x + 3y + 2 = 0 and 6x + 5y + 6 = 0 and the point of intersection oflines 4x – 3y – 17 = 0 and 2x + 3y + 5 = 0. (5 marks)

Sol. Let point P be the point of intersection of lines 4x + 3y + 2 = 0 and6x + 5y + 6 = 04x + 3y + 2 = 0

4x + 3y = – 2 ......(i)Multiplying throughout by 3 we get,

12x + 9y = – 6 .....(ii)6x + 5y + 6 = 0

6x + 5y = – 6 .....(iii)

X X

Y

Y

45ºA

P

O

•B

•C

2 2


SCHOOL SECTION278

Multiplying throughout by – 2 we get,– 12x – 10y = 12 ......(iv)Adding (ii) and (iv), 12x + 9y = – 6– 12x – 10y = 12

– y = 6 y = – 6

Substituting y = – 6 in equation (i),4x + 3 (– 6) = – 2

4x – 18 = – 2 4x = – 2 + 18 4x = 16 x =

16

4 x = 4 P (4, – 6)

Let Q be the point of intersection of lines 4x – 3y – 17 = 0 and2x + 3y + 5 = 04x – 3y – 17 = 0

4x – 3y = 17 .....(v)2x + 3y + 5 = 0

2x + 3y = – 5 .....(vi)Multiplying throughout by – 2 we get,

– 4x – 6y = 10 .....(vii)Adding (v) and (vii),

4x – 3y = 17– 4x – 6y = 10

– 9y = 27 y = – 3

Substituting y = – 3 in equation (v),4x – 3 (– 3) = 17

4x + 9 = 17 4x = 17 – 9 4x = 8 x = 2 Q (2, – 3)

The equation of line PQ by two point from,

y – y

y – y1

1 2=

x – x

x – x1

1 2

y – (–6)

–6 – (–3) =x – 4

4 – 2

y 6

–6 3

=

x – 4

2

y 6

–3

=

x – 4

2 2 (y + 6) = – 3 (x – 4) 2y + 12 = – 3x + 12 3x + 2y + 12 – 12 = 0 3x + 2y = 0

The required equation of line is 3x + 2y = 0.


SCHOOL SECTION 279

MCQ’s1. What is the slope of a line having inclination 45º ?

(a) 3 (b)1

3(c) 2 (d) 1

2. If the slope of a line joining the points (2, k) and (–3, 0) is 2

5 then what is

the value of k ?(a) 2 (b) – 2(c) – 5 (d) 5

3. If the slope of a line joining the points (k, – 5) and (– 2, – 4) is 1

2 then what

is the value of k ?(a) – 1 (b) 1(c) – 4 (d) 4

4. For what value of x will be points (x, – 1) (2, 1) and (4, 5) lie on a line.(a) 1 (b) – 1(c) 4 (d) 2

5. A (1, 2), B (– 2, – 1) and C (2, – 2) are the vertices of ABC. What is the slopeof line AC ?(a) 4 (b) – 4

(c)1

4(d)

–1

4

6. If (– 4, 3) is a point on the line 3y = mx – 1 then what is the value of m ?

(a)–5

2(b)

5

2(c) – 2 (d) 2

7. The slope of a line is 4 and y-intercept is – 3, then its equation is(a) y = 4x – 3 (b) y = 4x + 3(c) y = – 4x – 3 (d) y = – 4x + 3

8. The slope of a line is –1

2 and its y-intercept is 5. What is the equation of

line ?(a) 2y = x – 10 (b) 2y = – x + 10(c) 2y = – x – 5 (d) 2y = x + 5

9. What are the values of m and n, if D (m, – 2) is the midpoint of the segmentjoining (– 3, n) and (2, – 5).

(a) m = 1, n = –1

2(b) m = – 1 n =

1

2

(c) m = –1

2, n = 1 (d) m =

–1

2, n = – 1


SCHOOL SECTION280

10. If the points (k, 2k), (3k, 3k) and (3, 1) are collinear then what is the valueof k ?

(a)–1

3(b)

1

3(c) – 3 (d) 3

11. What is the slope of line x y

– 23 4

?

(a)– 4

3(b)

4

3

(c)– 3

4(d)

3

4

12. The vertices of ABC are A (3, – 4), B (5, 7) and C (– 4, 5). Then the slope ofseg BC is :

(a)2

9(b)

9

2

(c)– 2

9(d)

– 9

2

13. The equation of line 2y – 3x + 5 = 0, in slope intercept form is ......... .

(a) y = –3 5

x –2 2

(b) y = –3 5

x2 2

(c) y = 3 5

x –2 2

(d) y = 3 5

x2 2

14. The point (4, – 3) lies on the 5x + 8y = c, then the value of c is .......... .(a) 4 (b) – 4

(c)1

4(d)

–1

4

15. The slope of a line is 3 and y intercept is – 4, then the equation of line is.............. .(a) y = 3x + 4 (b) y = 3x – 4(c) y = – 3x + 4 (d) y = – 3x – 4

16. Slope of the line 3 (x + 3) = y – 1, .......... .

(a) 3 (b)1

3

(c) – 3 (d)–1

3

17. The equation of a line passing through origin with slope m = 3

2, is ........ .

(a) 3x – 2y = 0 (b) 3x + 2y = 0(c) 2x + 3y = 0 (d) 2x – 3y = 0

18. y – 2 = 4 (x + 3) passes through the point ............ .(a) (– 3, 2) (b) (2, – 3)(c) (3, – 2) (d) (– 2,3)


SCHOOL SECTION 281

19. The slope of the line y – 3x – 6 = 0 is ............... .

(a)1

3(b)

–1

3(c) 3 (d) – 3

20. The equation of line 4x – y – 7 = 0 in the double intercept form is ......... .

(a)

x y1

7 – 74

(b)

x y1

7 74

(c) x y

1–7 – 74

(d)

x y1

–7 74

21. Equation of X-axis is ........... .(a) x = 0 (b) y = 0(c) x = k (d) y = k

22. The equation of a line parallel to y = – 3x + 2 and passing through point(3, 5) is ............. .(a) y – 3 = – 3 (x – 5) (b) y – 5 = – 3 (x – 3)(c) y – 5 = 2 (x – 3) (d) y – 3 = 2 (x – 5)

23. Equation of a line parallel to X-axis and passing through (– 3, 4) is ............ .(a) x = – 3 (b) x = 3(c) y = – 4 (d) y = 4

24. The equation of a line passing through the point (2, 7) and whose y-interceptis 3 is ............... .(a) 2x – y + 3 = 0 (b) 2x – y – 3 = 0(c) 2x + y + 3 = 0 (d) 2x + y – 3 = 0

25. The line y = – 3x lies in the ................. quadrant.(a) I and II (b) II and IV(c) III and IV (d) IV and III

: ANSWERS :

1. (d) 1 2. (a) 23. (c) – 4 4. (d) 2

5. (b) – 4 6. (a) –5

27. (a) y = 4x – 3 8. (b) 2y = – x + 10

9. (c) m = –1

2, n = 1 10. (a)

–1

3

11. (b)4

312. (a)

2

9

13. (c) y = 3 5

x –2 2

14. (b) – 4

15. (b) y = 3x – 4 16. (a) 3

17. (a) 3x – 2y = 0 18. (a) (– 3, 2)


SCHOOL SECTION282

19. (c) 3 20. (a)

x y1

7 – 74

21. (b) y = 0 22. (b) y – 5 = – 3 (x – 3)

23. (d) y = 4 24. (a) 2x – y + 3 = 0

25. (b) II and IV

1 Mark Sums1. Find the slope of a line whose inclination is 45º.Sol. Inclination of the line () = 45º


Slope of the line is 1.

2. Write the slope of line 2y = 3x – 5.Sol. 2y = 3x – 5

y = 3 5

x –2 2


Comparing the above equation with slope intercept form y = mx + c we

get m = 3

2

Slope of line is 3

2

3. Write the y-intercept of the line 3y = 2x + 7Sol. 3y = 2x + 7

y = 2 7

x3 2


Comparing the above equation with slope intercept form y = mx + c we

get c = 7

2

y-intercept of line is 7

2

4. If m = 5 and c = – 3, then write the equation of the line.Sol. m = 5, c = – 3

By slope point form, the equation of line isy = mx + c

y = 5x – 3

5x – y – 3 = 0

5. Write the equation of X-axis.

Sol. Equation of X-axis is y = 0

6. Write the equation of a line parallel to Y-axis and passing through point(3, 0).Sol. Equation of a line parallel to Y-axis and passing through point (3, 0) is

x = 3.


SCHOOL SECTION 283

7. Find the slope of a line having inclination 60º.Sol. Inclination of the line () = 60º

Slope of the line = tan = tan 60= 3

Slope of the line is 3 .

8. State the slope of X-axis.Sol. Slope of X-axis = tan 0 = 0

Slope of X-axis = 0

9. If (– 3, – 2) lies on the line 2y = mx + 5, find m.Sol. (– 3, – 2) lies on the line 2y = mx + 5

Its co-ordinates will satisfy the equation 2 (– 2) = m (– 3) + 5 – 4 = – 3m + 5 3m = 5 + 4 3m = 9

m =9

3

m = 3

10. Write the equation of a line parallel to Y-axis and passing through thepoint (3, 4).

Sol. Equation of a line parallel to Y-axis and passing through the poitn (3, 4)is x = 3.

11. If the slope of a line is 2 and its y-intercept is 5, write its equation.Sol. Slope of the line (m) = 2

Its y-intercepts (c) = 5 Equation of the line by slope-intercept form,

y = mx + c

y = 2x + 5

12. The equation of a line passing through the origin is y = 3x. Find thex-coordinate of a point on the line, if its y-coordinate is 3.

Sol. Equation of the line passing through origin y = 3xy-co-ordinate of a point lying on this line is 3 [Given]

3 = 3x

x = 3

3 x = 1

13. A line has the equation y = 3x – 2. State its y-intercept.Sol. Equation of the line is y = 3x – 2

Comparing the given equation with slope-intercept form y = mx + c,c = – 2

y intercept of the line is – 2.

14. Write the equation of a line parallel to X-axis and passing through thepoint (– 2, 5).

Sol. Equation of a line parallel to X-axis and passing through the point (– 2, 5)is y = 5.


SCHOOL SECTION284

15. Write the x-intercept and the y-intercept of the line represented by the

equation x y

+ = 12 3

.

Sol. Comparing the equation x y

12 3 with the double intercept form,

x y1

a b we get a = 2 and b = 3

x intercept = 2 and y intercept = 3

16. What are the co-ordinates of the midpoint of the line segment joiningA (2, 5) and B (4, 1) ?

Sol. A (2, 5) and B (4, 1) Co-ordinates of the midpoint of line segment AB by midpoint formula is,

2 4 5 1,

2 2

=6 6

,2 2

= (3, 3)

17. What is the x-intercept of line 3x – 4y = 12 ?Sol. Substituting y = 0 in the equation 3x + 4y = 12

3x – 4(0) = 12 3x – 0 = 12 3x = 12

x =12

3 x = 4

x intercept is 4.

18. What is the equation of a line whose slope is – 2 and y-intercept is 3 ?Sol. Slope (m) = – 2

y intercept (c) = 3 Equation of the line by slope-intercept form is y = mx + c

y = – 2x + 3

2x + y – 3 = 0

19. What is the y-intercept of line 2x – 3y = 4 ?Sol. 2x – 3y = 4

3y = 2x – 4Dividing throughout by 3,

y = 2

3x –

4

3Comparing the above equation with slope intercept form y = mx + c, we get

y intercept = – 4

3

20. The slope of line PQ is – 1 and the slope of line QR is – 1. What can yousay about the points P, Q and R ?

Sol. Slope of line PQ = – 1Slope of line QR = – 1


SCHOOL SECTION 285

Slopes of line PQ and line QR are equal and point Q is a common pointfor both the lines

Points P, Q and R are collinear

21. What is the equation of a line parallel to X-axis and passing through thepoint (5, – 7) ?

Sol. Equation of a line parallel to X-axis and psssing through the point (5, –7)is y = – 7

22. What is the slope of line whose inclination is 0º ?Sol. Inclination of line () = 0º

Its slope = tan Its slope = tan 0

Its slope = 0

23. What is the x-intercept of the line 3x – 4y = 7 ?Sol. Substituting y = 0 in the equation,

3x – 4y = 7 3x – 4 (0) = 7 3x – 0 = 7 3x = 7

x = 7

3

x intercept is 7

3

24. The slope of line AB is 23

. What is the slope of line DE which is parallel

to line AB ?Sol. line DE || line AB [Given]

Slope of line DE = slope of line AB

But, slope of line AB = 2

3[Given]

Slope of line DE = 2

3

25. What is the equation of a line whose slope is – 3 and the y-intercept 52

?

Sol. Slope of the line (m) = – 3

It’s y intercept (c) = 5

2 Equation of the line by slope-intercept form,

y = mx + c

y = – 3x + 5

2 2y = – 6x + 5 [Dividing throughout by 2]

6x + 2y – 5 = 0

SCHOOL SECTION286

` Introduction :Mensuration is a special branch of mathematics that deals with themeasurement of geometric figures.In previous classes we have studied certain concepts related to areas ofplane figures (shapes) such as triangles, quadrilaterals, polygons and circles.

Now we will study how to find some measurements related to circle andthe surface area and the volume of solid figures.

` Circle : Arc, Sector, Segment : Area of sector :

Sector of a circle is the partof the circle enclosed by tworadii of the circle and theirintercepted arc. (i.e. arc betweenthe two ends of radii)

Area of the sector (A) = × r360

2

Length of an arc :Length of an arc of a circle (arc length) is the distance along the curvedline making up the arc.

Length of the arc (l) = × 2 r360

Relation between the area of the sector and the length of an arc :

Area of the sector =r

× length of arc2


1. The diameter of a circle is 10 cm. Find the length of the arc, when thecorresponding central angle is as given below : ( = 3.14)

(i) 144º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

l =144

360 × 2 × 3.14 × 5

l =144

360 × 3.14 × 10

l = 12.56 cm

The length of the arc is 12.56 cm.

Mensuration6.

Major arc

Central angle

Minor arc

rO

A

B

X

Y


SCHOOL SECTION 287

(ii) 45º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

l =45

360 × 2 × 3.14 × 5

l =45

360 × 3.14 × 10

l =5

4 ×

314

100

l =785

2 100

l =392.5

100 l = 3.925 l = 3.93 cm


(iii) 270º (2 marks)Sol. Diameter of a circle = 10 cm

It radius (r) =10

2= 5 cm

l =360

× 2r

l =270

360 × 2 × 3.14 × 5

l =3

4 × 3.14 × 10

l = 23.55 cm


(iv) 180º (2 marks)Sol. Diameter of a circle = 10 cm

Its radius (r) =10

2= 5 cm

l =360

× 2r

=180

360 × 2 × 3.14 × 5

= 15.70 cm



SCHOOL SECTION288


2. Find the angle subtended at the centre of a circle by an arc, given thefollowing :

(i) radius of circle = 5.5 m, length of arc = 6.05 m ( = 227

) (2 marks)

Sol. l =360

× 2r

6.05 =360

× 2 ×

22

7 × 5.5

605

100=

360

×

22

7 × 11

605 × 360 × 7

100 × 22 ×11 =

= 63º

Measure of the arc is 63º.

(ii) radius of circle = 20 m length of arc = 78.50 m ( = 3.14) (2 marks)

Sol. l =360

× 2r

78.50 =360

× 2 × 3.14 × 20

785

10=

360

×

314

100 × 40

785 × 9 ×100

10 × 314 =

= 225º

Measure of an arc is 225º.


3. The radius of the circle is 7 cm and m (arc RYS) = 60º, with the help ofthe figure, answer the following questions : (3 marks)

(i) Name the shaded portion.Sol. P-RYS

(ii) Find the area of the circle.Sol. Area of circle = r2

=22

7 × 7 × 7

= 154 cm2

Area of a circle is 154 cm2.

(iii) Find A (P-RYS)

Sol. Area of the sector =360

× r2

=60

360 ×

22

7 × 7 × 7

=77

3= 25.67

Area of the sector P -RYS is 25.67 cm2

S

Y

R

PX 60º


SCHOOL SECTION 289

(iv) Find A (P-RXS)Sol. Area of sector P-RXY = Area of circle – Area of sector P-RYS

= 154 – 25.67= 128.33 cm2

Area of sector P-RXY is 128.33 cm2.


4. The radius of a circle is 7 cm. Find area of the sector of this circle if theangle of the sector is :

(i) 30º (2 marks)


× r2

=30

360 ×

22

7 × 7 × 7

=77

6= 12.83

Area of the sector is 12.83 cm2.

(ii) 210º (2 marks)


× r2

=210

360 ×

22

7 × 7 × 7

=539

6= 89.83


(iii) 3 rt. angles (2 marks)


× r2

=270

360 ×

22

7 × 7 × 7

=231

2= 115.50



5. An arc of a circle having measure 36 has length 176 m. Find thecircumference of the circle. (2 marks)

Sol. Length of arc (l) = 176 m measure of arc () = 36º

l =360

× 2r

176 =36

360× 2r


SCHOOL SECTION290

176 =1

10 × 2r

176 × 10 = 2r 2r = 1760

But, circumference = 2r

Circumference of the circle is 1760 m.


6. An arc of length 4 cm subtends an angle of measure 40º at the centre.Find the radius and the area of the sector formed by this arc.(2 marks)

Sol. Length of arc (l) = 4 cm measure of arc () = 40º

l =360

× 2r

4 =40

360 × 2 × × r

4 × 9

2= r

r = 18 cm.

Area of the sector =× r

2

l

=4 ×18

2

= 36cm2

Radius of the circle is 18 cm and Area of the sector is 36cm2.


7. If the area of the minor sector is 392.5 sq. cm and the correspondingcentral angle is 72º, find the radius. ( = 3.14) (2 marks)

Sol. Measure of arc () = 72ºArea of the sector = 392.5 cm2

Area of the sector =360

× r2

392.5 =72

360 × 3.14 × r2

3925

10=

72

360 ×

314

100 × r2

3925 × 360 ×100

10 × 72 × 314 = r2

r2 = 625 r = 25 [Taking square roots]

Radius of the circle is 25 cm.


8. Find the area of sector whose arc length and radius are 10 cm and 5 cmrespectively. (2 marks)

Sol. Length of arc (l) = 10 cmRadius of a circle (r) = 5 cm


SCHOOL SECTION 291


2 × l

=5

2 × 10

= 25 cm2

Area of the sector is 25 cm2.


9. If the area of minor sector of a circle with radius 11.2 cm is 49.28cm2,find the measure of the arc. (2 marks)

Sol. Radius of a circle = 11.2 cmArea of the sector = 49.28 cm2

Area of the sector =360

× r2

49.28 =360

×

22

7 × 11.2 × 11.2

4928

100=

360

×

22

7 ×

112

10 ×

112

10

4928 × 360 × 7 ×10 ×10

100 × 22 ×112 ×112 =

= 45º

Measure of arc is 45º.


10. Find the length of the arc of a circle with radius 0.7 m and area of thesector is 0.49 m2. (2 marks)

Sol. Radius of a circle = 0.7 cmArea of the sector = 0.49 m2


2 × l

0.49 =0.7

2 × l

49

100=

7

20 × l

49 20

100 7

= l

l = 1.4

The length of the arc is 1.4 m.


11. Two arcs of the same circle have their lengths in the ratio 4:5. Find theratio of the areas of the corresponding sectors. (2 marks)

Sol. Ratio of lengths of two arcs is 4 : 5.Let the common multiple be ‘x’

Lengths of two arcs are (4x) units and (5x) units respectivelyLet the lengths of two arcs be ‘l1’, and ‘l2’ and Areas of their correspondingsectors be A1 and A2.


SCHOOL SECTION292

l1 = (4x) units l2 = (5x) units

Both Arcs are of the same circle. Their radii are equal

Now,

A1 =r

2 × l1 .......(i)

A2 =r

2 × l2 ......(ii)

Dividing (i) and (ii) we get,

1

2

A

A =1

2

l

l

1

2

A

A =4x

5x A1 : A2 = 4 : 5

Ratio of the areas of sectors is 4 : 5.


12. Adjoining figure depicts a racing trackwhose left and right ends are semicircular.The distance between two inner parallelline segments is 70 m and they are each105 m long. If the track is 7 m wide, findthe difference in the lengths of the inneredge and outer edge of the track. (4 marks)

Sol. Diameter of inner circular edge (d1) = 70 mWidth of the track = 7 m

Diameter of outer circular edge (d2) = 70 + 7 + 7= 84 m

The inner and outer edges of the racing tracks comprises of twosemicircles and parallel segments of length 105 m each

Length of outer edge =1

2d2 + 105 +

1

2d2 + 105

= d2 + 210

= (84 + 210) m

Length of inner edge =1

2d1 + 105 +

1

2d1 + 105

= d1 + 210= (70 + 210) m

Difference in the lengths of= (84 + 210) – (70 + 210)

inner and outer edge= 84 + 210 – 70 – 210= 14

= 14 × 22

7= 44 m

The difference in the lengths of inner edge and outer edge of thetrack is 44 m.

105 m

105 m

70 m70 m

7 m

7 m


SCHOOL SECTION 293


13. In the adjoining figure,A horse is tied to a pole fixedat one corner of a 30 m × 30 m squarefield of grass by means of a 10 m long rope.( = 3.14). (3 marks)

(i) Find the area of that part of thefield in which the horse can graze.

Sol. Side of a square = 30 m.Length of the rope = radius of the sector

Radius of the sector (r) = 10 mMeasure of arc () = 90º [Angle of a square]

Area of field that can be grazed = Area of sector

=360

× r2

=90

360 × 3.14 × 10 × 10

=1

4 × 314

Area of field that can be grazed = 78.5 m2

(ii) In the adjoining figure,What will be the area of the partof the field in which the horse cangraze, if the pole was fixed on a sideexactly at the middle of the side?

Sol. If the pole is fixed at the middle ofthe middle of the side of asquare, thenArea of field that can be grazed = 2 × Area of sector

= 2 × 78.5

Area of field that can be grazed = 157 m2


2. The area of a circle is 314 sq.cm and area of its sector is 31.4 sq.cm.Find the area of its major sector. (2 marks)

Sol. Area of a circle = 314 cm2

Area of major sector = Area of a circle – Area of its minor sector= 314 – 31.4= 282.6

Area of major sector is 282.6 cm2


3. Prove A = 12

C r, for a circle having radius, circumference and area r, C,

A respectively. (2 marks)Sol. L.H.S. = A

L.H.S. = r2 ........(i)

30 m

10 m

10 m

10 m 10 m

30 m


SCHOOL SECTION294

R.H.S. =1

Cr2

R.H.S =1

2 × 2r × r

R.H.S. = r2 ........(ii)

L.H.S. = R.H.S [From (i) and (ii)]

A =12

Cr


4. The radius of a circle is 3.5 cm and area of the sector is 3.85 cm2. Findthe length of the corresponding arc and the measure of arc. (3 marks)

Sol. Radius of a circle (r) = 3.5 cm

Area of the sector = 3.85 cm2

Area of sector =r

2 × l

3.85 =3.5

2 × l

3.85 × 2

3.5= l

3.85 × 2 ×10

100 × 35 = l

l =22

10 l = 2.2 cm

Area of sector = r360

2

3.85 =22

3.5 3.5360 7

3.85 =22 35 35

360 7 10 10

385

100=

3511

360 10

385 × 360 ×10

100 ×11 × 35 =

= 36º

Length of arc is 2.2 cm and measure of an arc is 36º.


6. Find the perimeter of each ofthese sectors. (Give your answersin terms of ) (3 marks)

Sol. Radius of the sector (r) = 8 cmMeasure of arc () = 40º

10 cm

120º40º8 cm


SCHOOL SECTION 295

Length of arc (l) = 2 r360

=40

360 × 2 × × 8

=16

9 cm

Perimeter of the sector = r + r + l

= 8 + 8 + 16

9

=16

169

= 16 1 +9

Perimeter of the sector =16 (9 + )

9

cm

(b) Radius of a sector (r) = 10 cmMeasure of arc () = 126º

Length of arc (l) = 2 r360

=126

360 × 2 × × 10

= 7 cmPerimeter of the sector = r + r + l

= 10 + 10 + 7 Perimeter of the sector = (20 + 7 ) cm.


7. Find the area of the shaded part. (Give your answers in terms of )

(3 marks)

Sol. (a) Area of shaded part = Area of sector I + Area of sector II

=30 90

3 3 4 4360 360

Area of shaded part =3

4 sq.units4

(b) Area of shaded part = Area of sector I + Area of sector II

=70 50

9 9 8 8360 360

Area of shaded part =63 80

sq. units4 9

3 cm

4 cm

30º

(a)

9 c

m

8 cm

70º

50º(b)


SCHOOL SECTION296


10. In the adjoining figure, seg QR isa tangent to the circle with centre O.Point Q is the point of contact.Radius of the circle is 10 cm.OR = 20 cm. Find the area of theshaded region. ( = 3.14, 3 = 1.73 ) (4 marks)

Sol. In OQR,m OQR = 90º [Radius is perpendicular to the tangent]

OQ2 + QR2 = OR2 [By Pythagoras theorem] 102 + QR2 = 202

QR2 = 400 – 100 QR2 = 300 QR = 300

QR = 100 × 3

QR = 10 3 QR = 10 (1.73) QR = 17.3 cm

Area of OQR =1

2 × Product of Perpendicular sides

=1

2 × OQ × QR

=1

10 17.32

= 86.5 cm2

In OQR, m OQR = 90ºOQ = 10 cmOR = 20 cm

OQ =1

2 OR

By converse of 30º - 60º - 90º triangle theorem. m ORQ = 30º m QOR = 60º [Remaining angle]

Now, For sector O-QXTMeasure of arc () = 60º

Radius (r) = 10 cm

Area of Sector O-QXT =360

× r2

=60

360 × 3.14 × 10 × 10

=157

3Area of sector O-QXT = 52.33 cm2

Area of shaded region = Area of OQR – Area of sector O-QXT= 86.5 – 52.33= 34.17 cm2

Area of the shaded region is 34.17 cm2

Q R

T

O

10 c

m


SCHOOL SECTION 297


11. In the adjoining figure,PR = 6 units and PQ = 8 units.Semicircles are draw takingsides PR, RQ and PQ as diametersas shown in the figure. Find out the areaof the shaded portion. ( = 3.14) (5 marks)

Sol. Diameter PR = 6 units Its radius (r1) = 3 units

Diameter PQ = 8 units Its radius (r2) = 4 units

In PQR,m RPQ = 90º ......(i) [Angle subtended by a semicircle]

QR2 = PR2 + PQ2 [By Pythagoras theorem] QR2 = 62 + 82 QR2 = 36 + 64 QR = 100 QR = 10 units [Taking square roots]

Diameter QR = 10 units Its radius (r3) = 5 units

PQR is a right angled triangle [From (i)]

A (PQR) =1

2× product of perpendicular sides

=1

2 × PR × PQ

=1

2 × 6 × 8

= 24 sq. units.

Area of shaded portion = Area of semicircle with diameter PR + Area

of semicircle with diameter PQ + Area of PQR

– Area of semicircle with diameter QR

=1

2r1

2 + 1

2r2

2 + 24 – 1

2r3

2

=1 1 1

r r – r 242 2 2

2 2 2

1 2 3

=1

2 (r1

2 + r22 – r3

2) + 24

=1

2 × 3.14 (32 + 42 – 52) + 24

=1

2 × 3.14 × (9 + 16 – 25) + 24

=1

2 × 3.14 (0) + 24

= 0 + 24= 24 sq. units

Area of shaded portion = 24 sq.units

P

R Q


SCHOOL SECTION298

` Area of segment of a circle :A segment of a circle is the regionbounded by a chord and an arc.

Area of segment = sin

r –360 2

2


1. In the adjoining figure,A (O-AXB) = 75.36 cm2 andradius = 12 cm, find the areaof the segment AXB. ( = 3.14). (2 marks)

Sol. Radius of a circle (r) = 12 cmA (O-AXB) = 75.36 cm2

Area of OAB =1

2 r2 sin

=1

2 × 12 × 12 × sin 60º

= 72 × 3

2= 36 3= 36 × 1.73= 62.28 cm2

Area of the segment AXB = A (O-AXB) – A (OAB)= 75.36 – 62.28= 13.08 cm2

Area of the segment AXB is 13.08 cm2.


2. Calculate the area of the shadedregion in the adjoining figure whereABCD is a square with side 8 cm each. (3 marks)

Sol. Mark point X as shown in the figureABCD is a square [Given]side = 8 cmRadius (r) = side of a square

r = 8 cmMeasure of arc () = 90º [Angle of a square]

Area of the segment AXC = r2sin

360 2

= 823.14 90 sin 90

360 2

= 64 1.57 1

2 2

= 64 1.57 1

2

A

O

P Q

R

A

X

B

O 60º

BA

D C8 cm

X


SCHOOL SECTION 299

=64 0.57

2

=36.48

2 cm2

Area of shaded region = 2 × Area of segment AXC

= 2 × 36.48

2= 36.48 cm2

Area of shaded region is 36.48 cm2.


3. In the adjoining figure,P is the centre of the circle withradius 18 cm. If the area of thePQR is 100 cm2 and area of thesegment QXR is 13.04 cm2.Find the central angle . ( = 3.14) (3 marks)

Sol. Radius of a circle (r) = 18 cmArea of PQR = 100 cm2

Area of the segment QXR = 13.04 cm2

Area of sector P-QXR = Area of PQR + Area of segment QXR= 100 + 13.04

Area of sector P-QXR = 113.04 cm2

Area of sector =360

× r2

113.04 =360

× 3.14 × 18 × 18

11304 =360

× 314 × 18 × 18

11304 × 360

314 ×18 ×18 =

= 40

Central angle is 40º.


4. In the adjoining figure, the centre of thecircle is A and ABCDEF is a regular hexagonof side 6 cm. Find the following :( 3 = 1.73, = 3.14)(i) Area of segment BPF(ii) Area of the shaded portion (5 marks)

Sol. Side of hexagon = radius of a circle Radius (r) = 6 cm

Measure of arc () = Angle of a regular hexagon = 120º

Area of sector A-BPF =360

× r2

=120

360 × 3.14 × 6 × 6

= 3.14 × 12= 37.68 cm2

P

Q R

X

18 c

m

A

P

B F

C E

D

M

P


SCHOOL SECTION300

In ABF,seg AB seg AF [Radii of the same circle]

ABF AFB ......(i) [Isosceles triangle theorem]mBAF + mABF + mAFB = 180º [Sum of measures of angles

of a triangle is 180º] 120 + mABF + mABF = 180 [Given, from (i)] 2mABF = 180 – 120 2mABF = 60

mABF = 60

2 mABF = 30º mABM = 30º .......(ii) [B - M - F]

In AMB,mAMB = 90º [By construction]mABM = 30º [From (ii)]

mBAM = 60º [Remaining angle] AMB is 30º - 60º - 90º triangle, By 30º - 60º - 90º triangle theorem,

AM =1

2 AB [Side opposite to 30º]

AM =1

2 × 6

AM = 3 cm

BM = 3

2 × AB [Side opposite to 60º]

BM = 3

2 × 6

BM = 3 3 cm

seg AM chord BF

BM = 1

2 BF

[The perpendicular drawn

from the centre of circle to

a chord, bisec ts the chord]

3 3 = 1

2 BF

BF = 6 3 BF = 6 (1.73) BF = 10.38 cm.

Area of ABF =1

2 × base × height

=1

2 × BF × AM

=1

2 × 10.38 × 3

= 15.57 cm2

Area of segment BPF = Area of sector A-BPF – Area of ABF= 37.68 – 15.57

Area of segment BPF = 22.11 cm2


SCHOOL SECTION 301

(ii) side = 6 cm

Area of regular hexagon ABCDEF =3 3

2 × (side)2

=3 3

2 × 6 × 6

= 54 3

= 54 × 1.73

= 93.42 cm2

Area of the shaded portion = Area of regular hexagon ABCDEF

– Area of ABF

= 93.42 – 15.57

= 77.85 cm2

Area of segment BPF is 22.11 cm2 and Area of shaded portion is

77.85 cm2.


8. Find the area of the shaded region.

( = 3.14, 3 = 1.73) (3 marks)

Sol. Radius of the sector (r) = 12 cmMeasure of arc () = 60º

Area of segment = r2sin

–360 2

=3.14 × 60 sin 60

12 –360 2

2

=3.14 3 1

144 –6 2 2

=3.14 3

144 –6 4

=6.28 – 3(1.73)

14412

=6.28 – 5.19

14412

=144 ×1.09

12= 12 × 1.09

= 13.08 cm2

Area of shaded region is 13.08 cm2.

12 cm

60º


SCHOOL SECTION302


9. In the adjoining figure,m POQ = 30º and radius OP = 12 cm.Find the following (Given = 3.14)(i) Area of sector O-PRQ (ii) Area of OPQ(iii) Area of segment PRQ (3 marks)

Sol. Radius of the circle (r) = 12 cmMeasure of arc () = 30º

Area of sector O - PRQ =360

× r2

=30

360 × 3.14 × 12 × 12

= 37.68 cm2

In OMP, m OMP = 90º [Given]m POM = 30º [Given and O - M - Q]

m OPM = 60º [Remaining angle] OMP is 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem

PM =1

2 OP

=1

2 × 12

PM = 6 cm.OP = OQ = 12 cm [Radii of same circle]

Area of OPQ =1

2 × base × height

=1

2 × OQ × PM

=1

2 × 12 × 6

= 36 cm2

Area of segment PRQ = Area of sector O-PRQ – Area of OPQ= 37.68 – 36= 1.68 cm2

(i) Area of sector O-PRQ is 37.68 cm2

(ii) Area of OPQ is 36 cm2

(iii)Area of segment PRQ is 1.68 cm2


12. In the adjoining figure,PR and QS are two diameters of the circle.If PR = 28 cm and PS = 14 3 cm, find(i) Area of triangle OPS(ii) The total area of two shaded segments.

( 3 = 1.73) (4 marks)Sol. Draw seg OM side PS

OP = 1

PR2 [Radius is half of diameter]

OP = 1

282

OP = 14 cm

O

12 cm 30º M

Q

P R•

P S

Q R

O

120º


SCHOOL SECTION 303

seg OM chord PS [By construction]

PM =1

PS2 [The perpendicular drawn from the

centre of a circle to a chord bisec ts

the chord]

PM =1

14 32

PM = 7 3 cmIn OMP,OMP = 90º [By construction]OM2 + PM2 = OP2 [By Pythagoras theorem]

OM2 + 7 32

= 142

OM2 = 196 – 147 OM2 = 49 OM = 7 cm [Taking square roots]

Area of OPS =1

2 × base × height

Area of OPS =1

2 × PS × OM

=1

2 × 14 3 × 7

= 49 3= 49 (1.73)

Area of OPS = 84.77 cm2

Area of sector OPS =360

× r2

=120 22

14 14360 7

=616

3= 205.33 cm2

Area of segment PS = Area of sector OPS – Area of OPS= 205.33 – 84.77= 120.56 cm2

Similarly we can prove,Area of segment QR= 120.56 cm2

Total area of two shaded segments = 120.56 + 120.56 = 241.12 cm2

Area of OPS is 84.77 cm2 and total area of two shaded segments is241.12 cm2.


13. In the adjoining figure,seg PQ is a diameter of semicircle PNQ.The centre of arc PMQ is O.OP = OQ = 10 cm and m POQ = 60º.Find the area of the shaded portion

(Given = 3.14, 3 = 1.73) (5 marks)

Sol. For a segment PMQ,radius (r) = 10 cmmeasure of arc () = 60º

P Q

M

Q•

O

10 c

m

10 cm

60º


SCHOOL SECTION304

Area of segment PMQ =sin

r –360 2

2

=3.14 × 60 sin 60

10 –360 2

2

=3.14 3 1

100 –6 2 2

=3.14 3

100 –6 4

=6.28 – 3(1.73)

10012

=6.28 – 5.19

10012

=100 1.09

12

=109

12 Area of segment PMQ = 9.08 cm2

In OPQ,seg OP seg OQ [Radii of same circle]

OPQ OQP [Isosceles triangle theorem]Let, m OPQ = m OQP = x

m OPQ + m OQP + m POQ = 180º [Sum of the measures ofangles of a triangle is 180º]

x + x + 60 = 180 2x = 180 – 60 2x = 120

x = 120

2 x = 60 m POQ = m OPQ = m OQP = 60º OPQ is an equilateral triangle [An equiangular triangle is an

equilateral triangle] OP = OQ = PQ = 10 cm [Sides of an equilateral triangle]

Diameter PQ = 10 cm

Radius (r) = 10

2= 5 cm

Area of semicircle =1

r2

2

=1

3.14 5 52

= 39.25 cm2

Area of the shaded portion = Area of semicircle – Area of segment PMQ= 39.25 – 9.08= 30.17 cm2

The area of shaded portion is 30.17 cm2.


SCHOOL SECTION 305

CUBOID [RECTANGULAR PARALLELOPIPED]

l

h

bA cuboid is a solid figure bounded by sixrectangular faces, where the opposite facesare equal.A cuboid has a length, breadth and heightdenoted as ‘l’, ‘b’ and ‘h’ respectively as shownin the figure,

In our day to day life we come across cuboidssuch as rectangular room, rectangular box,brick, rectangular fish tank, etc.The following are the formulae for the surface area of cuboid :

FORMULAE

1. Total surface area of a cuboid = 2 (lb + bh + lh)2. Vertical surface area of a cuboid = 2 (l + b) × h3. Volume of a cuboid = l × b × h


1. The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total surfacearea. (2 marks)

Sol. Length of a cuboid (l) = 16 cmits breadth (b) = 14 cmits height (h) = 20 cm

Total surface area of a cuboid = 2 (lb + bh + lh)= 2 (16 × 14 + 14 × 20 + 16 × 20)= 2 (224 + 280 + 320)= 2 × 824= 1648 cm2

Total surface area of a cuboid is 1648 cm2.


4. The cuboid water tank has length 2 m, breadth 1.6 m and height 1.8m.Find the capacity of the tank in litres. (2 marks)

Sol. Length of the cuboidal water tank (l) = 2 mits breadth (b) = 1.6 m

and its height (h) = 1.8 m.Volume of cuboidal water tank = l × b × h

= 2 × 1.6 × 1.8= 5.76 m3

= 5.76 × 1000 litres [l m3 = 1000 litres]

Volume of cuboidal water tank is 5760 litres.


6. A fish tank is in the form of a cuboid whose external measures are80cm × 40cm × 30cm. The base, side faces and back faces are to be coveredwith a coloured paper. Find the area of the paper needed. (2 marks)

Sol. Length of cuboidal fish tank (l) = 80 cmits breadth (b) = 40 cmits height (h) = 30 cm


SCHOOL SECTION306

Cuboid is make up of 6 rectangular facesArea of the base of the fish tank = l × b

= 80 × 40= 3200 cm2

Area of two side faces = 2 × b × h= 2 × 40 × 30= 2400 cm2

Area of back face = l × h= 80 × 30= 2400 cm2

Area of the paper needed = 3200 + 2400 + 2400= 8000 cm2

The area of the paper needed is 8000 cm2.


7. Find the total cost of white washing the 4 walls of a cuboidal room atthe rate Rs. 15 per m2. The internal measures of the cuboidal room arelength 10 m, breadth 4 m and height 4 m. (3 marks)

Sol. Length of the cuboidal room (l) = 10mIts breadth (b) = 14mIts height (h) = 4m

Vertical Surface area of the room = 2 (l + b) × h= 2 (10 + 4) × 4= 2 × 14 × 4= 112m2

Area of white washing = 112m2

Rate of white washing = Rs 15 per m2

Total cost = Area of white washing × rate ofwhite washing

= 112 × 15= 1680

Total cost of white washing is Rs. 1680.


9. A beam 4m long, 50cm wide and 20cm deep is made of wood whichweighs 25kg per m3. Find the weight of the beam. (3 marks)

Sol. Length of the beam (l) = 4mits breadth (b) = 50cm

=50

100 m

=5

10 m

its height (h) = 20 cm

=20

100m

=2

10m


SCHOOL SECTION 307

CUBE

l

Volume of the beam = l × b × h

= 4 × 5

10 ×

2

10

=40

100m3

Weight of the beam = 25kg per m3

Total weight of the beam = 25 × 40

100= 10 kg

The weight of the beam is 10 kg.


5. The length, breadth and height of a cuboid are in the ratio 5:4:2. If thetotal surface area is 1216 cm2, find the dimensions of the solid. (3 marks)

Sol. Ratio of the length, breadth and height of a cuboid is 5 : 4 : 2Let the common multiple be ‘x’

Length of a cuboid = 5x cmits breadth = 4x cm

and its height = 2x cmTotal surface area of a cuboid = 1216 cm2

Total surface area of a cuboid = 2 (lb + bh + lh) 1216 = 2 [(5x) (4x) + (4x) (2x) + (5x) (2x)]

1216

2= 20x2 + 8x2 + 10x2

608 = 38x2

608

38= x2

x2 = 16 x = 4 [Taking square roots]

Length of a cuboid = 5x= 5 (4)= 20 cm

its Breadth = 4x= 4 (4)= 16cm

and its height = 2x= 2 (4)= 8 cm

Dimensions of a cuboid are 20 cm, 16 cm and 8 cm.

A cube is a cuboid bounded by six equal square faces.Hence its length, breadth and height are equal.

The edge of the cube = length = breadth = heightThe edge of the cube is denoted as ‘l’A dice is an example of cube.The following are the formulae for the surface area of the cube :

FORMULAE 1. Total surface area of a cube = 6l2

2. Vertical surface area of a cube = 4l2

3. Volume of cube = l3


SCHOOL SECTION308


2. The side of a cube is 60 cm. Find the total surface area of the cube.Sol. Side of a cube (l) = 60 cm

Total surface area of a cube = 6l2

= 6 (60)2

= 6 × 60 × 60= 21600 cm2

Total surface area of a cube is 21600 cm2.


3. The perimeter of one face of a cube is 24 cm.Find (i) the total area of the 6 faces (ii) the volume of the cube.

Sol. Perimeter of one face of a cube = 24 cmPerimeter of one face of a cube = 4l

4l = 24

l =24

4 l = 6 cm.


= 6 (6)2

= 6 × 6 × 6= 216 cm2

Volume of the cube = l3

= 63

= 216 cm3.

Total area of the 6 faces is 216 cm2 and volume of the cube is 216 cm3.


5. The volume of a cube is 1000 cm3. Find its total surface area.Sol. Volume of a cube = 1000 cm3

Volume of a cube = l3

l3 = 1000 l = 10 cm [Taking cube roots]


= 6 × 102

= 6 × 10 × 10= 600 cm2

Total surface area of a cube is 600 cm2.


8. A solid cube is cut into two cuboids exactly at middle. Find the ratio ofthe total surface area of the given cube and that of the cuboid.

Sol. Side of a cube = l Total Surface of a cube = 6l2

Length of cuboid (l1) = side of a cubel1 = l

Its Breadth (b1) =2

l

Its height (h1) = l2

l 2

l

l

l


SCHOOL SECTION 309

RIGHT CIRCULAR CYLINDER

h

r

Total surface area of a cuboid= 2 (l1b1 + b1h1 + l1h1)

= 2 × + × + ×2 2

l ll l l l

= 2 + +2 2

2 22l l

l

= 2 2

2

2l

+ 2l

= 2 × 2l2

Total surface area of a cuboid= 4l2

Ratio of the total surface area of a cube and cuboid=T.S.A of a cube

T.S.A of a cuboid

=6

4

2

2

l

l

=3

2= 3 : 2

The ratio of the total surface area of the given cube and thatof the cuboid is 3 : 2.

A right circular cylinder (Cylinder) is a solid figurebounded by two flat circular surfaces and a curvedsurface.The perpendicular distance between the two base facesis called height of the cylinder and is denoted by ‘h’.The radius of the base of the cylinder is denoted by ‘r’.The cylinders which we see regularly are drum,pipe, road roller, coins, test tube, refill of a ball pen, syringe etc.The following are the formulae for the surface area of a right circularcylinder :

FORMULAE 1. Curved surface area of a right circular cylinder = 2rh 2. Total surface area of a right circular cylinder = 2r (r + h) 3. Volume of a right circular cylinder = r2h


1. The radius of the base of a right circular cylinder is 3 cm height is 7cm.Find (i) curved surface area (ii) total surface area (iii) volume of the

closed right circular cylinder.

22Given =

7 (3 marks)

Sol. Radius of a right circular cylinder = 3cmits height (h) = 7cm

(i) Curved surface area of a cylinder = 2 rh

= 2 × 22

7 × 3 × 7

Curved surface area of a cylinder = 132 cm2


SCHOOL SECTION310

(ii) Total Surface area of a cylinder = 2r (r + h)

= 2 × 22

7 × 3 (3 + 7)

= 2 × 22

7 × 3 × 10

=1320

7= 188.57 cm2

(iii) Volume of the cylinder = r2h

=22

7 × 3 × 3 × 7

= 198 cm3

Curved surface area is 132 cm2 Total surface area is 188.57cm2 andvolume of the cylinder is 198 cm3


2. The volume of a cylinder is 504 cm3 and height is 14 cm. Find its curvedsurface area and total surface area. Express answer in terms of .(3 marks)

Sol. Volume of a cylinder = 504 cm3

Its height (h) = 14 cmVolume of a cylinder = r2h

504 = × r2 × 14

504

14= r2

r2 = 36 r = 6 cm [Taking square roots]

Curved surface area of a cylinder = 2rh= 2 × × 6 × 14= 168 cm2

Total surface area of a cylinder = 2r (r + h)= 2 × × 6 (6 + 14)= 2 × × 6 × 20= 240 cm2

Curved surface area of a cylinder is 168cm2

and Total surface area of a cylinder is 240cm2


3. The radius and height of a cylinder are n same ratio 3:7 and its volumeis 1584 cm3. Find its radius. (3 marks)

Sol. The ratio of radius and height of a cylinder is 3 :7Let the common multiple be ‘x’

Radius of cylinder (r) = ‘3x’ cmand its height (h) = ‘7x’ cm

Volume of a cylinder = 1584 cm3

Volume of a cylinder = r2h

1584 =22

7 × (3x) × (3x) × (7x)

1584 = 22 × 9 × x3

1584

22 × 9 = x3

x3 = 8 x = 2 [Taking cube roots]


SCHOOL SECTION 311

Radius (r) = 3x= 3(2)= 6cm

Radius of the cylinder is 6 cm.


4. Keeping the height same, how many times the rod of the given cylindershould be made to get the cylinder of double the volume of given cylinder ?

(3 marks)Sol. Let the radius and the volume of the given cylinder be r1 and v1 respectively.

The radius and the volume of the required cylinder be r2 and v2 respectively.Let the heights of the cylinder be h [ their heights are same]From the given condition,

v2 = 2v1

r h 22 = r h 2

1

r22 = 2r2

1

r2 = 2 r1 [Taking square roots]

The radius of the required cylinder should be 2 times the radiusthe given cylinder


1. A cylindrical hole of diameter 30 cm is boredthrough a cuboid wooden block with side 1meter. Find the volume of the object so formed( = 3.14) (4 marks)

Sol. side of cubical wooden block = 1 m

= 100 cm

Volume of cubical wooden block = l3

= (100)3

= 1000000 cm3

A cylindrical hole is bored through the cubical wooden block

Height of cylindrical hole (h) = 1m

= 100 cm

Diameter of cylindrical hole = 30 cm

Its radius (r) =30

2= 15 cm

Volume of cylindrical hole = r2h

= 3.14 × 15 × 15 × 100

= 70650 cm3

Volume of the object so formed = Volume of cubical wooden block

– Volume of cylindrical hole

= 1000000 – 70650

= 929350 cm3

Volume of the object so formed is 929350 cm3.


SCHOOL SECTION312

EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185)4. An ink container of cylindrical shape is filled with ink upto 91%. Ball

pen refills of length 12 cm and inner diameter 2 m are filled upto 84%.If the height and radius of the ink container are 14 cm and 6 cmrespectively, find the number of refills that can be filled with this ink.

(4 marks)Sol. Height of the cylindrical container (h) = 14cm

Its radius (r) = 6 cmVolume of cylindrical container = r2h

= × 6 × 6 × 14= 504 cm3

But, volume of ink filledin the cylindrical container = 91% of 504

=91

× 504100

cm3

Length of ball pen refill (h1) = 12mits inner diameter = 2 mm

Its radius (r1) = 1 mm

=1

10cm

Volume of the refill = r12 h1

= × 1

10 ×

1

10 × 12

=12

100

cm3

But, volume of ink filled = 84% of 12

100

=84 12

×100 100

cm3

Number of refills that can be filled with ink

=Volume of ink filled in the cylindrical container

Volume of ink filled in each refill

=

91× 504

10084 ×12

100 ×100

=91× 504 100 ×100

×100 84 ×12

= 4550

Number of refills that can be filled with this ink is 4550.


15. A building has 8 right cylindrical pillars whose cross sectional diameteris 1 m and whose height is 4.2 m. Find the expenditure to paint thosepillars at the rate of Rs. 24 per m2. (3 marks)

Sol. Diameter of a pillar = 1 m

Its radius (r) =1

2 m

Its height (h) = 4.2 mCurved surface area of a pillars = 2rh

= 2 × 22

7 ×

1

2 × 4.2

= 13.2 m2


SCHOOL SECTION 313

Curved surface area of 8 pillars = 8 × 13.2= 105.6 m2

Rate of painting = Rs. 24 per m2

Total expenditure = Area to be painted × Rate of painting= 105.6 × 24= 2534.40

Total expenditure to paint the pillars is Rs. 2534.40.


16. A 10 m deep well of diameter 1.4 m us dug up in a field and the earthfrom digging is spread evenly on the adjoining cuboid field. The lengthand breadth of that filled are 55m and 14 m respectively. Find thethickness of the earth layer spread. (4 marks)

Sol. Diameter of well = 1.4 m

Its radius (r) =1.4

2= 0.7 m

Its depth (h) = 10 mVolume of cylindrical well = r2h

=22

0.7 0.7 107

=22 7 7

107 10 10

=154

10= 15.4 m3

Volume of earth dug is 15.4 m3

Now, Earth dug from the well is spread evenly on the adjoining cuboid fieldVolume of cuboid = Volume of earth dug

= 15.4 m3

Length of a cuboid (l) = 55 mIts breadth (b) = 14 m

Volume of cuboid = l × b × h 15.4 = 55 × 14 × h

154

10 × 55 ×14 = h

h =1

50m

h = 0.02 m

The thickness of the earth layer spread is 0.02 cm.


18. A roller of diameter 0.9 m and length 1.8 m is used to press the ground.Find the area of ground pressed by it in 500 revolutions. (Given = 3.14)

(3 marks)Sol. Diameter of the roller = 0.9 m

its radius (r) =0.9

2= 0.45 m

its length (h) = 1.8 m


SCHOOL SECTION314

RIGHT CIRCULAR CONE

hl

rA P

O

Curved surface area of the roller = 2rh= 2 × 3.14 × 0.45 × 1.8= 6.28 × 0.81= 5.0868 m2

Area of the ground pressed by the roller in 1 revolution = curved surfacearea of roller

Area of the ground pressed in one revolution= 5.0868 m2

Area of the ground pressed in 500 revolution= 500 × 5.0868

=50868

50010000

= 2543.4 m2

Area of the ground pressed by the roller = 2543.4 m2.

An ice-cream cone, a clown’s hat, a funnel are examplesof cones. A cone has one circular flat surface and onecurved surface.In the diagram alongside,seg OA is the height of the cone denoted by ‘h’.seg AP is the radius of the base denoted by ‘r’.seg OP is the slant height of the cone denoted by ‘l’.The h, r and l of a cone represents the sides of a right angled trianglewhere l is the hypotenuse.

l2 = r2 + h2.FORMULA

Volume of a right circular cone = 13

× r2h


1. The curved surface area of a cone is 4070 cm2 and its diameter is70 cm. What is its slant height ? (2 marks)

Sol. Diameter of a cone = 70 cm.

Its radius (r) =70

2= 35 cm.

Curved surface area of a cone = 4070 cm2

Curved surface area of a cone = r l

4070 =22

357

l

4070

22 × 5 = l

l = 37

Slant height of a cone is 37 cm.


2. The base radii of two right circular cones of the same height are in theratio 2:3. Find the ratio of their volumes. (3 marks)

Sol. Let the radii of two right circular cone be r1 and r2 and their volumesbe v1 and v2 respectively


SCHOOL SECTION 315

r

r1

2=

2

3.......(i) (Given)

v

v1

2=

1r h

31

r h3

21

22

v

v1

2=

r

r

2122

v

v1

2=

r

r

2

1

2

v

v1

2=

2

3

2

[From (i)]

v

v1

2=

4

9 v1 : v2= 4 : 9

Ratio of volumes of two right circular cone is 4 : 9


3. A cone of height 24 cm has a plane base of surface area 154 cm2. Findits volume. (2 marks)

Sol. Height of a cone (h) = 24 cmSurface area of base = 154 cm2

Volume of a cone =1

3 × Surface area of base × height

=1

3 × r2 × h

=1

3 × 154 × 24

= 1232 cm3

Volume of the cone is 1232 cm3.


4. Curved surface area of a cone with base radius 40 cm is 1640 sq.cm.Find the height of the cone. (3 marks)

Sol. Curved surface area of a cone = 1640cm2

its radius (r) = 40 cm.Curved surface area of a cone = r l

1640 = × 40 × l

1640

40= l

l = 41 cmNow,

r2 + h2 = l2

402 + h2 = 412

h2 = 412 – 402

h2 = 1681 – 1600 h2 = 81 h = 9 cm [Taking square roots] Height of a cone is 9 cm.


SCHOOL SECTION316


17. The total surface area of cone is 71.28 cm2. Find the volume of thiscone if the diameter of the base is 5.6 cm. (3 marks)

Sol. Diameter of the base = 5.6 cm

Its radius (r) =5.6

2= 2.8 cm

Total surface area of a cone = 71.28 cm2

Total surface area of cone = r (r + l)

71.28 =22

7 × 2.8 (2.8 + l)

7128

100=

22

7 ×

28

10 (2.8 + l)

7128 10

100 22 4

= 2.8 + l

81

10= 2.8 + l

8.1 – 2.8 = l l = 5.3 cm

r2 + h2 = l2

(2.8)2 + h2 = (5.3)2

h2 = (5.3)2 – (2.8)2

h2 = (5.3 + 2.8) (5.3 – 2.8) h2 = 8.1 × 2.5

h2 =81 × 25

10 ×10

h = 9 × 5


h = 45

10 h = 4.5 cm

Volume of a cone =1

r h3

2

=1 22

2.8 2.8 4.53 7

=1 22 28 28 45

3 7 10 10 10

=36960

1000

Volume of a cone is 36.96 cm3.


19. The diameter of the base of metallic cone is 2 cm and height is 10 cm.900 such cones are molten to form 1 right circular cylinder whose radiusis 10 cm. Find total surface area of the right circular cylinder so formed.(Given = 3.14) (4 marks)

Sol. Diameter of the base of metallic cone = 2 cm

Its radius (r) = 2

2 = 1 cm

Its height (h) = 10 cm


SCHOOL SECTION 317

Volume of a metallic cone =1

r h3

2

=1

3 × × 1 × 1 × 10

=10

3

cm3

Volume of 900 metallic cones =10

9003

= 3000 cm3

900 cones are melted to form a right circular cylinder Volume of a cylinder = 3000

For a cylinder, Radius (r2) = 10 cm and height be h2

Volume of a cylinder = r h21 1

3000 = × 10 × 10 h2

h1 = 30 cmTotal surface area of cylinder = 2r1 (r1 + h1)

= 2 × 3.14 × 10 (10 + 30)= 6.28 × 40= 2512 cm2

Total surface area of the right circular cylinder is 2512 cm2.


20. The volume of a cone of height 5 cm is 753.6 cm3. This cone and a cylinderhave equal radii and height. Find the total surface area of cylinder.(Given = 3.14) (3 marks)

Sol. Height of a cone (h) = 5 cmVolume of a cone = 753.6 cm3

Volume of a cone =1

r h3

2

753.6 =1

× 3.14 × r × 53

2

7536

10=

1 314× × r × 5

3 1002

7536 × 3 ×100

10 × 314 × 5 = r2

r2 = 144

r = 12 cm [Taking square roots]

Cone and cylinder have equal radii and height

Radius of a cylinder = 12 cm and its heights = 5 cm.

Total surface area of cylinder = 2r (r + h)

= 2 × 3.14 × 12 (12 + 5)

= 75.36 × 17

= 1281.12 cm2

Total surface area of a cylinder is 1281.12 cm2.


SCHOOL SECTION318

FRUSTUM OF THE CONE :If the cone is cut off by a plane parallel to the base notpassing through the vertex, two parts are formed as(i) cone (a part towards the vertex)(ii) frustum of cone (the part left over on the

other side i.e. towards base of the original cone)

FORMULAE

1. Slant height (l) of the frustum = 221 2h + r – r

2. Curved surface area = p (r1 + r2) l

3. Total surface area of the frustum = 2 21 2 1 2r + r + r + r l

4. Volume of the frustum = 2 21 2 1 2

1r + r + r × r h

3


5. The curved surface area of the frustum of a cone is 180 sq. cm and theperimeters of its circular bases are 18 cm and 6 cm respectively. Findthe slant height of the frustum of a cone. (3 marks)

Sol. Curved surface area of the frustum of a cone = 180 cm2

Perimeters of circular bases are 18 cm and 6 cm 2r1 = 18 ........(i)

2r2 = 6 ........(ii)Adding (i) and (ii), we get2r1 + 2r2 = 18 + 6

2 (r1 + r2) = 24

(r1 + r2) =24

2 (r1 + r2) = 12 .......(iii)

Curved surface area of the frustum of a cone = (r1 + r2) l 180 = (r1 + r2) l 180 = 12 × l [From (iii)] l = 15 cm

Slant height of the frustum of a cone is 15 cm.


21. The height of a cone is 40 cm. A small cone is cut off at the top of a

plane parallel to its base. If its volume be 164

of the volume of the given

cone, at what height above the base is the section cut ? (5 marks)

Sol. Let the radius, height and volume of thesmaller cone be r1, h1 and v1 respectively.The radius height and volume of thegiven figure cone be r2, h2 and v2 respectively.

h2 = 40 m [Given]Consider points A, B, C, E and F as shown in the figure,In AEF and ABC,A A [Common angle]AEF ABC [Each is 90º]

AEF ~ ABC [By AA test of similarity]

hl

r1

r2

hl

r2

r1

A

F

B

E

C


SCHOOL SECTION 319

SPHERE

AE

AB=

EF

BC[c.s.s.t.]

h

h1

2=

r

r1

2......(i)

V1 =1

v64 2 [Given]

V

V1

2=

1

64......(ii)

V

V1

2=

1 r h31 r h3

21 1

22 2

1

64=

r h×

r h

2

1 1

2 2

1

64=

h h×

h h

2

1 1

2 2[From (i)]

1

64=

h

h

3

1

2

h

h1

2=

1

4

h

401 =

1

4

h1 =40

4 h1 = 10 cm The height above the base in the section cut = h2 – h1

= 40 – 10= 30 cm

The height above the base in the section cut is 30 cm.

The set of all points of space which are at a fixeddistance from a fixed point is called a sphere.The fixed point is called the centre and the fixeddistance is called the Radius of the sphere.In the adjoining figure, point O is the centre of thesphere and seg OA is the radius of the sphere which isdenoted as ‘r’.Since the entire surface of the sphere is curved, itsarea is called as curved surface area or simply surfacearea of the sphere.Some common examples of a sphere are cricket ball, football, globe,spherical soap bubble etc.The following are the formulae for surface area of the sphere :

FORMULAESurface area (curved surface area) of a sphere = 4r2

Volume of a sphere = 43

× r3

h•O Ar


SCHOOL SECTION320


1. Find the volume and surface area of a sphere of radius 4.2 cm.

22=

7

(3 marks)Sol. Radius of a sphere (r) = 4.2 cm

Volume of a sphere =4

3r3

=4

3 ×

22

7 × 4.2 × 4.2 × 4.2

=4

3 ×

22

7 ×

42

10 ×

42

10×

42

10

=310464

1000= 310.464= 310.46 cm3

Surface area of a sphere = 4r2

= 4 × 22

7 × 4.2 × 4.2

= 4 × 22

7 ×

42

10 ×

42

10

=22176

100= 221.76 cm2

Volume of sphere is 310.46 cm3 and surface area of a sphere is 221.76 cm2.


2. The volumes of two spheres are in the ratio 27 : 64. Find their radii ifthe sum of their radii is 28 cm. (3 marks)

Sol. Let the radii of two spheres be r1 and r2 and their volumes be v1 and v2.v

v1

2=

27

64........(i) [Given]

v

v1

2=

4r

34

r3

31

32

27

64=

4r

34

r3

31

32

r

r

3132

=27

64

r

r1

2=

3

4[Taking cube roots]

Let the common multiple be x r1 = 3x and r2 = 4x

r1 + r2 = 28 [Given] 3x + 4x = 28 7x = 28

x =28

7 x = 4


SCHOOL SECTION 321

r1 = 3x r2 = 4x r1 = 3 (4) r2 = 4 (4) r1 = 12 cm r2 = 16 cm

Radii of two spheres are 12 cm and 16 cm.


3. The surface area of a sphere is 616cm2. What is its volume ?

22=

7

(3 marks)Sol. Surface area of sphere = 616 cm2

Surface area of a sphere = 4r2

616 = 4 × 22

7 × r2

616 7

4 22

= r2


Volume of a sphere =4

3r3

=4 22

7 7 73 7

=4312

3= 1437.33 cm3

The volume is 1437.33 cm3.


4. If the radius of a sphere is doubled, what will be the ratio of its surfacearea and volume as to that of the first sphere.

Sol. Let the original radius be ‘r1’ Original surface Area (A1)= 4r1

2

New radius (r2) = 2r1 New surface Area (A2) = 4r2

2

= 4 × × (2r1)2

= 4 × × 4r12

= 16r12

Ratio of New surface area to the original surface area =A

A2

1

=

16 r

4 r

2121

=4

1 Ratio of New surface area to the original surface area is 4 : 1

Now, original volume (v1) =4

3r1

3

New volume (v2) =4

3r2

3

=4

3(2r1)

3

=4

3 ×× 8r1

3

v2 =32

3 r13


SCHOOL SECTION322

HEMISPHERE

Ratio of New volume to the original volume =v

v2

1

=

32r

34

r3

31

31

=32

4

=8

1

Ratio of New volume to the original volume is8 : 1

Half of a sphere is called as hemisphere.Any hemisphere is made up of a curved surface and a planecircular surface.The following are the formulae for the surface area of a hemisphere :

FORMULAE 1. Curved surface area of a hemisphere = 2r2

2. Total surface area of a hemisphere = 3r2

3. Volume of a hemisphere = × r323


5. The curved surface area of a hemisphere is 90517

cm2, what is its volume?

(3 marks)

Sol. Curved surface area of a hemisphere = 9051

7cm2

Curved surface area of a hemisphere = 2r2

9051

7= 2 ×

22

7 × r2

6336

7= 2 ×

22

7 × r2

6336 7

7 2 22

= r2


Volume of a hemisphere =2

3r3

=2

3 ×

22

7 × 12 × 12 × 12

=25344

7= 3620.57 cm3

Volume of a hemisphere is 3620.57 cm3.


SCHOOL SECTION 323


14. The following shapes are made up of cones and hemispheres. Find theirvolume. (3 marks)

Sol. (a) For a hemisphere,Height = radius (r) = 3.5 cmVolume of the figure = Volume of cone + Volume of hemisphere

=1

3r1h +

2

3r3

=1

3r2 [h + 2r]

=1 22

3.5 3.5 [12.3 2(3.5)]3 7

=1 22 35 35

(12.3 7)3 7 10 10

=1 385

19.33 10

=385 193

3 10 10

=74305

3 100

=24768.33

100= 247.68 cm3

Volume of the given figure is 247.68 cm3.

(b) Diameter of smaller hemisphere = 3 cm

Its radius r1 = 3

2 = 1.5 cm

Diameter of bigger figure = 10 cm

Its radius r2 = 10

2 = 5 cm

Volume of the figure = Volume of smaller hemisphere +volume of bigger hemisphere

=2 2

r r3 3

3 31 2

= 2r r

33 3

1 2

=2 22

(1.5) 53 7

3 3

=2 22

(3.375 125)3 7

=2 22

128.3753 7

=5648.5

21= 268.98 cm3

Volume of the given figure is 268.98 cm3.

12.3 cm

3.5 cm

3 cm

10 cm


SCHOOL SECTION324


2. A toy is a combination of a cylinder, hemisphere and a cone, each with radius10cm. Height of the conical part is 10 cm and total height is 60cm.Find the total surface area of the toy. ( = 3.14, 2 = 1.41) (5 marks)

Sol.

A toy is a combination of cylinder, hemisphere and cone, each withradius 10 cm

r = 10 cm Height of the conical part (h) = 10 cm

Height of the hemispherical part = its radius = 10cmTotal height of the toy = 60cm

Height of the cylindrical part (h1) = 60 – 10 – 10 = 60 – 20 = 40 cml2 = r2 + h2

l2 = 102 + 102

l2 = 100 + 100l2 = 200

l = 200 [Taking square roots]

l = 10 2 cm

Slant height of the conical part (l) = 10 2= 10 × 1.41= 14.1 cm

Total surface area of the toy = Curved surface area of the conicalpart + Curved surface area of thecylindrical part + Curved surfacearea of the hemispherical part

= rl + 2rh1 + 2r2

= r (l + 2h1 + 2r)= 3.14 × 10 (14.1 + 2 × 40 + 2 × 10)= 31.4 (14.1 + 80 + 20)= 31.4 × 114.1= 3582.74 cm2

Total surface area of the toy is 3582.74 cm2.


3. A test tube has diameter 20 mm and height is15 cm. The lower portion is a hemisphere inthe adjoining figure. Find the capacity of thetest tube. ( = 3.14) (5 marks)

Sol. Diameter of a test tube = 20 mm

its radius (r) =20

2= 10 mm= 1 cm

Its height (h) = 15 cmHeight of hemispherical part (h1) = radius of hemisphere

= 1 cm Height of cylindrical part (h2) = h – h1

= 15 – 1= 14 cm

60 cm

10 cm10 cm

10 c

m

15 cm


SCHOOL SECTION 325

Volume of test tube = Volume of cylindrical part +Volume of hemispherical part

= r2h2 + 2

3r3

= r22

h + r3

2

= 3.14 (1) 2

14 +3

= 3.14 × 44

3

=138.16

3Volume of test tube = 46.05 cm3

Capacity of a test tube is 46.05 cm3


5. A cylinder of radius 12 cm contains water upto depth of 20 cm. Aspherical iron ball is dropped into the cylinder and thus water level israised by 6.75 cm. what is the radius of the ball ? (5 marks)

Sol. Radius of the cylinder (r) = 12 cmA spherical iron ball is dropped into the cylinder and the water levelrises by 6.75 cm

Volume of water displaced = volume of the iron ballHeight of the raised water level (h) = 6.75 mVolume of water displaced = r2h

= × 12 × 12 × 6.75 cm3

Volume of iron ball = × 12 × 12 × 6.75 cm3

But, Volume of iron ball =4

r3

3

× 12 × 12 × 6.75 =4

3 × × r3

12 ×12 × 6.75 × 3

4= r3

r3 = 3 × 12 × 6.75 × 3 r3 = 3 × 3 × 3 × 4 × 6.75 r3 = 3 × 3 × 3 × 27 r = 3 × 3 × 3 × 3 × 3 × 33 [Taking cube roots] r = 3 × 3 r = 9

Radius of the iron ball is 9 cm.


22. A piece of cheese is cut in theshape of the sector of a circle ofradius 6 cm. The thickness of thecheese is 7 cm. Find(i) The curved surface area of the cheese.(ii) The volume of the cheese piece. (4 marks)

Sol. For a sector,Measure of arc () = 60ºRadius (r) = 6 cm

6.75 cm

20 cm

60º6 cm

7 c

m


SCHOOL SECTION326

(i) Curved surface area of the cheese = Length of arc × height

= 2 r h360

=60 22

2 6 7360 7

= 44 cm2

The curved surface area of the cheese is 44 cm2.

(ii) Volume of the cheese piece = A (sector) × height

= r h360

2

=60 22

6 6 7360 7

= 132 cm3

The volume of the cheese piece 132 cm3.


1. Which are polyhedrons from the following ?

Sol. (i) No (ii) Yes (iii) Yes (iv) Yes (v) No


2. Using Euler’s formula, find V, if E = 30, F = 12. If the solid figure is aprism, how many sides the base polygon has. (1 mark)

Sol. F + V = E + 2 12 + V= 30 + 2 V = 32 – 12 V = 20

Number of sides of base polygon = 1

202 = 10


3. Verify Euler’s formula for these solids :(i) (2 marks)Sol. F = 8, V = 12, E = 18

L.H.S. = F + V= 8 + 12

L.H.S. = 20 ......(i)R.H.S. = E + 2

= 18 + 2 R.H.S. = 20 ......(ii) L.H.S. = R.H.S. [From (i) and (ii)] F + V = E + 2

Diamond Test tubeTileUnsharpenedPencil

Nail

(i) (ii) (iii) (iv) (v)


SCHOOL SECTION 327

(ii) (2 marks)Sol. F = 8, V = 6, E = 12

L.H.S. = F + V= 8 + 6

L.H.S. = 14 ......(i)R.H.S. = E + 2


(iii) (2 marks)Sol. F = 8, V = 12, E = 18

L.H.S. = F + V= 8 + 12

L.H.S. = 20 ......(i)R.H.S. = E + 2


(iv) (2 marks)Sol. F = 6, V = 6, E = 10

L.H.S. = F + V= 6 + 6

L.H.S. = 12 ......(i)R.H.S. = E + 2



47. A sphere and a cube have the same surface area. Show that the ratio of

the volume of the sphere to that of the cube is 6 : . (4 marks)

Proof : Surface are of sphere = surface area of cube 4r2 = 6l2 ......(i)

r2

2l=

6

4

r2

2l=

3

2

r

l=

3

2 × ......(ii)

Volume of sphere

Volume of cube =4 r3 3

3l

=4 r

3

3

3

l


SCHOOL SECTION328

=4 r × r

3 ×

2

2l l

=4 r r

×3

2

2l l

=6 r

×3

2

2

l

l l[From (i)]

=r

2 ×l

=3

2 ×2 ×

=2 × 2 × 3

2 ×

Volume of sphere

Volume of cube =6

Ratio of the volume of the sphere to that of the cube is 6

48. Marbles of diameter 1.4 cm are dropped into a beaker containing somewater and are fully submerged. The diameter of the beaker is 7 cm. Findhow many marbles have been dropped in it if the water rises by 5.6 cm.

(5 marks)Sol. Diameter of marble = 1.4 cm

its radius (r) =1.4

2= 0.7 cm

Volume of a marble =4

r3 3

=4 7 7 7

× × × ×3 10 10 10

cm3

Marbles are submerged fully in the water, water level rises by 5.6 cm Height of water displaced (h) = 5.6 cm

Diameter of beaker = 7 cm

Its radius (r1) =7

2cm

Volume of water displaced = r h 21

=7 7 56

× × × cm2 2 10

3

Number of marbles =Volume of water displaced

Volume of marble

=7 7 56 4 7 7 7

×2 2 10 3 10 10 10

=7 7 56 3 1 10 10 10

× × × × × × × ×2 2 10 4 7 7 7

= 150 Number of marbles is 150.

49. Water flows at the rate of 10 m per minute through a cylindrical pipehaving its diameter is 20 mm. How much time will it take to fill aconical vessel of base diameter 40 cm and depth 24 cm ? (5 marks)

Sol. Diameter of conical vessel = 40 cm

Its radius (r) = 40

2 = 20cm


SCHOOL SECTION 329

Its depth (h) = 24 cm

Volume of conical vessel =1

r h3 2

=1

× × 20 × 20 × 243

= 20 × 20 × 8 × = 3200 cm3

Diameter of cylindrical pipe= 20 mm

Its radius (r1) =20

2= 10 mm

=10

10 cm [ 1 cm = 10 mm]

= 1 cmWater flowing in 1 minute (h) = 10 m

= 10 × 100 cm [ 1 m = 100 cm]= 1000 cm

Volume of water flowing in 1 minute through a cylindrical pipe= r h 2

1

= × 1 × 1 × 1000= 1000 cm3

Time taken to fill conical vessel =Volume of conical vessel

Volume of water flowing in 1 minute

=3200

1000

=32

10mins

=32

× 6010

secs [1 minute = 60 seconds]

= 192 seconds= 3 minutes and 12 seconds

The time taken to fill the conical vessel is 3 minutes and 12 seconds.

50. Find the length of 13.2 kg. copper wire of diameter 4 mm, when 1 cubiccm of copper weighs 8.4 gm. (4 marks)

Sol. Volume of 8.4 gm of copper = 1 cm3

Volume of 13.2 kg i.e. 13200 gm of copper = 13200

8.4

=13200 ×10

84

=11000

cm7

3

Diameter of copper wire = 4 mm

Its radius (r) =4

2= 2 mm

=2

10 cm


SCHOOL SECTION330

Volume of copper wire = r2h

11000

7=

22 2 2× × × h

7 10 10

11000 × 7 ×10 ×10

7 × 22 × 2 × 2 = h

h = 12500 cm h = 125 m [ 1 metre = 100 cm]

Length of wire is 125 m.

51. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4cm thick iron. Find the volume of the iron. (4 marks)

Sol. Girth of garden roller = 440 cmGirth of garden roller = 2 r

2r = 440

22

2 × × r7

= 440

r =440 × 7

2 × 22 r = 70 cm Radius of outer cylinder (r) = 70 cm

Width of the roller (h) = 63 cmThickness of the roller = 4 cm

Radius of inner cylinder (r1) = 70 – 4= 66 cm

Volume of iron= Volume of outer cylinder – Volume of inner cylinder

= r2h – r h 21

= h r – r 2 21

=22

× 63 × 70 – 667

2 2

= 198 (70 66) (70 – 66)= 198 × 136 × 4= 107712 cm3

The volume of the iron is 107712 cm3.

52. A semi-circular sheet of metal of diameter 28 cm is bent into an open

conical cup. Find the depth and capacity of cup. ( 3 = 1.73) (5 marks)

Sol. Diameter of semicircle sheet = 28 cm

Its radius (r) =28

2= 14 cm

A semicircular sheet is bent to form a open cone Slant height of a cone (l) = radius of a semicircular sheet l = 14 cm

Circumference of a base of a cone = length of semicircle= r

=22

×147

= 44 cm


SCHOOL SECTION 331

Let the radius of a cone be r1

Circumference of a base of a cone = 2r1

44 = 2r1

44 =22

2 × × r7 1

44 × 7

2 × 22 = r1

r1 = 7 cm

l2 = r h2 21

142 = 72 + h2

h2 = 142 – 72

h2 = 196 – 49 h2 = 147

h = 147

h = 49 × 3

h = 7 3 h = 7 × 1.73 h = 12.11 cm

Volume of conical cup =1

r h3 2

1

=1 22

× × 7 × 7 ×12.113 7

=1864.94

3= 621.646= 621.65 cm3

Depth of a conical cup is 12.11 cm and volume of conical cup is 621.65 cm3.

53. A cone and a hemisphere have equal bases and equal volumes. Find theratio of their heights. (3 marks)

Sol. A cone and a hemisphere have equal bases their radii are equal

Height of hemisphere = radius of hemisphereVolume of a cone = Volume of hemisphere

1

r h3 2

=2

r3 3

h = 2r

h

r=

2

1 Ratio of heights of a cone and a hemisphere is 2 : 1

54. A bucket is in the form of a frustum of a cone and holds 28.490 litres ofwater. The radii of the top and bottom are 28 cm and 21 cm respectively.Find the height of the bucket. (4 marks)

Sol. Radii of circular ends are 25 cm and 21 cm r1 = 28 cm and r2 = 21 cm

Volume of bucket = 28.490 litres= 28.490 × 1000 cm3 [ 1 litre = 1000 cm3]= 28490 cm3


SCHOOL SECTION332

Volume of bucket = 1r r r r × h

3 2 2

1 2 1 2

28490 = 1 22× 28 21 28 × 21 × h

3 7 2 2

28490 = 22784 441 588 × h

21

28490 =22

×1813 × h21

28490 × 21

22 ×1813 = h

h = 15 cm

The height of the bucket is 15 cm.

55. A oil funnel of tin sheet consists of a cylindrical portion 10 cm longattached to a frustum of a cone. If diameter of the top and bottom ofthe frustum is 18 cm and 8 cm respectively and the slant height of thefrustum of cone is 13 cm. Find the surface area of the tin required tomake the funnel. (Express your answer in terms of ) (4 marks)

Sol. Diameters of circular ends of frustum are 18 cm and 8 cm

r1 = 18

2 = 9 cm and r2 =

8

2 = 4 cm

Slant height (l) = 13 cmCurved surface area of frustum of frustum = (r1 + r2) l

= (9 + 4) × 13= × 13 × 13= 169 cm2

Radius of a cylinder (r2) = 4 cmIts height (h) = 10 cm

Curved surface area of a cylinder = 2r2h= 2 × × 4 × 10= 80 cm2

Surface area of tin required to make the funnel= Curved surface area of frustum + curved surface area of cylinder= 169 + 80= 249 cm2

The surface area of the tin required to make the funnel is 249 cm2.

56. There are 3 stair-steps as shown in the figure. Each stair-step has width25 cm, height 12 cm and length 50 cm. How many bricks have beenused in it if each brick is 12.5 cm × 6.25 cm × 4 cm. (5 marks)

Sol. Length of a stair-step (l) = 50 cmits breadth (b) = 25 cmits height (h) = 12 cm

Volume of a stair-step = l × b × h= 50 × 25 × 12= 15000 cm3

Volume of 3 stair-step = 6 × 15000= 90000 cm3

Length of a brick (l1) = 12.5 cmits breadth (b2) = 6.25 cmits height (h1) = 4 cm


SCHOOL SECTION 333

Volume of a brick = l1 × b1 × h1

= 12.5 × 6.25 × 4

= 312.5 cm3

Number of bricks required =Volume of 3 stair-steps

Volume of each brick

=90000

312.5

=6 × 50 × 25 ×12

12.5 × 6.25 × 4= 288

Number of bricks required is 288.

57. If V is the volume of a cuboid of dimensions a × b × c and S its surface

area, then prove that

1 2 1 1 1= + +

V S a b c . (3 marks)

Proof : L.H.S. =1

V

L.H.S. =1

abc.......(i)

R.H.S. =2 1 1 1

S a b c

=2 bc ac ab

2 (ab bc ac) abc

=1 (ab + bc + ac)

×ab + bc + ac abc

R.H.S. =1

abc......(ii)

L.H.S. = R.H.S. [From (i) and (ii)]

1V

=

2 1 1 1+ +

S a b c

MCQ’s1. An arc of a circle having measure 45º has length 25 cm. What is the

circumference of the circle ?(a) 200 cm (b) 100 cm(c) 50 cm (d) 45 cm

2. The measure of arc of circle is 90º. If the radius of the circle is 7 cm. Whatis the area of sector ?(a) 78.5 cm2 (b) 77 cm2

(c) 35.8 cm2 (d) 38.5 cm2

3. P is the centre of a circle, m (arc RYS) = 60º. If the radius of the circle is4.2 cm, what is the perimeter of P-RYS ?(a) 4.4 m (b) 8.4 m(c) 12.8 m (d) 17.2 m


SCHOOL SECTION334

4. The radius of a circle is 3.5 cm. What is the measure of arc whose length is5.5 cm ?(a) 30º (b) 45º(c) 60º (d) 90º

5. A cylinder with base radius 8 cm and height 2 cm is melted to form a coneof height 6 cm. What is the radius of the cone ?(a) 2 cm (b) 4 cm(c) 8 cm (d) 16 cm

6. Which of the following represents Euler’s formula ?(a) F – V = E + 2 (b) F + V = E + 2(c) F + E = V + 2 (d) F – V = E – 2

7. What is the curved surface area of a cone of height 15 cm and base radius8 cm ?(a) 60 cm2 (b) 68 cm2

(c) 120 cm2 (d) 136 cm2

8. How many solid metallic spheres each of diameter 6 cm are required to bemelted to form a solid metallic cylinder of height 45 cm and diameter 4 cm.(a) 3 (b) 4(c) 15 (d) 6

9. The radius and slant height of a cone are 5 cm and 10 cm respectively. Whatis its curved surface area ?(a) 314 cm2 (b) 157 cm2

(c) 78.5 cm2 (d) 100 cm2

10. The radii of the circular ends of a bucket which is 24 cm high are 14 cmand 7 cm respectively. What will be its slant height ?(a) 12 cm (b) 21 cm(c) 45 cm (d) 25 cm

11. The diameter of a sphere 6 cm is melted and drawn into a wire of diameter2 mm. What will be the length of the wire ?(a) 12 m (b) 18 m(c) 24 m (d) 36 m

12. Two cubes each with 12 m edge are joined end to end. What is the differencein surface area of the resulting cuboid and the surface area of two cubes ?(a) 288 m2 (b) 144 m2

(c) 1440 m2 (d) 770 m2

13. Area of a sector with central angle 60º will be ............ of area of a circle.

(a)2

3

rd

(b)1

6

th

(c)1

2(d)

1

4

th

14. If area of semicircle is 77 cm2 its perimeter is ................. .(a) 72 cm (b) 120 cm(c) 36 cm (d) 40 cm


SCHOOL SECTION 335

15. Area of minor segment AXB is ................ cm2.(a) 38.5 cm2 (b) 24.5 cm2

(c) 14.0 cm2 (d) 154 cm2

16. The capacity of a bowl is 144cm3. Find the radius.(a) 8 cm (b) 4 cm(c) 7 cm (d) 6 cm

17. A solid metallic ball of radius 14 cm is melted and recasted into small ballsof radius 2 cm. Find how many such balls can be made ?(a) 434 (b) 343(c) 433 (d) 344

18. Find the capacity of swimming pool of length 20 m breadth 5 m and depth 4 m ?(a) 40000 l (b) 400000 l(c) 4000 l (d) 4000000 l

19. A cube of side 40 cm is divided into 8 equal cubes. Then its surface areawill increase .................. times.(a) 4 (b) 8(c) 2 (d) 5

20. Find the number of coins 2.4 cm in diameter and 2 mm thick to be meltedto form a right circular cylinder of height 12 cm and diameter 6 cm ?(a) 350 (b) 370(c) 400 (d) 375

21. The curved surface area of a right cone is double that of another right cone.If the ratio of their slant heights is 1 : 2, find the ratio of their radii ?(a) 1 : 4 (b) 2 : 3(c) 3 : 2 (d) 4 : 1

22. The area swept out by a horse tied in a rectangular grass field with a rope8 m long is ............... .(a) 16 cm2 (b) 64 cm2

(c) 48 cm2 (d) 32 cm2

23. The angle swept by the minute hand of a clock of length 9 cm in 15 mins is................. .(a) 90 (b) 45(c) 30 (d) 60

24. A (sector) = 1

12 A (circle). Hence, measure of the corresponding central

angle will be ................ .(a) 30 (b) 45(c) 60 (d) 90

25. A cylinder and a cone have equal radii and equal heights. If the volume ofthe cylinder is 300 cm3, then what is the volume of the cone ?(a) 100 cm3 (b) 10 cm3

(c) 110 cm3 (d) 300 cm3

O 7 B

x

A


SCHOOL SECTION336

: ANSWERS :1. (a) 200 cm 2. (d) 38.5 cm2

3. (c) 12.8 m 4. (d) 90º

5. (c) 8 cm 6. (b) F + V = E + 2

7. (d) 136 cm2 8. (c) 15

9. (b) 157 cm2 10. (d) 25 cm

11. (d) 36 m 12. (a) 288 m2

13. (b)1

6

th

14. (c) 36 cm

15. (c) 14.0 cm2 16. (d) 6 cm

17. (b) 343 18. (b) 400000 l

19. (c) 2 20. (d) 375

21. (d) 4 : 1 22. (a) 16 cm2

23. (a) 90 24. (a) 30

25. (a) 100 cm2

1 Mark Sums1. Find the length of the arc when the corresponding central angle is 270º

and circumference is 31.4 cm.Sol. Measure of central angle () = 270º

Circumfernce (2r) = 31.4 cm

Length of the arc = 2 r360

l =270

× 31.4360

l =3 31.4

4

l =94.2

4

l = 23.55 cm

2. If length of an arc is 7 cm, 2r = 36, then find the angle subtended atthe centre by the arc.

Sol. Length of the arc (l) = 7 cm2r = 36

l = 2 r360

7 = 36360

=7 360

36

= 7 × 10

= 70º

The angle subtended at the centre by the arc is 70º.


SCHOOL SECTION 337

3. Find the area of a circle with radius 7 cm.Sol. Radius of circle (r) = 7 cm

Area of the circle = r2

=22

7 77

= 154 cm2

The area of a circle is 154 cm2.

4. Using Euler’s formula, write the value of V, if E = 30 and F = 12.Sol. F + V = E + 2

12 + V = 30 + 2 12 + V = 32 V = 32 – 12

V = 20

5. The area of a circle is 314 cm2 and the area of its minor sector is 31.4 cm2.Find the area of its major sector.

Sol. Area of major sector = Area of circle – Area of minor sector= 314 – 3.14= 282.6 cm2

The area of the major sector is 282.6 cm2.

6. Perimeter of one face of a cube is 24 cm. Find the length of its side.Sol. Perimeter of one face of a cube = 24 cm

Perimeter of one face of a cube = 4l 24 = 4l

l = 24

4 l = 6

The length of the side of a cube is 6 cm.

7. A cubical tank has each side of length 2 m. Find the capacity of the tankin cubic metres.

Sol. Side of a cubical tank (l) = 2 m

Volume of cubical tank = l3

= 23

= 8 m3

Capacity of the cubical tank is 8 m3.

8. If the radius is 2 cm and length of corresponding arc is 3.14 cm, find thearea of a sector.

Sol. Radius (r) = 2 cmLength of arc (l) = 3.14 cm

Area of sector = r

2l

= 3.14 × 2

2= 3.14 cm2

The area of a sector is 3.14 cm2.


SCHOOL SECTION338

9. The dimensions of a cuboid are 5 cm, 4 cm and 3 cm. Find its volume.Sol. Length of a cuboid (l) = 5 cm

Its breadth (b) = 4 cmIts height (h) = 3 cm

Volume of a cuboid = l × b × h= 5 × 4 × 3= 60 cm3

Volume of cuboid is 60 cm3.

10. Find the total surface area of a cube with side 1 cm.Sol. Length of side of cube (i) = 1 cm


= 6 (l)2

= 6 (1)= 6 cm2

The total surface area of cube is 6 cm2.

11. Volume of a cube is 1000 cm3, find the length of its side.Sol. Volume of a cube = 1000 cm3

Volume of a cube = l3

l3 = 100 l = 10 [Taking cube roots]

Length of the side of cube is 10 cm.

12. What is the volume of a cube with side 5 cm ?Sol. Side of a cube (l) = 5 cm

Volume of cube = l3

= (5)3

= 125 cm3

Volume of the cube is 125 cm3.

13. The area of a circle with radius R is equal to the sum of the areas ofcircles with radii 6 cm and 8 cm. What is the value of R ?

Sol. According to given condition,R2 = × (6)2 + × (8)2

R2 = [62 + 82] R2 = 36 + 64 R2 = 100 R = 10 [Taking square roots]

The value of R is 10.

14. The radius of the base of a cone is 7 cm and its height is 24 cm. What isits slant height ?

Sol. Radius of base of cone (r) = 7 cmIts height (h) = 24 cm

l2 = r2 + h2

l2 = 72 + 242

l2 = 49 + 576 l2 = 625 l = 25 [Taking square roots]

Slant height of cone is 25 cm.


SCHOOL SECTION 339

15. Using Euler’s formula, find F, if V = 6 and E = 12.Sol. F + V = E + 2

F + 6 = 12 + 2 F + 6 = 14 F = 14 – 6

F = 8

16. The area of a circle is 1368 cm2. What is the area of the sector of thecircle whose corresponding central angle is 120º ?

Sol. Area of circle = 1368 cm2

Area of sector of the circle whose corresponding central is 120º

=1

13683

= 456 cm2

Area of the sector is 456 cm2.

17. The corresponding central angle of an arc is 90º. What is the length ofthis arc, if the radius of the circle is 14 cm ?

Sol. Measure of central angle () = 90ºRadius (r) = 14 cm

Length of the arc (l) = 2 r360

l =90 22

2 14360 7

l = 22

Length of the arc is 22 cm.

18. Radius of a circle is 10 cm. The length of an arc of this circle is 25 cm.What is the area of the sector ?

Sol. Radius of circle (r) = 10 cmLength of arc (l) = 25 cm

Area of sector =r

2l

= 25 × 10

2= 25 × 2= 125 cm2

The area of the sector is 125 cm2.

19. A cylinder and a cone have equal radii and equal heights ? If the volumeof the cylinder is 300 cm3, what is the volume of the cone ?

Sol. A cylinder and cone have equal height and equal radii

Volume of cone =1

volume of cylinder3

=1

3003

= 100 cm3

Volume of the cone is 100 cm3.

20. What is the corresponding angle of a sector whose area is one-forth ofthe area of the circle ?

Sol. The corresponding angle of a sector whose area is one-forth of the areaof the circle is 90º.


SCHOOL SECTION340

21. The area of a sector with corresponding angle 45º is 8cm2. What is thearea of the circle ? ( = 3.14)

Sol. Area of sector = 8cm2

Measure of the arc ()= 45º

Area of sector = r360

2

8 = 45

r360

2

r2 = 8 × 360

45

r2 = 360

8 3.1445

r2 = 3.14 × 64 r2 = 200.96

Area of the circle is 200.96 cm2.

22. The dimensions of a cuboid are 3 cm × 9 cm × x cm. The volume of this cuboidis equal to the volume of a cube with side 6 cm. What is the value of x ?

Sol. Volume of cuboid = Volume of cube [Given] 3 × 9 × x = (6)3

3 × 9 × x = 6 × 6 × 6

x =6 6 6

3 9

x = 8

23. The area of a circle with radius 17 cm is equal to the sum of the areas ofcircles with radii r cm and 15 cm respectively. What is the value of r ?

Sol. According to given condition, (17)2 = r2 + (15)2

(17)2 = (r2 + 152) 172 = r2 + 152

r2 = 172 – 152

r2 = 289 – 225 r2 = 64

r = 8 [Taking square roots]

24. The radius of the base of a cone is 7 cm and its height is 24 cm. What isits slant height ?

Sol. Radius of base of cone (r) = 7 cmits height (h) = 24 cm

l2 = r2 + h2

l2 = 72 + 212

l2 = 49 + 576 l2 = 625 l = 25 [Taking square roots]

Slant height of cone is 25 cm

25. What is the corresponding central angle of a sector whose area is one-tenth the area of the circle ?

Sol. The corresponding central angle of a sector whose area is one tenththe area of the circle is 36º.


SCHOOL SECTION 341

CHAPTER : 1 - SIMILARITY

1. Bisectors of B and C in ABC meet each other at P. Line AP cuts the

side BC at Q. Then prove that : APPQ =

AB + ACBC

(5 marks)

Proof : In ABQ,ray BP bisects ABQ [Given]

AP

PQ = AB

BQ ........(i)By property of an angle

bisector of a triangle

In ACQ,ray CP bisects ACQ [Given]

AP

PQ =AC

CQ ........(ii)By property of an angle

bisector of a triangle

AP

PQ =AB

BQ = AC

CQ [From (i) and (ii)]

AP

PQ =AB + AC

BQ + CQ [By Theorem on equal ratios]

APPQ =

AB + ACBC

[ B - Q - C]

2. In PQR, PQR = 900, As shownin figure, seg QS side PR.seg QM is angle bisector of PQR.

Prove that :PM²MR²

= PSSR

(5 marks)

Proof : In PQR,seg QM bisects PQR [Given]

PM

MR=

PQ

QR

[Property of an angle

bisec tor of a triangle]

PM

MR

2

2 = PQ

QR

²² .........(i) [Squaring both sides]

In PQR,m PQR = 90º [Given]seg QS hypotenuse PR [Given]

PQR ~ PSQ ~ QSR .........(ii) [Theorem on similarity ofright angled triangles]

PSQ ~ PQR [From (ii)]

PQ

QR =PS

PQ

[Corresponding sides

of similar triangles] PQ² = PR × PS .......(iii)

Also, QSR ~ PQR [From (ii)]

QR

PR=

SR

QR

[Corresponding sides

of similar triangles] QR² = PR × SR ........(iv)

PM

MR

2

2 =PR × PS

PR × SR [From (i), (iii) and (iv)]

2

2

PMMR

= PSSR

•

Q

xB C

A

P

• x

R

S

P

M

Q

EXTRA HOTS SUMS


SCHOOL SECTION342

3. Find the radius of a circle drawn by a compass when angle between twoarms of compass is 1200 and length of each arm is 24cm. (5 marks)

Sol.

In the adjoining figure,seg AB and seg BC represents thearms of compass.In ABC,side AB side BC [Given]

BAC BCA ........(i) [Isosceles triangle theorem]In ABC,m ABC + m BAC + m BCA = 180º [ Sum of the measures of the

angles of a triangles is 180º] 120º + m BAC + m BAC = 180º [From (i)] 2 m BAC = 180º – 120º 2 m BAC = 60º m BAC = 30º .......(ii)

Draw seg BD side AC, A - D - C.In ABD,m BAD = 30º [From (ii) and A - D - C]m ADB = 90º [Given]

m ABD = 60º [Remaining angle] ABD is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem.

AD =3

2AB [Side opposite to 60º]

AD =3

2× 24

AD = 12 3 cmSimilarly, we can getDC = 12 3 cm

AC = AD + DC [ A - D - C]

AC = 12 3 + 12 3 AC = 24 3 cm.

The radius of the circle is 24 3 cm.

4. In BAC, BAC = 90º, segments AD, seg BE and seg CF are medians.Prove : 2 (AD² + BE² + CF² ) = 3BC². (5 marks)

Proof :

In BAC,seg AD is median on side BC. [Given]

AB² + AC² = 2AD² + 2BD² [By Appollonius theorem] AB² + AC² – 2BD² = 2AD²

AB² + AC² – 21

BC2

2

= 2AD² [ D is the midpoint of seg BC]

AB² + AC² – 1

24

BC2 = 2AD²

AB² + AC² – 1

2 BC² = 2AD² .........(i)

B

AD

C

24 cm

24 cm 1200

A

B

C

F D

E


SCHOOL SECTION 343

Multiplying throughout by 2, we get 2AB2 + 2AC2 – BC2 = 4AD2

Similarly, we can prove that2AB² + 2BC² – AC² = 4BE² .........(ii)2AC² + 2BC² – AB² = 4CF² ........(iii)Adding (i), (ii) and (iii) we get2AB2 + 2AC2 – BC2 + 2AB² + 2BC² – AC² + 2AC² + 2BC² – AB² = 4AD2 +4BE2 + 4CF2

3AB2 + 3AC2 + 3BC2 = 4AD2 + 4BE2 + 4CF2

3 (AB2 + AC2 + BC2) = 4 (AD2 + BE2 + CF2) ........(iv)In BAC,m BAC = 90º [Given]

BC2 = AB2 + AC2 ........(v) [By Pythagoras theorem]3 (BC2 + BC2) = 4 (AD2 + BE2 + CF2) [From (iv) and (v)]

3 × 2 BC2 = 4 (AD2 + BE2 + CF2) 6BC2 = 4 (AD2 + BE2 + CF2) 3BC2 = 2 (AD2 + BE2 + CF2) [Dividing throughout by 2] 2 (AD² + BE² + CF²) = 3BC²

5. In ABCD points P, Q, R and S lies on sides AB, BC, CD and AD respectivelysuch that seg PS || seg BD || seg QR and seg PQ || seg SR. Then provethat seg PQ || seg AC. (5 marks)

Proof : seg PS || seg BD [Given] On transversal AB,

APS ABD .......(i) [Converse of corresponding angles test]In APS and ABD,

PAS BAD [Common angles]APS ABD [From (i)]

APS ABD [By A-A test of similarity]

AP

AB =

PS

BD =

AS

AD......(ii) [c.s.s.t.]

seg QR || seg BD [Given]On transversal BC,CQR CBD .......(iii) [Converse of corresponding angles test]In CQR and CBD,QCR BCD [Common angles]CQR CBD [From (iii)]

CQR CBD [By AA test of similarity]

CQ

CB =

QR

BD =

CR

CD.......(iv) [c.s.s.t.]

In PQRS,seg PQ || seg RS [Given]seg PS || seg QR [Given]

PQRS is a parallelogram [By definition] PS = QR ........(v)

CQ

CB =

PS

BD =

CR

CD......(vi) [From (iv) and (v)]

AP

AB =

CQ

CB[From (ii) and (vi)

AB

AP =

CB

CQ [By Invertendo]

S D

R

CQB

P

A


SCHOOL SECTION344

AB – AP

AP =

CB – CQ

CQ [By Dividendo]

BP

AP =

BQ

CQ ......(vii) [ A - P - B and B - Q - C]

In ABC,BP

AP =

BQ

CQ [From (vii)]

seg PQ || seg AC [By converse of B.P.T.]

6. In ABC, m BAC = 90º.seg DE side AB, seg DF side AC,

prove A (AEDF) = AE × EB × AF × FC (5 marks)Proof : In ADB,

m ADB = 90º [Given]seg DC side AB [Given]

DE2 = AE × EB ......(i) [By property of geometric mean]In ADC,m ADC = 90º [Given]seg DF side AC [Given]

DF2 = AF × FC .......(ii) [By property of geometric mean]Multiplying (i) and (ii),DE2 × DF2 = AE × EB × AF × FC

DE × DF = AE × EB × AF × FC ......(iii)In AEDF,m EAF = m AED = m AFD = 90º [Given]

m EDF = 90º [Remaining angle] AEDF is a rectangle [By definition]

A (AEDF) = DE ×DF......(iv)From (iii) and (iv),

A (AEDF) = AE × EB × AF × FC [From (iii) and (iv)]

CHAPTER : 2 - CIRCLE1. From the end points of a diameter of circle perpendiculars are drawn to

a tangent of the same circle. Show that their feet on the tangent areequidistant from the centre of the circle. (5 marks)

Given : (i) A circle with centre O.(ii) seg AB is the diameter of the circle.(iii) Line l is tangent to the circle at point C.(iv) seg AD line l .(v) seg BE line l .

To Prove : OD = OE.Construction : Draw seg OC.Proof : seg AD line l [Given]

seg OC line l [Radius is perpendicular to the tangent]

seg BE line l [Given] seg AD || seg OC || seg BE [Perpendiculars drawn to the same

line are parallel to each other] On transversal AB and DE,

AO

OB=

DC

CE.........(i) [By property of intercepts made by

three parallel lines]

A

E

F

B D C

AD

O

B

C

E

l


SCHOOL SECTION 345

But, AO = OB [Radii of the same circle]

AO

OB= 1 .........(ii)

DC

CE= 1 [From (i) and (ii)]

DC = CE .........(iii)In OCD and OCE ,seg OC seg OC [Common side]OCD OCE [Each is a right angle]seg DC seg CE [From (iii)]

OCD OCE [By SAS test of congruence] seg OD seg OE [c.s.c.t.] OD = OE

2. The bisectors of the angles A,B of ABCintersect in I, the bisectors of the correspondingexterior angles intersect in E.Prove that AIBE is cyclic. (5 marks)

Proof :

Take points P and Q as shownin the figure.m CAB + m BAP = 180º [Linear pair axiom]

1

2m CAB +

1

2 m BAP =

1

2 × 180º[Multiplying throughout by

1

2]

m IAB + m BAE = 90º [ Ray AI and ray AE bisects CABand BAP respectively]

m IAE = 90º ......(i) [Angle addition property]Similarly,m IBE = 90º .....(ii)m IAE + m IBE = 90º + 90º [Adding (i) and (ii)]

m IAE + m IBE = 180º AIBE is cyclic [If opposite angles of a quadrilateral

are supplementary then quadrilateral is cyclic]

3. In the adjoining figure,line AP is a tangent to a circlewith centre O at point A.seg AF is angle bisector of BAC.Prove that : seg AP seg PE. (5 marks)

Proof :

BAE CAE [Ray AE bisects BAC]Let, m BAE = m CAE = x .........(i)PAC PAC [Angles in alternate segment]Let, mPAC = mABC = y .........(ii)m PAE = m PAC + m CAE [Angle addition property]

m PAE = (y + x) ........(iii) [From (i) and (ii)]PEA is a exterior angle of ABE,

m PEA = m ABE + m BAE [Remote interior angle theorem] m PEA = m ABC + m BAE [ B - E - C] m PEA = (y + x) ........(iv) [From (i) and (ii)]

In PAE, PAE PEA [From (iii) and (iv)]

seg AP seg PE [Converse of isosceles triangle theorem]

I

E

AQP

•

C

B•

××

))

A

F

C

P0

BE

O

0


SCHOOL SECTION346

4. In the adjoining figure,AB is diameter of a circle with centre O,seg AC is tangent to the circle at point A.line JD touches circle at point D,and intersects segment AC in point J.Prove that : seg AJ seg CJ. (5 marks)

Proof :Take a point M on line DJ such that M - D - J.seg AJ seg DJ ........(i) [The lengths of two tangent segments

from an external point to a circleare equal]

m ODM = 900 .......(ii) [Radius is perpendicular to the tangent]In OBD,seg OB seg OD [Radii of the same circle]

OBD ODB [Isosceles triangle theorem]Let, m OBD = m ODB = xº ......(iii)

m ODM = m ODB + m BDM [Angle addition property] 90 = x + m BDM [From (ii) and (iii)] mBDM = (90 – x)º ......(iv)But, BDM JDC .......(v) [Vertically opposite angles] m JDC = (90 – x)º ......(vi) [From (iv) and (v)]

In BAC,m BAC = 90º [Radius is perpendicular to tangent]m ABC = xº [From (iii) and A - O - B, B - D - C]

m ACB = (90 – x)º [Remaining angle] m JCD = (90 – x)º .......(vii) [B - D - C and A - J - C]In JDC,JCD JDC [From (vi) and (vii)]

seg DJ seg CJ ......(viii) [Converse of Isosceles triangle theorem] seg AJ seg CJ [From (i) and (viii)]

5. If four tangents of a circle determine a rectangle then show that itmust be a square. (5 marks)

Given :(i) Lines AB, BC, CD and AD

are the tangents to the circle at points P, Q, R and S respectively(ii) ABCD is a rectangle.

To Prove : ABCD is a squareProof : AP = AS ..........(i)

BP = BQ .........(ii)CR = CQ .........(iii)DR = DS .........(iv)Adding (i), (ii), (iii) and (iv), we get AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC ..........(v) [A - P - B, B - Q - C, C - R - D,

A - S - D]ABCD is a rectangle [Given]

AB = CD ..........(vi)[Opposite sides of a rectangle arecongruent] AD = BC .........(vii)

B

A J C

DO

M•

[The lengths of the twotangent segments from anexternal point to a circleare equal]

A B

R C

P

Q

D

S


SCHOOL SECTION 347

AB + AB = BC + BC [From (v), (vi) and (vii)] 2 AB = 2 BC AB = BC ABCD is a square [A rectangle in which adjacent sides

are congruent, is a square]

6. Two concentric circles with centre O. Seg AB, seg BC and seg AC arethe tangents to the smaller circle at points P, Q and R respectively andalso they are chords of the bigger circle.

Prove that seg PQ || seg AC , PQ = 1

AC2

. (5 marks)

Proof :

With respect to smaller circle,seg OP seg AB .......(i) [Radius is perpendicular to theseg OQ seg BC .......(ii) tangent]With respect to bigger circle,seg OP chord AB [From (i)]

AP = BP .....(iii) [Perpendicular drawn from centreof circle to chord bisects the chord]

seg OQ chord BC [From (ii) BQ = QC .....(iv) Perpendicular drawn from centre

of circle to chord bisects the chord]In ABC,P and Q are midpoints of sides [From (iii) and (iv)]AB and BC respectively.

seg PQ || seg AC[By Midpoint theorem]

PQ = 1

AC2

CHAPTER : 3 - GEOMETRIC CONSTRUTION1. Point I is the incentre of ABC, BIC = 120º, BC = 4 cm, median AP = 3 cm.

Draw ABC. (5 marks)Analysis :Let m ABI = m IBC = x and m ACI = m ICB = y

m ABC = 2x and m ACB = 2y [Angle addition property]In BIC,

m BIC + m IBC + m ICB = 180º 120 + x + y = 180 x + y = 180 – 120 x + y = 60 .......(i)

In ABC,m BAC + m ABC + m ACB = 180

m BAC + 2x + 2y = 180 m BAC + 2 (x + y) = 180 m BAC + 2 (60) = 180 [From (i)] m BAC + 120 = 180 m BAC = 180 – 120 m BAC = 60º

Now, ABC can be constructed with base BC, vertical angle BAC andmedian AP.

A

P O

Q

C

B

R


SCHOOL SECTION348

2. Point O is the orthocentre of ABC, m BOC = 105º, seg AD seg BC.BC = 6.5 cm and AD = 3.5 cm. Draw ABC. (5 marks)

Analysis : BOC EOF [Vertically oppsoite angles] m BOC = m EOF = 105º

In AFOE,m FAE + m AFO + m EOF + m AEO = 360º

[Angle sum property of a quadrilateral] m FAE + 90 + 105 + 90 = 360 m FAE + 285 = 360 m FAE = 360 – 285 m FAE = 75º m BAC = 75º [A - F - B, A - E - C]

Now,ABC can be constructed with base BC,vertical angle BAC and altitude AD.

A

B C

A

O60º

120º30º 30º

3 cm

4 cmP

A

B CP

I

120º

•• ××

(Rough Figure)

A

75º

3.5 cm

150º15º 15º

B C

A

P

Y

X

3.5 cm

6.5 cm

A

EO

F

B CD

6.5 cm

105º

(Rough Figure)


SCHOOL SECTION 349

A

7 cm

X

B50º

O

I7 cm

(Rough Figure)

3. Construct ABC such that BC = 8.8 cm, B = 50º, radius of incircle ofABC is 2.2 cm. (5 marks)

Steps of construction :1. Draw seg BC = 8.8 cm.2. At B, draw m ABC = 50º.3. Draw bisector of B, as incentre lies on angle bisector.4. Draw a line parallel to side BC at a distance of 2.2 cm from BC.5. Point of intersection of line parallel to BC and angle bisector is incentre.

Let incentre be I.6. From I, draw seg IM side BC. seg IM is in-radius.7. Draw incircle with IM as radius to touch sides AB and BC.8. From C draw the tangent to the circle to meet ray BA at A.

4. Draw a sector O-AXB with radius 7 cm and m (arc AXB) = 50º. Draw acircle touching the sides OA and OB and also the arc. (5 marks)

I

2.2 cm50º

8.8 cm

CB

A(Rough Figure)

MA

QP

B C

5 cm4.8 cm

O5.8 cm


SCHOOL SECTION350

Steps of construction :1. Draw m O = 50º and arc AB of radius 7 cm.2. Draw bisector of O. It intersects arc AB at X.3. At X, draw the PQ ray OX to cut ray OA at P and ray OB at Q.4. Draw bisector of Q. It intersects ray OX at I.5. Draw incircle with I as centre and IX as radius.

This circle touches the ray OA, ray OB and also arc AXB.

5. In ABC, BC = 5.8 cm, seg BP seg AC, seg CQ seg AB, BP = 5 cm ,CQ = 4.8 cm. Construct ABC. (5 marks)

I

AP

X

50ºO B Q

•• ×

×

7 cm

I

50ºB C X

Y

2.2 cm2.2 cm

A

••

8.8 cmM

A

QP

B C5.8 cm

5 cm4.8 cm

(Rough Figure)


SCHOOL SECTION 351

Steps of construction :1. Draw seg BC of length 5.8 cm.2. Draw a semicircle with seg BC as the diameter.3. Taking B as the centre and radius 5 cm cut an arc on the semicircle to get

point P and draw seg BP.4. Taking C as the centre and radius 4.8 cm cut an arc on the semicircle to get

point Q and draw seg CQ.5. Extend seg BQ and seg CP to intersect at point A.

ABC is the required triangle.

6. Draw a line l. Take a point P at a distance 5cm from line l. Draw a circlewith radius 3cm such that the circle touches the line l and passesthrough point P. (5 marks)

Steps of construction :1. Draw line l.2. Take a point M on line l and draw a perpendicular to line l at point M.3. With point M as the centre, cut an arc of radius 5 cm on the perpendicular

to get point P.4. With point M as the centre and radius 3 cm cut an arc on seg PM to get point N.5. Draw a line m perpendicular to line PM at point N.6. With point P as the centre cut an arc of radius 3 cm on line m to get point O.7. With point O as the centre and seg OP as the radius, draw the required circle.8. Draw a perpendicular from point O to line l to get point T.

CHAPTER : 4 - TRIGONOMETRY

1. If 21 + x sin = x, prove that tan2 + cot2 = x2 + 2

1x

. (5 marks)

Proof : 1 + x2 sin = x

sin =x

1 + x2

sin2 =x

1 + x

2

2 [Squaring both sides]

(Rough Figure)

P

N O

M T

3 cm

3 cm

5 c

m

P

N O

lM T

3 cm

3 cm

5 c

m


SCHOOL SECTION352

sin2 + cos2 = 1 cos2 = 1 – sin2

cos2 = 1 – x

1 + x

2

2

cos2 =1 x – x

1 + x

2 2

2

cos2 =1

1 + x2

tan2 =sin

cos

2

2

=x 1

1 x 1 x

2

2 2

=x 1 x

11 x

2 2

2

= x2

cot2 =1

tan2

=1

x2

L.H.S. = tan2 + cot2

= x2 + 1

x2

= R.H.S.

tan2 + cot2 = x2 + 2

1x

2. Prove : tan cot

+ = 1 + tan + cot1 – cot 1 – tan

(5 marks)

Proof : L.H.S. =tan cot

+1 – cot 1 – tan

=sin cos cos sin

1 – 1 –cos sin sin cos

=sin sin – cos cos cos – sin

cos sin sin cos

=sin sin cos cos

cos sin – cos sin cos – sin

=sin cos

cos (sin – cos ) sin (cos – sin )

2 2

=sin cos

–cos (sin – cos ) sin (sin – cos )

2 2

=1 sin cos

–(sin – cos ) cos sin

2 2

=1 sin – cos

(sin – cos ) cos × sin

3 3

=1 (sin – cos ) (sin sin . cos cos )

(sin – cos ) cos sin

2 2


SCHOOL SECTION 353

B

A E C

DF30º60º

G

28.5 m

1.5 m 1.5 m1.5 m

x

=sin + sin . cos + cos

cos . sin

2 2

=sin sin . cos cos

cos . sin cos . sin cos . sin

2 2

=sin cos

1cos sin

= tan + 1 + cot = 1 + tan + cot = R.H.S.

tan cot

+ = 1 + tan + cot1 – cot 1 – tan

3. Prove : sin8 – cos8 = (sin2 – cos2 ) (1 – 2 sin2 cos2 ). (5 marks)Proof : L.H.S. = sin8 – cos8

= (sin4 )2 – (cos4 )2

= (sin4 – cos4 ) (sin4 + cos4 )= (sin2 – cos2 ) (sin2 + cos2 ) (sin4 + cos4 )= (sin2 – cos2 ) (sin4 + cos4 ) [ sin2 + cos2 = 1]= (sin2 – cos2 ) (sin4 + cos4 + 2sin2 cos2 – 2sin2 cos2 )= (sin2 – cos2 ) [(sin2 + cos2 )2 – 2 sin2 cos2 ]= (sin2 – cos2 ) (1 – 2sin2 cos2 ) [ sin2 + cos2 = 1]= R.H.S.

sin8 – cos8 = (sin2 – cos2 ) (1 – 2 sin2 cos2 ).

4. A 1.5 m tall boy is standing at some distance from a 30 m tall building.The angle of elevation from his eyes to the top of building increasesfrom 30º to 60º as he walks towards the building. Find the distance hewalked towards the building. (5 marks)

Sol. Let the distance he walkedtowards the building x mHeight of tower (AB) = 30 mHeight of boy (CD) = 1.5 mBut CD = EF = AGAG = 1.5 mBG = AB – AGBG = 30 – 1.5BG = 28.5 mIn right angled BGF,

tan 60º =BG

GF[By definition]

3 =28.5

GF

GF =28.5

3

GF =28.5 3

3 3

GF =28.5 3

3

GF = 9.5 3 m


SCHOOL SECTION354

In right angled BGD,

tan 30º =BG

GD[By definition]

1

3=

28.5

9.5 3 x

9.5 3 x = 28.5 3

x = 28.5 3 – 9.5 3

x = 19 3

Distance he walked towards the building is 19 3 m.

5. Prove : (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A. (5 marks)Proof : L.H.S = (sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + 2sin A . cosec A + cosec2 A + cos2A + 2cos A . sec A +sec2 A

= (sin2 A + cos2 A) + (cosec2 A) + (sec2 A) + 2 sin A . cosec A +2 cos A . sec A

= 1 + (1 + cot2 A) + (1 + tan2 A) + 2 × sin A 1

sin A

+ 2cos A 1

cos A

= 1 + 1 + cot2 A + 1 + tan2 A + 2 + 2= 7 + tan2 A + cot2 A= R.H.S.

(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.

6. Prove that : . 2 2

3 3

1 – sin cos sin – cos× = sin

cos (sec – cosec ) sin + cos

(5 marks)

Proof : L.H.S. =1 – sin . cos sin – cos

×cos (sec – cosec ) sin + cos

2 2

3 3

=

1 – sin . cos (sin cos ) (sin – cos )×

(sin cos ) (sin – sin . cos cos )1 1cos –

cos sin

2 2

=

(1 – sin . cos ) (sin – cos )×

(1 – sin . cos )sin – coscos

sin . cos

[ sin2 + cos2 = 1]

=

1× sin – cos

sin – cossin

=sin

sin –cossin – cos

= sin = R.H.S.

. 2 2

3 3

1 – sin cos sin – cos× = sin

cos (sec – cosec ) sin + cos

7.tan A

sec A -1 + tan A

sec A +1 = 2 cosec A (5 marks)

Proof : L.H.S. =tan A

sec A -1 +tan A

sec A +1


SCHOOL SECTION 355

A

D300 450

80 m

E

B

300

450

C

=

sinA

cosA1

-1cos A

+

sinA

cosA1

+1cos A

=

sinA

cosA(1 cosA )

cos A

– +

sinA

cosA(1+ cosA )

cos A

=sinA

1 – cosA + sinA

1+ cosA

= sin A 1 1

+1 – cos A 1+ cos A

= sin A 1+ cosA + 1– cosA

(1–cosA) (1+cosA)

= sin A 2

1 – cos A

²

= sin A 2

sin A

²

sin A + cos A =1

1- cos A = sin A

² ²

² ²

=2

sinA= 2 cosec A= R.H.S.

tan A

sec A -1 + tan A

sec A +1 = 2 cosec A

6. From the top of a light house, 80 metres high, two ships on same side oflight house are observed . The angles of depression of the ships as seenfrom the lighthouse are found to be of 450 and 300. Find the distancebetween the two ships (Assume that the two ships and the bottom ofthe lighthouse are in a line). (5 marks)

Sol. In the adjoining figure,seg AB represents the lighthouse.A is the position of the observerD and C are the position of the ships.Draw ray AE || seg BD.EAD and EAC are theangles of depression.

m EAD = 30º and m EAC = 45ºOn transversal ADm EAD = m ADB = 30º [Converse of alternate angles test]On transversal ACm EAC = m ACB = 45º [Converse of alternate angles test]In right angled ABD,

tan 30 =AB

DB[By definition]

1

3=

80

DB DB = 80 3 m ......(i)


SCHOOL SECTION356


tan 45 =AB

CB[By definition]

1 =80

CB.......(ii)

CB = 80 mBD = BC + CD [ D - C - B]

80 3 = 80 + CD [From (i) and (ii)]

CD = 80 3 – 80

CD = 80 3 – 1 m

The distance between the two ships is 80 3 – 1 m.

CHAPTER : 5 - CO-ORDINATE GEOMETRY1. A (8, 5), B (9, – 7), C (– 4, 2) and D (2, 6) are the vertices of a ABCD. If P,

Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.Show that PQRS is a parallelogram, using the slopes.

Sol. P, Q, R and S are the midpoints of side AB, BC, CD and AD of ABCD.A (8, 5), B (9, – 7), C (– 4, 2) and D (2, 6)By midpoint formula,

P x x y y

,2 2

1 2 1 2

8 9 5 – 7,

2 2

17

, –12

Q 9 – 4 –7 2

,2 2

5 –5

,2 2

R – 4 2 2 6

,2 2

(– 1, 4)

S 8 2 5 6

,2 2

11

5 ,2

Slope of PQ = y – y

x – x2 1

2 1 =

5– – (–1)

25 17

–2 2

=

–32

–122

= 1

4

Slope of QR =

54 – –

25

–1 –2

=

132

– 72

= –13

7

Slope of RS =

11– 4

25 – (– 1)

=

326

= 1

4

Slope of PS =

11– (–1)

217

5 –2

=

132–72

= –13

7

Slope of PQ = Slope of RS seg PQ || seg RS .......(i)

Slope of QR = slope of PS seg QR || seg PS .......(ii) PQRS is a parallelogram. [From (i), (ii) and by definition]


SCHOOL SECTION 357

2. Find the equations of the lines which through the point (3, 4) and thesum of whose intercepts on the axes is 14.

Sol. Let the intercepts made by the lines on the co-ordinate axes be a andb respectively.

a + b = 14 ........(i)

The equation of the line is x y

1a b

Since the line passes through the point (3, 4)3 4

1a b

3b + 4a = ab .......(ii)From (i), a = 14 – bSubstituting a = 14 – b in (ii) we get,3b + 4 (14 – b) = (14 – b)b

3b + 56 – 4b = 14b – b2

b2 – 15b + 56 = 0 b2 – 8b – 7b + 56 = 0 b (b – 8) – 7 (b – 8) = 0 (b – 8) (b – 7) = 0 b = 8 OR b = 7

By (i) when b = 8, c = 14 – 8 = 6and when b = 7, c = 14 – 7 = 7

Equations of the required lines arex y

16 8 and

x y1

7 7

4x + 3y = 24 and x + y = 7

4x + 3y – 24 = 0 and x + y – 7 = 0

3. Find the equation of a line which passes through the point (– 3, 7) andmakes intercepts on the co-ordinate axes which are equal in magnitudebut opposite in sign.

Sol. Let the intercepts made by the line on the co-ordinate axes be a and b. a = – b .......(i)


1a b

x y

1– b b

– x + y = bThis line passes through the point (– 3, 7)

– (– 3) + 7 = b b = 10 The equation of the line is – x + y = 10

x – y + 10 = 0

4. Find the equation of a line which contains the point (4, 1) and whosex-intercept is twice its y-intercept.

Sol. Let the intercepts made by the line on the co-ordinate axes be a andb respectively.

a = 2b


1a b

x y

12b b


SCHOOL SECTION358

x + 2y = 2bSince this line contains the point (4, 1)4 + 2 (1) = 2b

6 = 2b b = 3 Equation of the required line is x + 2y = 6

x + 2y – 6 = 0

5. Find the equation of side AC of an isosceles ABC, if the equation of side ABis x – y – 4 = 0 and B (4, 0) and C (6, 4) are the extremities of the base.

Sol. Let A (h, k)Since A lies on side AB i.e. onx – y – 4 = 0

h – k – 4 = 0 h = k + 4 A (k + 4, k)

Since ABC is an isosceles triangle with BC as base, l (AB) = l (AC)

(k 4 – 4) (k – 0) 2 2 = (k 4 – 6) (k – 4) 2 2

On squaring both sides,k2 + k2 = (k – 2)2 + (k – 4)2

k2 + k2 = k2 – 4k + 4 + k2 – 8k + 16 0 = – 12k + 20 12k = 20

k =20

12

k = 5

3

k + 4 =5

43 =

17

3

A 17 5

,3 3

Equation of side AC by two point form is

x – x

x – x1

1 2 =

y – y

y – y1

1 2

x – 617

6 –3

=

y – 45

4 –3

x – 6

y – 4 =

176 –

35

4 –3

x – 6

y – 4 =

13

73

x – 6

y – 4 =1

7 7 (x – 6) = y – 4 7x – 42 = y – 4

7x – y – 38 = 0


SCHOOL SECTION 359

6. Find the equations of the line which cut off intercepts on the axes whosesum is 1 and product is – 6.

Sol. Let the intercepts made by the line on the co-ordinates axes be a andb respectively.

a + b = 1 ........(i)and ab = – 6 .......(ii)From (ii)

b = – 6

aSubstituting this in (i) we get,

a – 6

a= 1

a2 – 6 = a a2 – a – 6 = 0 a2 – 3a + 2a – 6 = 0 a (a – 3) + 2 (a – 3) = 0 (a – 3) (a + 2) = 0 a = 3 or a = – 2

By (i) when a = 3, b = 1 – 3 = – 2and when a = – 2, b = 1 – (– 2) = 3Now, equation of the line making intercepts a and b isx y

1a b

Equations of the required lines arex y

13 –2 and

x y1

–2 3

2x – 3y = 6 and – 3x + 2y = 6

2x – 3y – 6 = 0 and 3x – 2y + 6 = 0

CHAPTER : 6 - MENSURATION

1. A tinmaker converts a cubical metallic box into 10 cylindrical tins. Sideof the cube is 50 cm and radius of the cylinder is 7 cm. Find the height ofeach cylinder so made if the wastage of 12% is incurred in the process.

(Given = 227

).

Sol. Side of the cubical metallic box (l) = 50 cmTotal surface area of cubical box = 6l2

= 6 × (50)2

= 6 × 2500= 15000 cm2

Wastage incurred in the process of making 10 cylindrical tins= 12% of 15000

=12

×15000100

= 1800cm2

Area of metal sheet used to make 10 cylindrical tins= Total surface area of cubical box – Wastage incurred in the process= 15000 - 1800= 13200 cm2

Area of metal sheet used to make each cylindrical tin

=13200

10= 1320 cm2


SCHOOL SECTION360

Radius (r) = 7 cmArea of metal sheet used to

make each cylindrical tin = Total surface area of cylinder

Total surface area of cylinder = 2r (r + h)

1320 = 2 22

7× 7 (7 + h)

1320 = 2 22 × (7 + h)

1320

2 × 22= 7 + h

30 = 7 + h h = 30 – 7 h = 23 cm

Height of each cylinder is 23 cm.

2. The three faces A, B, C of a cuboid have surface area 450 cm2, 600 cm2

and 300 cm2 respectively. Find the volume of the cuboid.Sol. Surface area of face A = 450 cm2

Surface area of face A = l × h l × h = 450 .....(i)

Surface area of face B = 600cm2

Surface area of face B = l × b l × b = 600 .....(ii)

Surface area of face C = 300 cm2

Surface area of face C = b × h b × h = 300 .....(iii)

Multiplying (i), (ii) and (iii),l × h × l × b × b × h = 450 × 600 × 300

l2 × b2 × h2 = 450 × 2 × 300 × 300

l × b × h = 900 300 300 [Taking square roots]

l × b × h = 30 × 300 l × b × h = 9000cm3

But, Volume of the cuboid = l × b × hVolume of the cuboid = 9000cm3

Volume of the cuboid is 9000 cm3.

3. Oil tins of cuboidal shape are made from a metallic sheet with length 8m and breadth 4 m . Each tin has dimensions 60 40 20 in cm and isopen from the top. Find the number of such tins that can be made.

Sol. Length of the metallic sheet (l) = 8 m= 8 × 100= 800 cm

its breadth (b) = 4 m= 4 × 100= 400 cm

Area of metallic sheet = l × b= 800 × 400= 320000 cm2

Length of the oil tin (l1) = 60 cmits breadth (b1) = 40 cmits height (h1) = 20 cm

Area of metallic sheet required for each tin= surface area of vertical faces + surface area of the base= [2 (l

1 + b

1) × h

1] + [l

1 b1]

A

B

Ch

bl


SCHOOL SECTION 361

= [2 (60 + 40) × 20] + [60 × 40]= (2 × 100 × 20) + (60 × 40)= 4000 + 2400= 6400 cm2

Number of tins that can be made

=Area of metallic sheet

Area of metal required for each tin

=320000

6400= 50

50 Oil tins can be made.

4. Plastic drum of cylindrical shape is made by melting spherical solid plasticballs of radius 1 cm. Find the number of balls required to make a drumof thickness 2 cm, height 90 cm and outer radius 30 cm. (5 marks)

Sol. Outer radius of the drum (r1) = 30cmIts thickness = 2 cm

inner radius of the drum (r2) = 30 – 2= 28 cm

Outer height of cylindrical plastic drum (h1) = 90cmInner height of cylindrical plastic drum (h2)

= Outer height – thickness of base= 90 – 2= 88cm

Volume of plastic required for the cylindrical drum= Volume of outer cylinder – Volume of inner cylinder= r1

2h1 – r22h2

= [(30)2 × 90 – (28)2 × 88]= × (900 × 90 – 784 × 88]= × (81000 – 68992)= 12008 cm3

Radius of spherical solid plastic ball (r) = 1cm

Volume of each plastic ball =4

3r3

=4

3 × × r × 1 × 1 × 1

=4

3 cm3

Number of balls required

to make the drum =Volume of plastic required for the drum

Volume of each plastic ball

=

1200843

=3

120084

= 9006

Number of plastic balls required to make the cylindrical drum is 9006.


SCHOOL SECTION362

5. Water drips from a tap at the rate of 4 drops in every 3 seconds. Volume ofone drop is 0.4 cm3. If dripped water is collected in a cylinder vessel ofheight 7 cm and diameter is 8 cm. In what time will the vessel be completelyfilled ? What is the volume of water collected ? How many such vesselswill be completely filled in 3 hours and 40 minutes ? (5 marks)

Sol. Diameter of the cylindrical vessel = 8cm Its radius (r) = 4 cm

its height (h) = 7 cm Volume of the cylindrical vessel = r2h

=22

4 4 77

= 22 × 16= 352 cm3

Volume of water collected = 352 cm3

Volume of one drop of water = 0.4 cm3

Volume of 4 drops of water = 4 × 0.4= 1.6 cm3

Water drips from the tap at the rate of 4 drops in every 3 seconds Volume of water collected in 3 seconds = 1.6 cm3

Volume of water collected in 1 seconds = 1.6

3cm3

Time required to fill the cylindrical vessel

=Volume of cylindrical vessel

Volume of water collected in each second

=3521.6

3

=3

3521.6

=352 3 10

1.6 10

=352 3 10

16

= 660 seconds= 11 minutes [ 1 minutes = 60 seconds]

3 hours and 40 minutes = 3 × 60 min + 40 min= 180 + 40= 220 minutes

Number of vessels that can be completely filled in 220 minutes

=220

11= 20

20 vessels can be filled in 3 hours and 40 minutes.

6. In the adjoining figure,ABCDEF is a regular hexagonwith each side 14 cm. From eachvertex, arcs with radius 7 cm are drawn.Find the area of the shaded portion. (5 marks)

Sol. Draw seg BN chord PQRadii of each arc = 7 cm [Given]i.e. PB = BQ = 7 cmABCDEF is a regular hexagon......

E D

C

Q

BA

F

P

X N•


SCHOOL SECTION 363

In BPQ, BP = BQ mBPQ = m BQP .......(i) [Isosceles triangle theorem] m BPQ + m BQP + m PBQ = 180º [Sum of the measures of angles

of a triangle is 180º] m BPQ + m BPQ + 120 = 180 [From (i) and angle of regular

hexagon] 2 mBPQ = 180 – 120 2 mBPQ = 60 m BPQ = 30º ......(ii)

In BNP,BNP = 90º [Construction]BPN = 30º [From (ii) and P - N - Q]

PBN = 60º [Remaining angle] BNP is 30º - 60º - 90º triangle

By 30º - 60º - 90º triangle theorem,

BN = 1

BP2


BN = 7

2 BN = 3.5 cm

PN = 3

BP2


PN =3

72

PN = 3.5 3 cmPQ = 2PN [Perpendicular drawn from the centre

of the circle to the chord bisects thechord]

PQ = 2 3.5 3

PQ = 7 3 cm

A (BPQ) =1

PQ BN2

=1

7 3 3.52

= 3.5 3 3.5= 3.5 × 3.5 × 1.73= 20.25 × 1.73= 21.1925= 21.19 cm2

A (B-PXQ) = r360

2

=120 22

7 7360 7

=154

3= 51.33 cm2

A (segment PXQ) = A (B – PXQ) – A (BPQ)= 51.33 – 21.19= 30.14 cm2


SCHOOL SECTION364

A (shaded portion) = 6 × A (segment PXQ)= 6 × 30.14= 180.84 cm2

The area of shaded portion is 180.84 cm2.

7. A metallic right circular cylindrical disc is of height 30 cm and the diameterof the base is one half time the height. This metallic disc is melted andmoulded into the sphere. Assuming that no metal is wasted duringmoulding, find the radius and total surface area of the sphere. (5 marks)

Sol. Height of cylindrical disc (h) = 30 cm

its diameter =1

12

times the height

=3

302

= 45 cm

Radius of cylindrical disc (r) =45

2 cm

Volume of cylindrical disc = r2h

=45 45

302 2

=60750

cm4

2

Let the radius of the sphere be ‘r1’ cmThe metallic disc is melted and moulded into the sphere [Given]

Volume metallic sphere = Volume cylindrical disc

4

r3

31 =

60750cm

43

4

r3

31 =

45 4530

2 2

r31 =

45 45 30 3

4 2 2

r31 =

45 45 45

2 2 2

r1 =45

2[Taking cube roots]

r1 = 22.5 cm Total surface area of sphere = 4r2

= 224 22.5

7 2

=4 22 506.25

7

=88 506.25

7

=4455000

7= 6364.28 cm2 (Approximately)

Radius of the sphere is 22.5 cm andtotal surface area of the sphere is 6364.28 cm2.

MAHESH TUTORIALS PVT. LTD. 1

Q.I. Solve the following : (4)(i) In the adjoining figure,

line l || line m ||line n.Lines p and q are transversals.From given informationfind ST.

(ii) In the adjoining figure,seg BE seg AB andseg BA seg AD.If BE = 6 and AD = 9

find A ( ABE)

A ( ABD)

Q.II. Attempt the following : (9)(i) In the adjoining figure,

RP : PK = 3 : 2 thenfind the values of :(a) A (TRP) : A (TPK)(b) A (TRK) : A (TPK)(c) A (TRP) : A (TRK)

(ii) In the adjoining figure,ML || BC and NL || DC.

Then prove that : AM AN

=AB AD

.

p q

R

TC

B S

Al

m

n

8

10

BE

A D

T

R P K

B

M

A L C

D

N

CHAPTER 1 : Similarity

SET - A

S.S.C.

GEOMETRY

Marks : 30

Duration : 1 hr. 15 min.

GEOMETRY MT EDUCARE PVT. LTD.

2 MAHESH TUTORIALS PVT. LTD.

(iii) In PQR, if QS is the angle bisectorof Q then, show that

A ( PQS)

A ( QRS)

=

PQ

QR

Q.III. Solve the following : (12)(i) In the adjoining figure,

AB || DC.Using the information givenfind the value of x.

(ii) Let X be any point on side BC of ABC,XM and XN are drawn parallel to BAand CA. MN meets produced CB in T.Prove that TX2 = TB . TC.

(iii) In the adjoining figure,DEFG is a square and BAC = 90ºProve that :(a) AGF ~ DBG(b) AGF ~ EFC(c) DBG ~ EFC(d) DE2 = BD . EC

Q.IV. Solve the following : (5)(i) Two poles of height ‘a’ meters and

‘b’ meters are ‘p’ meters apart.Prove that the height ‘h’ drawn fromof the point of intersection N of thelines joining the top of each pole tothe foot of the opposite pole is

ab

a + b meters.

Best of Luck

P

Q

S

R

D C

A

O

B

3

x – 33x – 19

x – 5

A

N

M

CXBT

S

N ab

h

x yTBA

R

p

A

G F

BD E C

MAHESH TUTORIALS PVT. LTD.

MT EDUCARE PVT. LTD. GEOMETRY

3

Q.I. Solve the following : (4)

(i) Find the side of square whose diagonal is 16 2 cm .

(ii) In ABC ~ DEF and A (ABC) = 9 cm2, A (DEF) = 64 cm2, DE = 5.6 cm,then find AB.

Q.II. Attempt the following : (9)(i) The ratio of the areas of two triangles with the common base is 6 : 5.

Height of the larger triangle is 9 cm. Then find the correspondingheight of the smaller triangle.

(ii) In the adjoining figure,seg ED || seg QO andseg DF || seg OR.Prove that seg EF || side QR.

(iii) A vertical stick 12 m long casts a shadow 8 m long on the ground. Atthe same time a tower casts the shadow 40 m on the ground.Determine the height of the tower.

Q.III. Solve the following : (12)(i) In ABC, PQ is a line segment intersecting AB at P and AC at Q such

that PQ || BC. If PQ divides DABC into two equal parts means equal

in area , find BP

AB.

(ii) In the adjoining figure,ABCD is a square. The BCE onside BC and ACF on the diagonal ACare similar to each other. Then show

that A (BCE) = 1

A (ΔACF)2


SET - B

S.S.C.

GEOMETRY

Marks : 30


P

DE F

O

Q R

F

CD

A B

E



(iii) ABC is a triangle in which AB = ACand D is any point on BC.Prove that : AB2 – AD2 = BD. CD.

Q.IV. Solve the following : (5)(i) G is the centroid of ABC.

GE and GF are drawn parallel

to AB and AC respectively.

Find A (GEF) : A (ABC).

(Hint : Draw the median AD.

Then GD : AD = 1 : 3.

A (GED) : A (ABD) = 1 : 9)

A

BD E

C

A

B E F

G

C

Best of Luck

MAHESH TUTORIALS PVT. LTD.

MT EDUCARE PVT. LTD. GEOMETRY

5

Q.I. Solve the following : (4)(i) In PQR, seg PM is the median. If PM = 9 and PQ2 + PR2 = 290. Find

QR.

(ii) D is a point on side BC of ABC such that ADC = BAC. Show thatAC2 = BC × DC.

Q.II. Attempt the following : (9)(i) ABD is a triangle in which A = 90º

and seg AC hypotenuse BD.Show that :(a) AB2 = BC . BD(b) AD2 = BD . CD(c) AC2 = BC . CD

(ii) In the adjoining figure,PQR = 90º.T is the mid point of the side QR.Prove that PR2 = 4PT2 – 3PQ2.

(iii) If the sides of a triangle measure 18 cm, 18 3 cm and 36 cm, show

that it is a 30º - 60º - 90º triangle.

Q.III. Solve the following : (12)(i) Seg AD is the median of ABC,

and AM BC. Prove that :

(a) AC2 = AD2 + BC × DM +

2BC

2

(b) AB2 = AD2 – BC × DM +

2BC

2


SET - C

S.S.C.

GEOMETRY

Marks : 30


A

B C D

P

QT

R

A

B M D C



(ii) In the adjoining figure,

if LK = 6 3 find MK, ML, KN, MN

and the perimeter of MNKL.

(iii) ABC is a triangle where C = 90º. Let BC = a, CA = b, AB = c and let‘p’ be the length of the perpendicular from C on AB. Prove that

(a) cp = ab, (b) 2 2 2

1 1 1= +

p a b

Q.IV. Solve the following : (5)(i) In ABC, seg DE || side BC. If 2A (ADE) = A (DBCE), find AB : AD

and show BC = 3 .DE

N

M

LK

45º

30º

6 3

Best of Luck

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