mahesh tutorials ssc question paper with solution -geometry

Paper - I

Q.1. Solve the following : 3

(i) Ray YM is the angle bisector

of XYZ, where XY = YZ.

Find the relation between XM and MZ.

(ii) O is the centre of the circle.If m ABC = 80º, the findm (arc AC) and m (arc ABC).

(iii) The circumcentre and incentre of ............... triangle are at the samepoint.


(i) ABC ~ DEF, if AB = 2.4 cm, DE = 1.6 cm, find the ratio of the areaof ABC and DEF.

(ii) In the adjoining figure,if m (arc APC) = 60ºand m BAC = 80ºFind (a) ABC (b) m (arc BQC).

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

S.S.C. Test - II

Batch : SB Marks : 30

Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3

MAHESH TUTORIALS

X

M

Y Z• •

B

O

A

C

B

A

C80º

•Q

•P

Paper - I


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

(ii) In PQR, if QS is the angle bisectorof Q then, show that

A ( PQS)

A ( QRS)

=

PQ

QR

(iii) Three congruent circles withcentres A, B and C and withradius 5 cm each, touch eachother in points D, E, G as shown in(a) What is the perimeter of ABC ?(b) What is the length of side DE of DEF ?


(i) 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

(ii) In a cyclic quadrilateral ABCD,the bisectors of opposite angles Aand C meet the circle at P and Qrespectively. Prove that PQ is adiameter of the circle.

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

P

Q

S

R

A B

C

D

EF

A

G F

B D E C

D P

Q

A

C

B

••

××

Best Of Luck

... 2 ...

Paper - I

A.1. Solve the following :(i) In XYZ,

ray YM bisects XYZ [Given]

XY

YZ =

XM

MZ[Property of ½

angle bisector of a triangle]

1 = XM

MZ[ XY = YZ]

XM = MZ ½

(ii) m ABC = 1

2 m (arc AC)

[By Inscribed angle theorem]

80º = 1

2 m (arc AC)

m (arc AC) = 160º ½

m (arc ABC)= 360º – m (arc AC)= 360 – 160

m (arc ABC)= 200º ½

(iii) The circumcentre and incentre of an equilateral triangle are 1at the same point.

A.2. Solve the following :(i) ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

AB

DE

2

2 [Areas of similar triangles] ½

A ( ABC)

A ( DEF)

=

(2.4)

(1.6)

2

2 [Given]

A ( ABC)

A ( DEF)

=

5.76

2.56½

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

X

M

Y Z• •

B

O

A

C

Paper - I... 2 ...

A ( ABC)

A ( DEF)

=

576

256½

A ( ABC)

A ( DEF)

=

9

4

A (ABC) : A (DEF) = 9 : 4 ½

(ii) (a) m ABC =1

2 m(arc APC) [Inscribed angle theorem] ½

m ABC =1

2 × 60

m ABC = 30º ½

(b) m BAC =1

2 m(arc BQC) [Inscribed angle theorem] ½

80 =1

2 m(arc BQC)

m(arc BQC) = 80 × 2

m(arc BQC) = 160º ½

(iii)

B

A

C80º

•Q

•P

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing ACB

A25º

6.3 cm B

C

65º

25º

130º

O

A 6.3 cm B

C

65º

(Rough Figure)

Paper - I... 3 ...

A.3. Solve the following :(i)

(ii) In PQRray QS bisects PQR [Given]

PQ

QR = PS

SR......(i)

[By property of angle

bisec tor of a triangle] 1

PQS and QRS have a common

vertex Q and their bases PS and

SR lie on the same line PR.

Their heights are equal

A ( PQS)

A ( QRS)

=

PS

SR[Triangles having equal heights] 1

A ( PQS)A ( QRS)

=

PQQR [From (i)] 1

½ mark for rough figure1 mark for drawing triangle1 mark for drawing perpendicular bisectors½ mark for drawing circle

R P6 cm40º

Q

O

R P6 cm40º

Q(Rough Figure)

P

Q

S

R

Paper - I

(iii)

(a) A - D - BB - E - C ½A - F - CAD = DB = BE = EC = AF = FC = 5cm .......(i)

[Radii of congruent circles and given]AB = AD + BD [ A - D - B]

AB = 5 + 5 [From (i)] ½ AB = 10 cm ......(ii)

Similarly,BC = 10 cm ......(iii)AC= 10 cm .......(iv)

Perimeter of ABC = AB + BC + AC ½= 10 + 10 + 10 [From (ii), (iii) and (iv)]

Perimeter of ABC = 30 cm ½

(b) In ABC,D and E are mid-points of sides AB and BC respectively. [From (i)]

DE =1

2AC [By mid-point theorem] ½

DE =1

2× 10 [From (iv)]

DE = 5 cm. ½


(a) DEFG is a square [Given]

DE = EF = GF = DG .......(i) [Sides of a square]

seg GF || seg DE [Opposite side of a square] ½ seg GF || seg BC [B - D - E - C)

On transversal AB,

AGF ABC .......(ii) [Converse of corresponding ½angles test]

On transversal AC,

A B

C

D

EF[If two circles are touching circlesthen the common point lies on theline joining their centres]

A

G F

B D E C

... 4 ...

Paper - I

AFG ACB ......(iii)

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] 1

(b) 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] 1(c) DBG ~ EFC [From (iv) and (v)]

(d) BD

EF =

DG

EC[c.s.s.t.]

EF × DG = BD × EC

DE × DE = BD × EC [From (i)]

DE2 = BD × EC 1

(ii) Proof :

DAP BAP [ ray AP bisects DAB] ½Let m DAP = m BAP = xº ....(i)DCQ BCQ [ ray CQ bisects DCB] ½Let, m DCQ = m BCQ = yº .....(ii)ABCD is cyclic [Given]

m DAB + m DCB = 180º [Opposite angles of a cyclicquadrilateral are supplementary]

m DAP + m BAP + DCQ + m BCQ = 180º[Angle addition property]

x + x + y + y = 180º [From (i) and (ii)] 1 2x + 2y = 180º .....(iii)

m DAP =1

2 m (arc DP) [Inscribed angle theorem]

x =1

2m (arc DP) [From (i)]

m (arc DP) = 2xº .....(iv) ½

D P

Q

A

C

B

••

××

... 5 ...

Paper - I

m DCQ =1

2 m (arc DQ) [Inscribed angle theorem]

y =1

2 m (arc DQ) [From (ii)]

m (arc DQ) = 2yº ......(v) ½m (arc DP) + m (arc DQ) = 2x + 2y [Adding (iv) and (v)]

m (arc PDQ) = 180º [Arc addition property and from (iii)] ½ Arc PDQ is a semicircle

seg PQ is a diameter of the circle. ½

(iii)

S B

A

7.6 cm

A

40º 40º

5 cm

D

O100º

S B

7.6 cm

A

5 cm

(Rough Figure)

D

50º

½ mark for rough figure½ mark for drawing centre O1 mark for drawing the perpendicular bisector1 mark for locating point A1 mark for drawing SAB

... 6 ...

Paper - II


(i) The sides of two similar triangles are 4 : 9. What is the ratio oftheir area ?

(ii) What is the relation betweenABC and ADC of cyclic ABCD ?

(iii) If the circumcentre lies in the exterior of the triangle, then it is.......... angled triangle.


(i) 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.

(ii) If two circles with radii 8 and 3 respectively touch internally thenshow that the distance between their centers is equal to thedifference of their radii, find that distance.

(iii) Construct an arc PQM such that seg PM of length 6.2 cm subtendsan angle of 40º on it.


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

(ii) 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 thesides of PQR.

S.S.C. Test - II



MAHESH TUTORIALS

AB

C

D

Paper - II

(iii) In the adjoining figure,in two chords AB and CDof the same circle are parallelto each other. P is the centreof the circle.Prove : m CPA = m DPB.


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

and show BC = 3 .DE

(ii) ABC is inscribed in a circle with centre O, seg AXis a diameter of the circle with radius r.seg AD seg BC. Prove that

(a) ABX ~ ADC, (b) A (ABC) = abc

4r.

(a is side opposite to A, ...)

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

Best Of Luck

A

B D C

O

X

... 2 ...

C DA B

P

Paper - II

A.1. Solve the following :(i) Let ABC ~ DEF,

AB

DE =

4

9[Given]

ABC ~ DEF

A ( ABC)

A ( DEF)

=

AB

DE

2


A ( ABC)

A ( DEF)

=

AB

DE

2

A ( ABC)

A ( DEF)

=

4

9

2

A ( ABC)

A ( DEF)

=

16

81

A (ABC) : A (DEF) = 16 : 81 ½

(ii) ABCD is cyclic [Given] m ABC + m ADC = 180º [Opposite angles

of quadrilateral are supplementary] ABC and ADC are supplementary. 1

(iii) If the circumcentre lies in the exterior of the triangle, then it 1is an obtuse angled triangle.

A.2. Solve the following :(i) In the adjoining figure,

seg AB and seg PQrepresents the vertical ½stick and the towerrespectively and seg BCand seg QR representsthe shadow cast by themrespectively. ½ABC ~ PQR

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

AB

C

D

P

Q R

?

40 m

A

B C8 m

12 m

Paper - II... 2 ...

AB

PQ =BC

QR [c.s.s.t.] ½

12

PQ =8

40[Given]

PQ =40 ×12

8 PQ = 60

Height of the tower is 60 m. ½

(ii) Let two circles with centres O andA touch each other internally at point T.

O - A - T [If two circles are touching 1circles then the commonpoint lies a the line joiningtheir centres]

OT = OA + AT [O - A - T] OA = OT – AT ½

OT = 8 units, AT = 3 units [Given] OA = 8 – 3 OA = 5 units

The distance between the centres is 5 units. ½

(iii)

OA T

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing PQM

MP 6.2 cm

40º

Q(Rough Figure)

MP 6.2 cm50º 50º

40º

Q

O

80º

Paper - II


(ii) In ABC,

AB = 5 unitsBC = 6 units [Given]AC = 7 unitsPerimeter of PQR = 360 units [Given] ½

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] ½

5

PQ =6

QR =7

PR=

18

360[From (i)]

5

PQ =6

QR =7

PR =

1

20.....(ii) ½

½ mark for rough figure½ mark for drawing RST1 mark for drawing the angle bisectors1 mark for drawing the incircle

... 3 ...

S7 cm

6 cm 6.5 cm

R

T××

••

O

S7 cm

6 cm 6.5 cm

R

T

(Rough Figure)

Paper - II

5

PQ = 1

20[From (ii)]

PQ = 100 units ½

6

QR = 1

20[From (ii)]

QR = 120 units ½

7

PR=

1

20[From (ii)]

PR = 140 units ½

(iii)

Construction : Draw seg BC. ½Proof : m CPA = m (arc CA) .......(i) ½m DPB = m (arc DB) ......(ii)

m ABC = 1

2 m (arc CA) ......(iii)

[Inscribed angle theorem] ½m BCD =

1

2 m (arc DB) ......(iv)

chord CD || chord AB [Given] On transversal BC,

ABC BCD ......(v) [Converse of alternate angles test] ½

1

2m (arc CA) =

1

2 m(arc DB) [From (iii), (iv) and (v)] ½

m (arc CA) = m (arc DB) ......(vi) ½ m CPA = m DPB [From (i), (ii) and (vi)]

A.4. Solve the following :(i) seg DE || side BC [Given]

On transversal AB,ABC ADE .....(i) [Converse of ½

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

ABC ~ ADE [By AA test of similarity] ½

[Definition of measureof minor arc]

A

D E

B C

... 4 ...

C DA B

P

Paper - II

AB

AD =

BC

DE =

AC

AE.....(ii) [c.s.s.t.] ½

A ( ABC)

A ( ADE)

=

AB

AD

2

2 .....(iii) [Areas of similar triangles] ½

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) ½

3

1=

AB

AD

2

2 [From (iii) and (iv)]

AB

AD=

3

1 ......(v) [Taking square roots] ½

AB : AD = 3 :1

BC

DE=

AB

AD[From (ii)] ½

BC

DE=

3

1[From (v)]

BC = 3 × DE ½

(ii) seg AX is a diameter [Given]

arc ACX is a semi circle.

m ABX = 90º [Angle subtended by ½a semicircle]

In ABX and ADC

ABX ADC [Each is 90º]

AXB ACD [Angles inscribed in the same arc

and C - D - B]

ABX ~ ADC [By AA test of similarity] 1BC = a, AB = c, AC = b [Given]

AB

AD=

AX

AC[c.s.s.t.] ½

c

AD=

2r

b

AD =bc

2r.......(i) ½

A

B D C

O

X

... 5 ...

Paper - II... 6 ...

A (ABC) =1

2 × base × height ½

=1

2 × BC × AD

=1

2 × a ×

bc

2r[From (i)] ½

A (ABC) =abc4r

½

(iii)

MD

E E

F30º

6.2 cm

60º

4.4

cm

30º

4.4

cmO

D F

E

60º

4.4

cm

M6.2 cm

(Rough Figure)

½ mark for rough figure1 mark for locating centre ‘O’1 mark for locating point E1 mark for drawing the altitude½ mark for drawing DEF

Paper - III


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

(ii) If two circles touch externally then show that the distance betweentheir centers is equal to the sum of their radii.

(iii) What is the point of concurrence of the angle bisectors of a trianglecalled ?


(i) 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.

(ii) In the adjoining figure,m (arc XAZ) = m (arx YBW).Prove that XY || ZW

(iii) Construct an arc DCV such that seg DV of length 9.5 cm subtendsan angle of 135º on it.

S.S.C. Test - II



MAHESH TUTORIALS

p q

R

TC

B S

Al

m

n

8

10

B C

E

D

A

X Y

A B

Z W

Paper - III


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

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

(iii) Prove that in a cyclic trapezium angles at the base are congruent.


(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.

(ii) In the adjoining figure,points P and Q are the centers ofthe circles. Radius QN = 3, PQ = 9.M is the point of contact of the circles.Line ND is tangent to the larger circle.Point C lies on the smaller circle.Determine NC, ND and CD.

(iii) SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and

SH

SV =

3

5; construct SVU.

D C

A

O

B

3

x – 33x – 19

x – 5

S

N ab

h

x yTBA

R

p

CD

Q PN M

... 2 ...

Paper - III


line l || line m || line n [Given] On transversals p and q,

AB

BC=

RS

ST

[By Property of Intercepts made

by threeparallel lines] ½

8

10=

12

ST[Given]

ST =12 ×10

8

ST = 15 units ½

(iii)

Let two circles with centres O and

A touch each other internally at point T. ½ O - A - T [If two circles are

touching circles then the common

point lies a the line joining their centres]

OT = OA + AT [O - A - T]

OA = OT – AT ½

(iii) The point of concurrence of the angle bisectors of a triangle 1called is called incircle.

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

p q

R

TC

B S

Al

m

n

8

10

O AT

Paper - III... 2 ...

A.2. Solve the following :(i) In ABC and AED,

BAC DAE [Common angle]ABC AED [ Each is 90º] 1

ABC ~ AED [By AA test of similarity] ½

AB

AE=

AC

AD[c.s.c.t.]

12

AE=

18

6[Given]

AE =12 × 6

18 AE = 4 units ½

(ii)

Construction : Draw seg XW ½

Proof : m XWZ = 1

2 m (arc XAZ) .......(i) ½

m WXY = 1

2 m (arc YBW) .......(ii)

But, m (arc XAZ) = m (arc YBW) .......(iii) [Given] ½ m XWZ = m WXY [From (i), (ii) and (iii)] XY || ZW [Alternate angles test] ½

(iii)

B C

E

D

A

X Y

A B

Z W

[Inscribed angletheorem]

D

C

V

9.5 cm 45º45º

135º D

C

V9.5 cm

135º

(Rough Figure)

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing DCV

Paper - III... 3 ...


(ii) seg AB || seg DC [Given]

On transversal BD,

CDB ABD [Converse of ½alternate angles test]

CDO ABO .......(i) [D - O - B]

In DOC and BOA,

CDO ABO [From (i)]

DOC BOA [Vertically opposite angles] 1 DOC ~ BOA [By AA test of similarity]

DO

BO=

OC

OA[c.s.s.t.] ½

3

x – 3=

x – 5

3x –19

K60º 55º

7 cm M

L

O

K60º 55º

7 cm M

L(Rough figure)

D C

A

O

B

3

x – 33x – 19

x – 5

½ mark for rough figure1 mark for drawing LKM1 mark for drawing perpendicular bisectors½ mark for drawing circle

Paper - III

3(3x – 19) = (x – 5) (x – 3)

9x – 57 = x2 – 3x –5x + 15

9x – 57 = x2 – 8x + 15 ½

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 ½

(iii)

Given : ABCD is a cyclic trapezium.seg AB || seg BC ½To prove : ABC DCBProof : ABCD is cyclic [Given]

m ABC + m ADC = 180º .....(i) [Opposite angles of a cyclic ½ quadrilateral are supplementary]

seg AD || seg BC [Given] On transversal DC,

m DCB + m ADC = 180º ......(ii) [Converse of interior angle ½test]

m ABC + m ADC = m DCB + m ADC [From (i) and (ii)] ½ m ABC = m DCB ABC DCB ½


AB = AT + TB [ A - T - B] AB = (x + y) = p ......(i) ½

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

ATN ABR [By AA test of similarity] ½

S

N ab

h

x yTBA

R

p

A D

B C

(½ mark for figure)

... 4 ...

Paper - III

AT

AB=

TN

BR[c.s.s.t]

x

p =h

a......(ii) [Given and from (i)] ½

In BTN and BAS,

TBN ABS [Common angle]

BTN BAS [ each is 90º] BTN BAS [By AA test of similarity] ½

BT

AB=

TN

AS[c.s.s.t]

y

p =h

b......(iii) [Given and from (i)] ½

Adding (ii) and (iii),

x y+

p p = h h

+a b

½

x + y

p = 1 1

h +a b

p

p =a + b

hab

[From (i)] ½

1 =a + b

hab

ab

a + b = h

h =ab

a +b metres ½

(ii) Construction : Draw seg CM and seg DP ½Sol. Q - M - P [If two circles are ½

touching circles thethe commonpoint lies on the linejoining their centre]

QN = 3 units[Given]

PQ = 9 unitsPN = PQ + QN [ P - Q - N]

PN = 9 + 3 PN = 12 units ½

QN = QM = 3 units [Radii of the same circle] MN = 6 units [ Diameter is twice the radius]

CD

Q PN M

... 5 ...

Paper - III

PM + QM = PQ [ P - M - Q]

PM + 3 = 9 [Given]

PM = 9 – 3

PM = 6 units

PM = PD = 6 units [Radii of the same circle] ½In NDP,

m NDP = 90º ......(i) [Radius is perpendiculer to

the tangent]

NP² = ND² + PD² [By Pythagoras theorem]

12² = ND² + 6²

144 = ND² + 36

ND² = 144 – 36 ND² = 108

ND = 108

ND = 36 3

ND = 6 3 units ½

Now, m NCM = 90º.....(ii) [Angle inscribed in a semicircleis a right angle]

NCM NDP [From (i) and (ii)]

seg CM || seg DP ....(iii) [By corresponding angles test] ½In NDP,

seg CM || seg DP, [From (iii)]

NC

CD=

NM

MP[By B.P.T.]

NC

CD=

6

6

NC

CD= 1

NC = CD ½ C is the midpoint of seg ND

NC = CD = 1

2ND

NC = CD = 1

6 32

NC = CD = 3 3 units. ½

... 6 ...

Paper - III

½ mark for SHR1 mark for constructing 5 congruent parts1 mark for constructing VS5S HS3S1 mark for constructing UVS RHS½ mark for required SVU

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)(iii)

... 7 ...

Paper - IV


(i) ABC ~ PQR, A = 47º, Q = 83º. What is the measure of C ?

(ii) A chord divides a circle into 2 arcs measuring 2x and 7x. What isthe measure of the minor arc ?

(iii) What is the point of concurrence of the altitudes of a triangle called ?


(i) ABC ~ DEF, if A (ABC) = 9 cm2, A (DEF) = 64 cm2, DE = 5.6 cmthen find AB.

(ii) If two circles with radii 8 and 3 respectively touch internally thenshow that the distance between their centers is equal to thedifference of their radii, find that distance.

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


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

(ii) 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

.

S.S.C. Test - II



MAHESH TUTORIALS

A

D NO M B

C

Paper - IV

(iii) Find the radius of the circle passing through the vertices of a rightangled triangle when lengths of perpendicular sides are 6 and 8.


(i) In ABC, PQ is a line segment intersecting AB at P and AC at Qsuch that PQ || BC. If PQ divides ABC into two equal parts means

equal in area, find BP

AB.

(ii) In a cyclic quadrilateral ABCD,the bisectors of opposite angles Aand C meet the circle at P and Qrespectively. Prove that PQ is adiameter of the circle.

(iii) LMN ~ XYZ, In LMN, LM = 6 cm, MN = 6.8 cm, LN = 7.6 cm and

LM

XY =

4

3; construct XYZ.

D P

Q

A

C

B

••

××

... 2 ...

Best Of Luck

Paper - IV

A.1. Solve the following :(i) ABC ~ PQR [Given]

m A = m P = 47º[c.a.s.t. and given]

½ m B = m Q = 83º

In ABC,m A + m B + m C = 180º [Angle sum property of a triangle]

47º + 83º + m C = 180º 130º + m C = 180º – 130º

m C = 50º ½

(ii) The measure of minor arc is 2x and measure of major arc is 7x2x + 7x = 360º [Measure of a circle is 360º] ½

9x = 360

x = 360

9 x = 40

Measure of minor arc = 2x = 2x × 40 = 80º. ½

(iii) The point of concurrence of the altitudes of a triangle is called 1orthocentre.

A.2. Solve the following :(i) ABC ~ DEF [Given]

A ( ABC)

A ( DEF)

=

AB

DE

2


9

64=

AB

(5.6)

2

2 [Given]

3

8=

AB

5.6[Taking square roots] ½

AB =3 5.6

8

AB = 3 × 0.7 ½

AB = 2.1 cm ½

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - IV... 2 ...

(ii) Let two circles with centres O andA touch each other internally at point T.

O - A - T [If two circles are touching 1circles then the commonpoint lies a the line joiningtheir centres]

OT = OA + AT [O - A - T] OA = OT – AT ½

OT = 8 units, AT = 3 units [Given] OA = 8 – 3 OA = 5 units

The distance between the centres is 5 units. ½

(iii)


OA T

M N35º

125º

P

35º 8.9 cm

110º

O

M N

125º

P

8.9 cm

(Rough Figure)

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing MPN

7 cmS

T

5 cm

U

120º

(Rough Figure)

Paper - IV

(ii)

ADB and CDB have a common base BD,

A( ADB)

A( CDB)

=

AN

CM....(i) [Triangles with common base] 1

In ANO and CMO,ANO CMO [ Each is 90º]AON COM [Vertically opposite angles] 1

ANO ~ CMO [By AA test of similarity]

AN

CM =

AO

CO......(ii) [c.s.s.t.] ½

A ( ADB)A ( CDB)

=

AOCO

[From (i) and (ii)] ½

7 cm ST

5 cm

U

120º• ×• ×

O

½ mark for rough figure½ mark for drawing STU1 mark for drawing the angle bisectors1 mark for drawing the incircle

A

D NO M B

C

... 3 ...

Paper - IV

(iii) Given : (a) In PQR, m PQR = 90º(b) PQ = 6 units, QR = 8 units.(c) Points P, Q and R lie on the circle. ½To Find : radius of the circle.Sol. In PQR,m PQR = 90º [Given]

PR² = PQ² + QR² [By Pythagoras theorem] PR² = 6² + 8² PR² = 36 + 64 PR² = 100 PR = 10 units [Taking square roots] ½

Let Y be a point on the circle as shown in the figurem PQR = 90º [Given]

m PQR = 1

2 m(arc PYR) [Inscribed angle theorem] ½

90º = 1

2m(arc PYR) ½

m (arc PYR) = 180º arc PYR is a semicircle seg PR is the diameter. Diameter = 10 units. Radius = 5 units [ Radius is half of the diameter] ½

Radius of the circle is 5 units.

A.4. Solve the following :(i) seg PQ divides ABC into two parts of equal areas [Given]

A (APQ) = 1

2 A (ABC)

A ( APQ)

A ( ABC)

=

1

2......(i) ½

seg PQ || side BC [Given]On transversal AC,AQP ACB ......(ii) [Converse of corresponding angles ½

test]In APQ and ABC,AQP ACB [From (ii)]PAQ BAC [Common angle]

APQ ~ ABC [By AA test of similarity] ½

A ( APQ)

A ( ABC)

=

AP

AB

2


1

2=

AP

AB

2

2 [From (i)]

8Y

R

Q

P6

•

(½ mark for figure)

B

P

A QC

... 4 ...

Paper - IV

1

2=

AP

AB[Taking square roots] ½

AP

AB=

1

2

AB – BP

AB=

1

2[ A - P - B] ½

AB BP

–AB AB

=1

2

BP

1 –AB

=1

2½

1

1 –2

=BP

AB

BP

AB=

2 – 1

2½

(ii) Proof :

DAP BAP [ ray AP bisects DAB] ½Let m DAP = m BAP = xº ....(i)DCQ BCQ [ ray CQ bisects DCB] ½Let, m DCQ = m BCQ = yº .....(ii)ABCD is cyclic [Given]

m DAB + m DCB = 180º [Opposite angles of a cyclicquadrilateral are supplementary]

m DAP + m BAP + DCQ + m BCQ = 180º[Angle addition property]

x + x + y + y = 180º [From (i) and (ii)] 1 2x + 2y = 180º .....(iii)

m DAP =1

2 m (arc DP) [Inscribed angle theorem]

x =1

2m (arc DP) [From (i)]

m (arc DP) = 2xº .....(iv) ½

m DCQ =1

2 m (arc DQ) [Inscribed angle theorem]

y =1

2 m (arc DQ) [From (ii)]

D P

Q

A

C

B

••

××

... 5 ...

Paper - IV... 6 ...

m (arc DQ) = 2yº ......(v) ½m (arc DP) + m (arc DQ) = 2x + 2y [Adding (iv) and (v)]

m (arc PDQ) = 180º [Arc addition property and from (iii)] ½ Arc PDQ is a semicircle seg PQ is a diameter of the circle. ½

(iii) 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 cmInformation 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)

2 marks for analysis1 mark for LMN1 mark for XZY

Paper - V


(i) ABC ~ PQR, AB = 6 cm, BC = 8 cm, CA = 10 cm and QR = 6 cm.What is the length of side PR ?

(ii) ABCD is cyclic quadrilateral such that 2m A = 3m C. What isthe measure of C ?

(iii) What is the point of concurrence of the medians of a triangle called ?


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

(ii) In the adjoining figure,m (arc XAZ) = m (arx YBW).Prove that XY || ZW

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


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

S.S.C. Test - II



MAHESH TUTORIALS

P

QT R

4 cm5 cm

5.6 cmx

• •

X Y

A B

Z W

Paper - V

(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 ( ACF)2

(iii) Chords AB and CD of a circleintersect in point Q in the interiorof a circle as shown in figure,if m (arc AD) = 25º circle andm (arc BC) = 31º, then find BQC.


(i) 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.

(ii) In the adjoining figure, two circles toucheach other internally in a point A. The radiusof the smaller circle with centre M is 5.The smaller circle passes through the centre Nof the larger circle. The tangent to the smallercircle drawn through C intersects the larger circlein point D. Find CD.

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

... 2 ...

D B

QAC

F

CD

A B

E

A

N

M

CXBT

A

DP

MNC

Best Of Luck

Paper - V

A.1. Solve the following :(i) ABC ~ PQR [Given]

AB

PQ =BC

QR = AC

PR[c.s.s.t.] ½

6

PQ =8

6 =

10

PR

8

6=

10

PR

PR =6 10

8

PR =15

2

PR = 7.5 cm ½

(ii) 2m A = 3m C [Given]

m A = 3

m C2

.......(i)

ABCD is cyclicm A + m C = 180º [Opposite angles of a cyclic ½

quadrilateral is supplementary]

3

2m C + m C = 180º [From (i)]

3m C 2m C

2

= 180

5m C

2

= 180

m C =180 2

5

m C = 72º ½

(iii) The point of concurrence of the medians of a triangle is called 1

Centroid.

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - V... 2 ...

A.2. Solve the following :(i) In PQR,

ray PT bisects QPR[Given]

PQ

PR=

QT

TR[Property of angle bisector of a triangle] ½

5.6

x=

4

5

x =5 5.6

4

x = 7 ½ PR = 7 cm

QR = QT + TR [ Q - T - R] QR = 4 + 5 QR = 9 cm ½

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

Perimeter of PQR = 19.6 cm ½

(ii)

Construction : Draw seg XW ½

Proof : m XWZ = 1

2 m (arc XAZ) .......(i) ½

m WXY = 1

2 m (arc YBW) .......(ii)

But, m (arc XAZ) = m (arc YBW) .......(iii) [Given] ½ m XWZ = m WXY [From (i), (ii) and (iii)] XY || ZW [Alternate angles test] ½

(iii)

P

QT R

4 cm5 cm

5.6 cmx

• •

X Y

A B

Z W

[Inscribed angletheorem]

S

80º

V

T5.6 cm10º 10º160º

OS

80º

V

T5.6 cm

(Rough Figure)

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing SVB

Paper - V... 3 ...


I

4.4 cm

M

125º

6.5 cm S

O

I

4.4 cmM

125º6.5 cm S

(Rough Figure)

½ mark for rough figure1 mark for drawing triangle1 mark for drawing perpendicular bisectors½ mark for drawing circle

Paper - V

(ii)

BCE ~ ACF [Given]

A ( BCE)

A ( ACF)

=

2

2

BC

AC......(i) [Areas of similar triangles] ½

ABCD is a square [Given] AB = BC = CD = AD ......(ii) [Sides of a square] ½

In ABC,m ABC= 90º [Angle of a square]

AC2 = AB2 + BC2 [By Pythagoras theorem] ½ AC2 = BC2 + BC2 [From (i)] AC2 = 2BC2 .....(iii) ½

A ( BCE)

A ( ACF)

=

BC

2BC

2

2 [From (i) and (iii)] ½

A ( BCE)

A ( ACF)

=

1

2

A (BCE) =12

A (ACF) ½

(iii) m(arc AD) = 25º [Given]

m ACD =1

2 m(arc AD) [Inscribed ½

angle theorem]

m ACD =1

2 × 25º

m ACD = 12.5º m ACQ = 12.5º [ D - Q - C] ½

m (arc BC) = 31º [Given]

m BAC =1

2 m(arc BC) [Inscribed angle theorem] ½

m BAC =1

2 × 31º

m BAC = 15.5º m QAC = 15.5º [ A - Q - B] ½

BQC is an exterior angle of AQC, m BQC = m QAC + m ACQ

[Remote interior angle theorem] ½ m BQC = 15.5º + 12.5º

m BQC = 28º ½

F

CD

A B

E

D B

QAC

... 4 ...

Paper - V


seg AB || seg MX [Given]seg NB || seg MX [ A - N - B]On transversal TX,TBN TXM ......(i) [Converse of corresponding ½

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.] 1

seg NX ll seg AC [Given]seg NX ll seg MC [ A - M - C]On transversal TC,TXN TCM ......(iii) [Converse of corresponding ½

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.] 1

TB

TX =

TX

TC[From (ii) and (iv)] ½

TX2 = TB × TC ½

(ii)

Construction : Draw seg AD. ½Sol. NM = MA = MP = 5 units [Radii of same circle]NA = 2 NM [Diameter is twice the radius]

NA = 2 × 5 NA = 10 units

A

N

M

CXBT

A

DP

MNC

... 5 ...

Paper - V

CN = NA = 10 units [Radii of same circle]CA = 2 CN [Diameter is twice the radius]CA = 2 × 10CA = 20 unitsCM = CN+ NM [C - N - M]

CM = 10 + 5 CM = 15 units ½

m CPM = 90º [Radius is perpendicular to tangent]m CDA = 90º [Angle subtended by a semicircle] ½

CPM CDA seg PM || seg DA ......(i) [By corresponding angles test]

In CPMCPM = 90º [Radius is perpendicular to ½

the tangent]CM2 = CP2 + PM2 [By Pythagoras theorem]152 = CP2 + 52

225 = CP2 + 25 CP2 = 225 – 25 CP2 = 200

CP = 100 × 2 [Taking square roots] CP = 10 2 units ½

In CDA,seg PM || seg DA [From (i)]

CP

PD=

CM

MA[By B.P.T.] ½

10 2

PD=

15

5

PD =10 2 × 5

15

PD =10 2

3units ½

CD = CP + PD [C - P - D]

CD =10 2 10 2

1 3

CD =30 2 10 2

3

CD =40 2

3 units ½

... 6 ...

Paper - V

(iii)

C X

V

130º

1.7

cm

D9.1 cm

(Rough Figure)

V V

130º

C X40º D

1.7 cm

9.1 cm 40º

O

100º

A

B

1.7 cm

½ mark for rough figure1 mark for locating centre ‘O’1 mark for locating point V1 mark for drawing the altitude½ mark for drawing CVX

... 7 ...

Paper - VI


(i) The areas of two similar triangles are 18 cm2 and 32 cm2 respectively.What is the ratio of their corresponding sides ?

(ii) Two circles with centres P and Q having diameter 25 cm and 15 cmrespectively touch each other externally at A, then what is thedistance between P and Q ?

(iii) If the circumcircle lies in the interior of the triangle, then it is........... angled triangle.


(i) From the information given in theadjacent figure, state whether thetriangles are similar with reasons.

(ii) In the adjoining figure,if m (arc APC) = 60ºand m BAC = 80ºFind (a) ABC (b) m (arc BQC).

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

Q.3. Solve the following : 9(i) Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm,

R = 95º.

S.S.C. Test - II



MAHESH TUTORIALS

A

P

Q R

B C

4.62.3 4

8

10

5

B

A

C80º

•Q

•P

Paper - VI

(ii) ABCD is a trapezium in which AB || DC and its diagonals intersect

each other at the point O. Show that AO CO

=BO DO

.

(iii) In the adjoining figure,A and B are centers of twocircles touching each other at M.Line AC and line BD are tangents.If AD = 6 cm and BC = 9 cm then findthe length of seg AC and seg BD.


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

(ii) ABC is inscribed in a circle with centre O, seg AXis a diameter of the circle with radius r.seg AD seg BC. Prove that

(a) ABX ~ ADC, (b) A (ABC) = abc

4r.

(a is side opposite to A, ...)

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

D

A B

C

M

A

C

X

BY

A

B D C

O

X

... 2 ...

Best Of Luck

Paper - VI

A.1. Solve the following :(i) Let ABC ~ PQR

A ( ABC)

A ( PQR)

=

AB

PQ

2


18

32 =

AB

PQ

2

2 [Given]

9

16 =

AB

PQ

2

2

AB

PQ = 3

4[Taking square roots]

AB : PQ = 3 : 4 ½

(ii) Diameter of the cicle with centre P = 25 cm Its radius (r1) = 12.5 cm

Diameter of the circle with centre Q = 15 cm ½ Its radius (r2) = 7.5 cm

Both the circles touch each other externally at point A Distance between P and Q = r1 + r2

= 12.5 + 7.5

Distance between P and Q = 20 cm ½

(iii) If the circumcircle lies in the interior of the triangle, then it is 1acute angled triangle.

A.2. Solve the following :

(i)AB

PQ =4.6

2.3

AB

PQ =2

1......(i) ½

BC

QR =10

5

BC

QR =2

1......(ii) ½

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

A

P

Q R

B C

4.62.3 4

8

10

5

Paper - VI... 2 ...

AC

PR=

8

4

AC

PR=

2

1......(iii) ½

In ABC and PQR,AB

PQ =BC

QR = AC

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

ABC PQR [By SSS test of similarity] ½

(ii) (a) m ABC =1

2 m(arc APC) [Inscribed angle theorem] ½

m ABC =1

2 × 60

m ABC = 30º ½

(b) m BAC =1

2 m(arc BQC) [Inscribed angle theorem] ½

80 =1

2 m(arc BQC)

m(arc BQC) = 80 × 2

m(arc BQC) = 160º ½

(iii)

B

A

C80º

•Q

•P

MP 6.2 cm

40º

Q(Rough Figure)

MP 6.2 cm50º 50º

40º

Q

O

80º

½ mark for rough figure½ mark for drawing base angles½ mark for drawing an arc½ mark for drawing PQM

Paper - VI... 3 ...


(ii) ABCD is a trapeziumside AB || side DC [Given]

On transversal AC,BAC DCA [Converse ½

of alternate angles test] BAO DCO ......(i) [ A - O - C] ½

In AOB and COD,BAO DCO [From (i)]AOB COD [Vertically opposite angles]

AOB ~ COD [By AA test of similarity] 1

AO

CO =

BO

DO[c.s.s.t.] ½

AO CO

=BO DO

[By Alternendo] ½

(iii)

A - M - B [If two circles are touching ½ circles then the common point

lies on the line joining their centres]

5.9 cm NR

4.9 cm

S

95º ××••

O

5.9 cmNR

4.9 cm

S

95º

(Rough Figure)

A B

O

D C

D

A B

C

M

½ mark for rough figure½ mark for drawing SRN1 mark for drawing the angle bisectors1 mark for drawing the incircle

Paper - VI

AM = AD = 6 cm ........(i) [Radii of the same circle]BM = BC = 9 cm ........(ii)AB = AM + MB [ A - M - B]

AB = 6 + 9 [From (i) and (ii)] ½ AB = 15 cm ........(iii)

In ABC,m ACB = 90º [Radius is perpendicular to the tangent] ½

AB² = AC² + BC² [By Pythagoras theorem] 15² = AC² + 9² [From (ii) and (iii)] 225= AC² + 81 AC² = 225 – 81 AC² = 144 AC = 12 cm [Taking square roots] ½

In ADB,m ADB = 90º [Radius is perpendicular to the tangent] ½

AB2 = AD2 + BD2 [By Pythagoras theorem] 152 = 62 + BD2 [From (i) and (iii)] 225= 36 + BD2

BD2= 225 – 36 BD2= 189

BD = 9 21

BD = 3 21 cm. [Taking square roots] ½

The lengths of seg AC and seg BD are 12 cm and 3 21cmrespectively.

A.4. Solve the following :(i) seg XY divides ABC in two

parts of equal areas

A(XYB) = 1

2 A(ACB)

A ( XYB)

A ( ACB)

=

1

2.......(i) ½

seg XY || side AC [Given]

On transversal BC,

XYB ACB .....(ii) [Converse of corresponding ½

angles test]

In XYB and ACB,

XYB ACB [From (ii)]

XBY ABC [Common angle]

XYB ~ ACB [By AA test of similarity] ½

A

C

X

BY

... 4 ...

Paper - VI

A ( XYB)

A ( ACB)

=

XB

AB

2


1

2=

2

2

XB

AB[From (i)]

1

2=

XB

AB[Taking square roots] ½

XB

AB=

1

2

AB – AX

AB=

1

2[ A - X - B] ½

AB

AB –

AX

AB=

1

2½

AX

1 –AB

=1

2

1

1 –2

=AX

AB

AX

AB=

2 – 1

2½

(ii) seg AX is a diameter [Given]

arc ACX is a semi circle.

m ABX = 90º [Angle subtended by ½a semicircle]

In ABX and ADC

ABX ADC [Each is 90º]

AXB ACD [Angles inscribed in the same arc

and C - D - B]

ABX ~ ADC [By AA test of similarity] 1BC = a, AB = c, AC = b [Given]

AB

AD=

AX

AC[c.s.s.t.] ½

c

AD=

2r

b

AD =bc

2r.......(i) ½

A

B D C

O

X

... 5 ...

Paper - VI

A (ABC) =1

2 × base × height ½

=1

2 × BC × AD

=1

2 × a ×

bc

2r[From (i)] ½

A (ABC) =abc4r

½

(iii)

X Y

Z

115º

3.3

cm

P9.5 cm

(Rough Figure)

ZZ

Y9.5 cm25ºX P

3.3 cm

115º

25º

O

130º

½ mark for rough figure½ mark for drawing centre O1 mark for drawing the perpendicular bisector1 mark for locating point Z1 mark for drawing XZY

... 6 ...

mahesh tutorials ssc question paper with solution -geometry

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