cbse ncert solutions for class 10 mathematics chapter 6 ... · class- x-cbse-mathematics triangles...

Class- X-CBSE-Mathematics Triangles

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CBSE NCERT Solutions for Class 10 Mathematics Chapter 6

Back of Chapter Questions

EXERCISE 6.1

1. Fill in the blanks using the correct word given in brackets:

(i) All circles are _________. (congruent, similar)

(ii) All squares are _________. (similar, congruent)

(iii) All _________ triangles are similar. (Isosceles, equilateral)

(iv) Two polygons of the same number of sides re similar, if (a) their

corresponding angles are _________ and (b) their corresponding sides are

_________. (equal, proportional)

Solution:

(i) similar, since size of circles may be different, but shape will be always

same.

(ii) similar, since size of squares may be different, but shape will be always

same.

(iii) All equilateral triangles are similar because of their same shape.

(iv) Two polygons of same number of sides are similar, if (a) their

corresponding angles are equal and (b) their corresponding sides are

proportional.

2. Give two different examples of pair of

(i) Similar figures.

(ii) Non-similar figures.

Solution:

(i) Two equilateral triangles with sides 1cmand 2cm.

Two squares with sides 1cmand 2cm.


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(ii) Trapezium and Square

Triangle and Parallelogram

3. State whether the following quadrilaterals are similar or not:

Solution:

Corresponding sides of two quadrilaterals are proportional i. e. 1: 2 but their

corresponding angles are not equal. Hence, quadrilateral PQRS and ABCD are not

similar.

EXERCISE 6.2

1. In Fig. (i) and (ii), DE ∥ BC. Find EC in (i) and AD in (ii).


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(i)

(ii)

Solution:

(i)

Let EC = 𝑥cm

Since DE ∥ BC

Hence, using basic proportionality theorem,

AD

DB=

AE

EC

⇒1.5

3=

1

𝑥

⇒ 𝑥 =3 × 1

1.5

⇒ 𝑥 = 2

Hence, EC = 2cm


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(ii)

Let AD = 𝑥

Since DE ∥ BC,


AD

DB=

AE

EC

⇒𝑥

7.2=

1.8

5.4

⇒ 𝑥 =1.8 × 7.2

5.4

⇒ 𝑥 = 2.4

Hence, AD = 2.4cm

2. Eand F are points on the sides PQ and PR respectively of a ∆PQR. For each of the

following cases, state whether EF ∥ QR:

(i) PE = 3.9cm, EQ = 3cm, PF = 3.6cmandFR = 2.4cm

(ii) PE = 4cm, QE = 4.5cm, PF = 8cmandRF = 9cm

(ii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cmandPF = 0.36cm

Solution:

(i)


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Given, PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4

Now,

PE

EQ=

3.9

3= 1.3

PF

FR=

3.6

2.4= 1.5

Since, PE

EQ≠

PF

FR

Hence, EF is not parallel to QR.

(ii)

Given, PE = 4, QE = 4.5, PF = 8, RF = 9

PE

EQ=

4

4.5=

8

9

PF

FR=

8

9

Since PE

EQ=

PF

FR

Hence,EF ∥ QR (using basic proportionality theorem)


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(iii)

Given, PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36

∴ 𝐸𝑄 = 𝑃𝑄 − 𝑃𝐸 = 1.28 − 0.18 = 1.1 and 𝐹𝑅 = 𝑃𝑅 − 𝑃𝐹 = 2.56 −0.36 = 2.2

PE

EQ=

0.18

1.1=

18

110=

9

55

PF

FR=

0.36

2.2=

9

55

Since PE

EQ=

PF

FR

Hence, EF ∥ QR (using basic proportionality theorem)

3. In Fig. if LM ∥ CB and LN ∥ CD, prove that

AM

AB=

AN

AD

Solution:


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In the given figure

Since LM ∥ CB.


AM

MB=

AL

LC … (i)

Again, since LN ∥ CD


AN

ND=

AL

LC … (ii)

From (i) and (ii)

AM

MB=

AN

ND

⇒ MB

AM=

ND

AN

⇒ MB

AM+ 1 =

ND

AN+

⇒MB + AM

AM=

ND + AN

AN

⇒AB

AM=

AD

AN

⇒AM

AB=

AN

AD

4. In Figure, DE ∥ AC andDF ∥ AE. Prove that

BF

FE=

BE

EC

Solution:


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In ΔABC

Since DE ∥ AC

Hence, BD

DA=

BE

EC … (i) (using basic proportionality theorem)

In ∆BAE,

Since DF ∥ AE

Hence, BD

DA=

BF

FE … (ii) (using basic proportionality theorem)

From (i) and(ii)

BE

EC=

BF

FE

5. In Figure, DE ∥ OQ andDF ∥ OR. Show thatEF ∥ QR.

Solution:


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In ∆POQ

Since DE ∥ OQ

PE

EQ=

PD

DO … (i)[Using basic proportionality theorem]

In ∆POR

Since DF ∥ OR

PF

FR=

PD

DO … (ii)[Using basic proportionality theorem]

From (i)and (ii)

PE

EQ=

PF

FR

Using converse of basic proportionality theorem

EF ∥ QR

6. In Figure A, B andC are points onOP, OQ andOR respectively such thatAB ∥ PQ

andAC ∥ PR. Show that BC ∥ QR.


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Solution:

In ∆POQ

Since AB ∥ PQ,

Hence, OA

AP=

OB

BQ … (i)[Using basic proportionality theorem]

In ∆POR

Since AC ∥ PR

Hence, OA

AP=

OC

CR … (ii)[Using basic proportionality theorem]

From (i)and (ii)

OB

BQ=

OC

CR

Hence, BC ∥ QR

(Using converse of basic proportionality theorem)


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7. Prove that a line drawn through the mid-point of one side of a triangle parallel to

another side bisects the third side. (Recall that you have proved it in ClassIX).

Solution:

Let in the given figure PQ is a line segment drawn through mid-point P of line AB

such that PQ ∥ BC

Hence, AP = PB

Now, using basic proportionality theorem

AQ

QC=

AP

PB

⇒AQ

QC=

AP

AP

⇒AQ

QC= 1

⇒ AQ = QC

Hence, Q is the mid-point of AC.

8. Prove that the line joining the mid-points of any two sides of a triangle is parallel

to the third side. (Recall that you have done it in Class IX).

Solution:


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Let in the given figure PQ is a line segment joining mid-points PandQ of line

ABand AC respectively.

Hence, AP = PBandAQ = QC

Now, since AP

PB=

AP

AP= 1and

AQ

QC=

AQ

AQ= 1

Hence, AP

PB=

AQ

QC

Now, using converse of basic proportionality theorem PQ ∥ BC

9. ABCD is a trapezium in whichAB ∥ DC and its diagonals intersect each other at the

point O. Show that

AO

BO=

CO

DO

Solution:

Let a line segment EF is drawn through point O such that EF ∥ CD

In ∆ADC

EO ∥ CD

Hence, using basic proportionality theorem

AE

ED=

AO

OC … (1)

Similarly in ∆BDC


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FO ∥ CD

Hence, using basic proportionality theorem

BF

FC=

BO

OD … (2)

Now consider trapezium ABCD

As FE ∥ CD

So, AE

ED=

BF

FC … (3)

Now from equation (1), (2), (3)

AO

OC=

BO

OD

⇒AO

BO=

OC

OD

10. The diagonals of a quadrilateralABCD intersect each other at the pointO such that AO

BO=

CO

DO. Show that ABCD is a trapezium.

Solution:

Draw a line segment OE ∥ AB

In ∆ABC

Since OE ∥ AB

Hence,AO

OC=

BE

EC

But by the given relation, we have:

AO

BO=

CO

DO

⇒AO

OC=

OB

OD


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Hence, OB

OD=

BE

EC

So, using converse of basic proportionality theorem, EO ∥ DC

Therefore AB ∥ OE ∥ DC

⇒ AB ∥ CD

Therefore, ABCD is a trapezium.

EXERCISE 6.3

1. State which pairs of triangles in Fig. are similar. Write the similarity criterion

used by you for answering the question and also write the pairs of similar

triangles in the symbolic form:

(i)

(ii)

(iii)

(iv)


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(v)

(vi)

Solution:

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Hence by AAA rule ∆ABC~∆PQR

(ii) AB

QR=

2

4=

1

2

BC

RP=

2.5

5=

1

2

CA

PQ=

3

6=

1

2

Since, AB

QR=

BC

RP=

CA

PQ

Hence, by SSS rule

∆ABC~∆QRP

(iii) Triangles are not similar as the corresponding sides are not proportional.

(iv) MN

PQ=

2.5

5=

1

2

ML

QR=

5

10=

1

2

∠M = ∠Q = 70o

Hence, by SAS rule

∆MNL~∆PQR


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(v) Triangles are not similar as the corresponding sides are not proportional.

(vi) In ∆DEF

∠D + ∠E + ∠F = 180o(Sum of angles of a triangle is 180o)

⇒ 70o + 80o + ∠F = 180o

⇒ ∠F = 30o

Similarly in ∆PQR

∠P + ∠Q + ∠R = 180o (Sum of angles of a triangle is 180o)

⇒ ∠P + 80o + 30o = 180o

⇒ ∠P = 70o

Now, since

∠D = ∠P = 70o

∠E = ∠Q = 80o

∠F = ∠R = 30o

Hence, by AAA rule

∆DEF~∆PQR

2. In Figure, ∆ODC~∆OBA, ∠BOC = 125oand ∠CDO = 70o. Find ∠DOC, ∠DCO

and∠OAB.

Solution:

Since DOB is a straight line

Hence, ∠DOC + ∠COB = 180o

⇒ ∠DOC = 180o − 125o = 55o

In∆DOC,

∠DCO + ∠CDO + ∠DOC = 180o

⇒ ∠DCO + 70o + 55o = 180o

⇒ ∠DCO = 55o


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Since, ∆ODC~∆OBA,

Hence, ∠OCD = ∠OAB [Corresponding angles equal in similar triangles]

Hence, ∠OAB = 55o

3. DiagonalsAC andBD of a trapeziumABCD withAB ∥ DC intersect each other at the

pointO. Using a similarity criterion for two triangles, show that OA

OC=

OB

OD

Solution:

In ∆DOCand∆BOA

AB||CD

Hence,∠CDO = ∠ABO[Atlernate interior angles]

∠DCO = ∠BAO[Atlernate interior angles]

∠DOC = ∠BOA [Vertically opposite angles]

Hence,∆DOC~∆BOA [AAArule]

⇒DO

BO=

OC

OA [Corresponding sides are proportional]

⇒OA

OC=

OB

OD

4. In Figure, QR

QS=

QT

PR and∠1 = ∠2. Show that ∆PQS~∆TQR.


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Solution:

In ∆PQR

∠PQR = ∠PRQ

Hence, PQ = PR … (i)

Given

QR

QS=

QT

PR

Using (i)

QR

QS=

QT

PQ… (ii)

Also, ∠RQT = ∠PQS = ∠1

Hence, by SAS rule

∆PQS~∆TQR

5. S andT are points on sidesPR andQR of ∆PQRsuch that∠P = ∠RTS. Show that

∆RPQ~∆RTS.

Solution:


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In ∆RPQand∆RTS

∠QPR = ∠RTS [Given]

∠R = ∠R [Common angle]

∠RQP = ∠RST [Remaining angle]

Hence, ∆RPQ~∆RTS [byAAA rule]

6. In Fig., if∆ABE ≅ ∆ACD, show that∆ADE~∆ABC.

Solution:

Since ∆ABE ≅ ∆ACD

Therefore AB = AC … (1)

AE = AD

⇒ AD = AE … (2)

Now, in ∆ADE and ∆ABC,

Dividing equation (2)by (1)

AD

AB=

AE

AC

∠A = ∠A [Common angle]

Hence,∆ADE~∆ABC [by SAS rule]

7. In Fig., altitudesAD andCEof∆ABC intersect each other at the pointP. Show that:

(i) ∆AEP~∆CDP

(ii) ∆ABD~∆CBE

(iii) ∆AEP~∆ADB

(iii) ∆PDC~∆BEC


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Solution:

(i)

In ∆AEP and ∆CDP

∠CDP = ∠AEP = 90o

∠CPD = ∠APE(Vertically opposite angles)

∠PCD = ∠PAE(Remaining angle)

Hence, by AAArule,

∆AEP~∆CDP

(ii)

In ∆ABDand∆CBE

∠ADB = ∠CEB = 90o

∠ABD = ∠CBE (Common angle)

∠DAB = ∠ECB(Remaining angle)

Hence, by AAArule,

∆ABD~∆CBE


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(iii)

In ∆AEPand∆ADB

∠AEP = ∠ADB = 90o

∠PAE = ∠DAB (Common angle)

∠APE = ∠ABD (Remaining angle)

Hence, by AAA rule,

∆AEP~∆ADB

(iv)

In ∆PDCand∆BEC

∠PDC = ∠BEC = 90o

∠PCD = ∠BCE (Common angle)

∠CPD = ∠CBE (Remaining angle)

Hence, by AAArule,

∆PDC~∆BEC

8. Eis a point on the sideAD produced of a parallelogram ABCD andBE intersectsCD

atF. Show that ∆ABE~∆CFB.

Solution:


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∆ABEand∆CFB

∠A = ∠C (Opposite angles of parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE ∥ BC)

∠ABE = ∠CFB (Alternate interior angles as AB ∥ DC)

Hence, by AAArule,

∆ABE~∆CFB

9. In Figure,ABC andAMP are two right triangles, right angled atB andMrespectively.

Prove that:

(i) ∆ABC~∆AMP

(ii) CA

PA=

BC

MP

Solution:

(i) In ∆ABCand∆AMP

∠ABC = ∠AMP = 90o


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∠A = ∠A(Common angle)

∠ACB = ∠APM(Remaining angle)

Hence, by AAArule,

∆ABC~∆AMP

(ii) Since, ∆ABC~∆AMP

Hence, CA

PA=

BC

MP (Corresponding sides are proportional)

10. CD andGH are respectively the bisectors of∠ACB and∠EGF such thatD andH lie

on sidesAB andFE of∆ABC and∆EFG respectively. If∆ABC~∆FEG, show that:

(i) CD

GH=

AC

FG

(ii) ∆DCB~∆HGE

(iii) ∆DCA~∆HGF

Solution:

Since ∆ABC~∆FEG

Hence, ∠A = ∠F

∠B = ∠E

∠ACB = ∠FGE

⇒∠ACB

2=

∠FGE

2

⇒ ∠ACD = ∠FGH(Angle bisector)

And ∠DCB = ∠HGE (Angle bisector)

In∆ACDand∆FGH

∠A = ∠F

∠ACD = ∠FGH (Angle bisector)


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∠ADC = ∠FHG (Remaining angle)

Hence, by AAA rule,

∆ACD~∆FGH

So, CD

GH=

AC

FG (Corresponding sides are proportional)

In∆DCBand∆HGE

∠B = ∠E

∠DCB = ∠HGE (Angle bisector)

∠BDC = ∠EHG (Remaining angle)

Hence, by AAA rule,

∆DCB~∆HGE

11. In Fig.,E is a point on sideCB produced of an isosceles triangleABC with AB =

AC. IfAD ⊥ BC andEF ⊥ AC, prove that∆ABD~∆ECF.

Solution:

In∆ABD and∆ECF,

Since, AB = AC (isosceles triangles)

So, ∠ABD = ∠ECF

∠ADB = ∠EFC = 90o

∠BAD = ∠CEF (Remaining angle)

Hence, by AAA rule,

∆ABD~∆ECF

12. SidesAB andBC and medianAD of a triangleABC are respectively proportional to

sidesPQ andQR and medianPM of ∆PQR(see Fig.). Show that ∆ABC~∆PQR.


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Solution:

Median divides opposite side.

So, BD =BC

2 andQM =

QR

2

Given that

AB

PQ=

BC

QR=

AD

PM

So, AB

PQ=

BD

QM=

AD

PM

Hence, by SSSrule

∆ABD~∆PQM

Hence, ∠ABD = ∠PQM (Corresponding angles of similar triangles)

Hence, ∠ABC = ∠PQR

And AB

PQ=

BC

QR

Hence, by SASrule

∆ABC~∆PQR

13. Dis a point on the sideBC of a triangleABC such that∠ADC = ∠BAC. Show that

CA2 = CB. CD.

Solution:

In ∆ACDand∆BAC

Given that ∠ADC = ∠BAC


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∠ACD = ∠BCA (Common angle)

∠CAD = ∠CBA (Remaining angle)

Hence, by AAA rule,

∆ADC~∆BAC

So, corresponding sides of similar triangles will be proportional to each other

CA

CB=

CD

CA

Hence CA2 = CB × CD

14. SidesAB andAC and medianAD of a triangleABC are respectively proportional to

sides PQand PR and medianPM of another trianglePQR. Show that∆ABC~∆PQR.

Solution:

Given that

AB

PQ=

AC

PR=

AD

PM

Let us extend ADandPM up to point EandL respectively such that AD = DE and

PM = ML. Now join Bto E, CtoE,QtoLandRto L.

We know that medians divide opposite sides.


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So, BD = DCand QM = MR

Also, AD = DE (by construction)

And PM = ML(By construction)

So, in quadrilateral ABEC, diagonals AEandBC bisects each other at point D.

Also,in quadrilateral PQLR, diagonals PLandQR bisects each other at point M.

So, quadrilaterals ABCD and PQLR are a parallelogram.

AC = BE and AB = EC (Since it is a parallelogram, opposite sides will be

equal)

Also PR = QL and PQ = LR (Since it is a parallelogram, opposite sides will be

equal)

In ∆ABEand∆PQL,

AB

PQ=

BE

QL=

AE

PL (

AC

PR=

BE

QL𝑎𝑛𝑑

AD

PM=

2AD

2PM=

AE

PL)

Hence, by SSS rule,

∆ABE~∆PQL

Similarly, ∆AEC~∆PLR

Hence, ∠BAE = ∠QPL and ∠EAC = ∠LPR

Hence, ∠BAC = ∠QPR

Now, in ∆ABCand∆PQR,

AB

PQ=

AC

PR and ∠BAC = ∠QPR

Hence, by SAS rule,

∆ABC~∆PQR

15. A vertical pole of length6m casts a shadow4m long on the ground and at the

same time a tower casts a shadow28m long. Find the height of the tower.

Solution:


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Let AB be a tower and CD be a pole

Shadow of ABisBE

Shadow of CDisDF

The sun ray will fall on tower and pole at same angle.

So, ∠DCF = ∠BAE and ∠DFC = ∠BEA

∠CDF = ∠ABE = 90o (Tower and pole are vertical to ground)

Hence, by AAA rule,

∆ABE~∆CDF

Therefore AB

CD=

BE

DF

⇒AB

6=

28

4

⇒ AB = 42

Hence, height of tower = 42meters

16. IfAD andPM are medians of trianglesABC andPQR, respectively

where∆ABC~∆PQR, prove that AB

PQ=

AD

PM

Solution:


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Since ∆ABC~∆PQR

So, their respective sides will be in proportion

Or, AB

PQ=

AC

PR=

BC

QR … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since, ADandPM are medians. So, they will divide their opposite sides equally.

Hence, BD =BC

2 and QM =

QR

2… (3)

From equation (1)and(3)

AB

PQ=

BD

QM

∠B = ∠Q (From equation2)

Hence, by SASrule

∆ABD~∆PQM

Hence, AB

PQ=

AD

PM (Corresponding sides are proportional)

EXERCISE 6.4

1. Let ∆ABC~∆DEF and their areas be, respectively,64cm2 and121cm2. IfEF =15.4cm, findBC.

Solution:

Since, ∆ABC~∆DEF

Hence, area(∆ABC)

area(∆DEF)= (

AB

DE)

2

= (BC

EF)

2

= (AC

DF)

2

Since EF = 15.4,area(∆ABC) = 64; area(∆DEF) = 121

Hence, 64

121=

BC2

15.42


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⇒BC

15.4=

8

11

⇒ BC =8×15.4

11= 8 × 1.4 = 11.2cm.

2. Diagonals of a trapeziumABCD withAB ∥ DC intersect each other at the pointO.

IfAB = 2CD, find the ratio of the areas of trianglesAOB andCOD.

Solution:

Since AB ∥ CD

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite angles)

Hence, by AAA rule,

∆AOB~∆COD

Hence, area(∆AOB)

area(∆COD)= (

AB

CD)

2

Since AB = 2CD

area(∆AOB)

area(∆COD)=

4

1= 4: 1

3. In Figure,ABC andDBC are two triangles on the same baseBC. IfAD intersectsBC

atO, show that

ar(ABC)

ar(DBC)=

AO

DO.

Solution:


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We know that area of a triangle =1

2× Base × height

Since ∆ABCand∆DBC are on same base,

Hence, ratio of their areas will be same as ratio of their heights.

Let us draw two perpendiculars APandDM on BC.

In ∆APO and ∆DMO

∠APO = ∠DMO = 90o ∠AOP = ∠DOM (Vertically opposite angles)

∠OAP = ∠ODM (Remaining angle)

Hence, by AAArule

∆APO ~ ∆DMO

Hence, AP

DM=

AO

DO

Hence, area(∆ABC)

area(∆DBC)=

AP

DM=

AO

DO

4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Let us assume that ∆ABC ~ ∆PQR

Now, area(∆ABC)

area(∆PQR)= (

AB

PQ)

2

= (BC

QR)

2

= (AC

PR)

2

Since area(∆ABC) =area(∆PQR)

Hence, AB = PQ

BC = QR

AC = PR

Since, corresponding sides of two similar triangles are of same length

Hence, ∆ABC ≅ ∆PQR (by SSSrule)


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5. D, E andF are respectively the mid-points of sidesAB, BC andCA of∆ABC. Find the

ratio of the areas of∆DEF and∆ABC.

Solution:

Since D and E are mid-points of AB and BC of ∆ABC

Hence, DE ∥ AC andDE =1

2AC (by mid-point theorem)

Similarly, EF =1

2AB and DF =

1

2BC

Now in ∆ABC and ∆EFD

AB

EF=

BC

FD=

CA

DE= 2

Hence, by SSSrule

∆ABC~∆EFD

Hence, area(∆ABC)

area(∆DEF)= (

AC

DE)

2

= 4

⇒area(∆DEF)

area(∆ABC)=

1

4= 1: 4

6. Prove that the ratio of the areas of two similar triangles is equal to the square of

the ratio of their corresponding medians.

Solution:


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Let us assume that ∆ABC~∆PQR. Let ADand PSbe the medians of these triangles.

So, AB

PQ=

BC

QR=

AC

PR… (1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

Since, AD and PS are medians

So, BD = DC =BC

2 and QS = SR =

QR

2

So, equation (1) becomes

AB

PQ=

BD

QS=

AC

PR

Now in ∆ABDand ∆PQS

∠B = ∠Q and, AB

PQ=

BD

QS

Hence, ∆ABD~∆PQS

Hence, AB

PQ=

BD

QS=

AD

PS… (2)

Since, area(∆ABC)

area(∆PQR)= (

AB

PQ)

2

⇒area(∆ABC)

area(∆PQR)= (

AD

PS)

2

[from equation (2)]

7. Prove that the area of an equilateral triangle described on one side of a square is

equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let ABCD be a square of side 𝑎. Therefore it's diagonal = √2𝑎

Let ∆ABEand∆DBF are two equilateral triangles.

Hence, AB = AE = BE = 𝑎 and DB = DF = BF = √2𝑎

We know that all angles of equilateral triangles are 60o.


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Hence, all equilateral triangles are similar to each other.

Hence, ratio of areas of these triangles will be equal to the square of the ratio

between sides of these triangles.

areaof∆ABE

areaof∆DBF= (

𝑎

√2𝑎)

2

=1

2

Hence, areaof∆ABE =1

2(areaof∆DBF)

8. ABC andBDE are two equilateral triangles such thatD is the mid-point ofBC. Ratio

of the areas of trianglesABC andBDE is

(A) 2: 1

(B) 1: 2

(C) 4: 1

(D) 1: 4

Solution: (C)

Since all angle of equilateral triangles are 60o. Hence, all equilateral triangles are

similar to each other. So ratio between areas of these triangles will be equal to the

square of the ratio between sides of these triangles.

Let side of ∆ABC = 𝑎

Hence, side of ∆BDE =𝑎

2

Hence, area(∆ABC)

area(∆BDE)= (

𝑎𝑎

2

)2

=4

1= 4: 1

9. Sides of two similar triangles are in the ratio4: 9. Areas of these triangles are in

the ratio

(A) 2: 3

(B) 4: 9

(C) 81: 16


Page - 35

(D) 16: 81

Solution: (D)

If, two triangles are similar to each other, ratio between areas of these triangles

will be equal to the square of the ratio between sides of these triangles.

Given that sides are in the ratio4: 9.

Hence, ratio between areas of these triangles = (4

9)

2

=16

81= 16: 81

EXERCISE 6.5

1. Sides of triangles are given below. Determine which of them are right triangles. In

case of a right triangle, write the length of its hypotenuse.

(i) 7cm, 24cm, 25cm

(ii) 3cm, 8cm, 6cm

(iii) 50cm, 80cm, 100cm

(iv) 13cm, 12cm, 5cm

Solution:

(i) Given that sides are 7cm, 24cm and 25cm.

Squaring the lengths of these sides we get 49,576 and 625.

Clearly, 49 + 576 = 625or 72 + 242 = 252.

Since, given triangle satisfies Pythagoras theorem. So, it is a right triangle.

As we know that the longest side in a right triangle is the hypotenuse.

Hence, length of hypotenuse = 25cm.

(ii) Given that sides are 3cm, 8cm and 6cm.

Squaring the lengths of these sides we may get 9,64 and 36.

Clearly, sum of squares of lengths of two sides is not equal to square of

length of third side. Hence, given triangle is not satisfying Pythagoras

theorem. So, it is not a right triangle

(iii) Given that sides are 50cm, 80cmand 100cm.

Squaring the lengths of these sides we may get 2500,6400and 10000.

Clearly, sum of squares of lengths of two sides is not equal to square of

length of third side. Therefore, given triangle is not satisfying Pythagoras

theorem. So, it is not a right triangle.

(iv) Given that sides are 13cm, 12cm and 5cm.

Squaring the lengths of these sides we may get 169,144 and 25.


Page - 36

Clearly, 144 + 25 = 169 or 122 + 52 = 132.

Since, given triangle is satisfying Pythagoras theorem. So, it is a right

triangle. As we know that the longest side in a right triangle is the

hypotenuse.

Hence, length of hypotenuse = 13cm.

2. PQR is a triangle right angled atP andM is apoint onQR such thatPM ⊥ QR. Show

that PM2 = QM ∙ MR.

Solution:

Let ∠MPR = 𝑥

In ∆MPR

∠MRP = 180o − 90o − 𝑥

⇒ ∠MRP = 90o − 𝑥

Similarly in ∆MPQ

∠MPQ = 90o − ∠MPR = 90o − 𝑥

∠MQP = 180o − 90o − (90o − 𝑥)

⇒ ∠MQP = 𝑥

Now in ∆MPQ and ∆MRP, we may observe that

∠MPQ = ∠MRP

∠PMQ = ∠RMP

∠MQP = ∠MPR

Hence, by AAA rule,

∆MPQ~∆MRP

Hence, QM

PM=

MP

MR

⇒ PM2 = QM ∙ MR

3. In Fig., ABD is a triangle right angled atA andAC ⊥ BD. Show that


Page - 37

(i) AB2 = BC ∙ BD

(ii) AC2 = BC ∙ DC

(iii) AD2 = BD ∙ CD

Solution:

(i) In ∆ABCand ∆ABD

∠CBA = ∠DBA (common angles)

∠BCA = ∠BAD = 90°

∠BAC = ∠BDA (remaining angle)

Therefore, ∆ABC~∆ABD (by AAA)

∴ AB

BD=

BC

AB

⇒ AB2 = BC ∙ BD

(ii) Let ∠CAB = 𝑥

In ∆CBA

∠CBA = 180o − 90o − 𝑥

∠CBA = 90o − 𝑥

Similarly in ∆CAD

∠CAD = 90o − ∠CAB = 90o − 𝑥

∠CDA = 180o − 90o − (90o − 𝑥)

∠CDA = 𝑥

Now in ∆CBAand ∆CAD,we may observe that

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA = 90o

Therefore ∆CBA~∆CAD(by AAA rule)

Therefore, AC

DC=

BC

AC


Page - 38

⇒ AC2 = DC × BC

(iii) In∆DCA&∆DAB

∠DCA = ∠DAB = 90°

∠CDA = ∠ADB (Common angle)

∠DAC = ∠DBA(remaining angle)

∆DCA~∆DAB (byAAAproperty)

Therefore, DC

DA=

DA

DB

⇒ AD2 = BD × CD

4. ABC is an isosceles triangle right angled atC. Prove that AB2 = 2AC2.

Solution:

Given that ∆ABC is an isosceles triangle.

Therefore, AC = CB

Applying Pythagoras theorem in ∆ABC (i.e. right angled at point C)

AC2 + CB2 = AB2

2AC2 = AB2 (as AC = CB)

5. ABC is an isosceles triangle withAC = BC. IfAB2 = 2AC2, prove thatABC is a

right triangle.

Solution:

Given that

AB2 = 2AC2


Page - 39

⇒ AB2 = AC2 + AC2

⇒ AB2 = AC2 + BC2 (as AC = BC)

Since triangle is satisfying the Pythagoras theorem

Therefore, given triangle is a right angled triangle.

6. ABC is an equilateral triangle of side2𝑎. Find each of its altitudes.

Solution:

Let AD be the altitude in given equilateral ∆ABC.

We know that altitude bisects the opposite side.

So, BD = DC = a

in ∆ADB

∠ADB = 90o

Now applying Pythagoras theorem

AD2 + BD2 = AB2

⇒ AD2 + a2 = (2a)2

⇒ AD2 + a2 = 4a2

⇒ AD2 = 3a2

⇒ AD = a√3

Since in an equilateral triangle, all the altitudes are equal in length.

So, length of each altitude will be √3a

7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of

the squares of its diagonals.

Solution:


Page - 40

In ∆AOB, ∆BOC, ∆COD, ∆AOD

Applying Pythagoras theorem

AB2 = AO2 + OB2

BC2 = BO2 + OC2

CD2 = CO2 + OD2

AD2 = AO2 + OD2

Adding all these equations,

AB2 + BC2 + CD2 + AD2 = 2(AO2 + OB2 + OC2 + OD2)

= 2 ((AC

2)

2

+ (BD

2)

2

+ (AC

2)

2

+ (BD

2)

2

) (diagonals bisect each other.)

= 2 ((AC)2

2+

(BD)2

2)

= (AC)2 + (BD)2

8. In Fig.,O is a point in the interior of a triangleABC, OD ⊥ BC, OE ⊥ AC and OF ⊥AB. Show that

(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2,

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Solution:

(i) In ∆AOF

Applying Pythagoras theorem

OA2 = OF2 + AF2

Similarly in ∆BOD

OB2 = OD2 + BD2

similarly in ∆COE

OC2 = OE2 + EC2


Page - 41

Adding these equations

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2

⇒ OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + EC2

(ii) As from above result

AF2 + BD2 + EC2 = (OA2 − OE2) + (OC2 − OD2) + (OB2 − OF2)

Therefore, AF2 + BD2 + EC2 = AE2 + CD2 + BF2

9. A ladder10m long reaches a window8m above the ground. Find the distance of

the foot of the ladder from base of the wall.

Solution:

Let OAbe the wall and ABbe the ladder

Therefore by Pythagoras theorem,

AB2 = OA2 + BO2

⇒ 102 = 82 + OB2

⇒ 100 = 64 + OB2

⇒ OB2 = 36

⇒ OB = 6

Therefore, distance of foot of ladder from of the wall = 6m

10. A guy wire attached to a vertical pole of height18m is24m long and has a stake

attached to the other end. How far from the base of the pole should the stake be

driven so that the wire will be taut?

Solution:


Page - 42

Let OB be the pole and AB be the wire.

Therefore, by Pythagoras theorem,

AB2 = OB2 + OA2

⇒ 242 = 182 + OA2

⇒ OA2 = 576 − 324

⇒ 𝑂𝐴 = √252 = √6 × 6 × 7 = 6√7

Therefore distance from base = 6√7m

11. An aeroplane leaves an airport and flies due north at a speed of 1000km per hour.

At the same time, another aeroplane leaves the same airport and flies due west at a

speed of1200 km per hour. How far apart will be the two planes after11

2 hours?

Solution:

Distance traveled by the plane flying towards north in 11

2hrs

= 1,000 × 11

2= 1,500km

Distance traveled by the plane flying towards west in 11

2hrs

= 1,200 × 11

2= 1,800km

Let these distances are represented by OA and OB respectively.

Now applying Pythagoras theorem

Distance between these planes after 11

2hrs, AB = √OA2 + OB2

= √(1,500)2 + (1,800)2 = √2250000 + 3240000

= √5490000 = √9 × 610000 = 300√61

So, distance between these planes will be 300√61km. after 11

2hrs.


Page - 43

12. Two poles of heights6m and11m stand on a plane ground. If the distance between

the feet of the poles is12m, find the distance between their tops.

Solution:

Let CD and AB be the poles of height 11𝑚and 6m.

Therefore, CP = 11 − 6 = 5m

From the figure we may observe that AP = 12m

In ∆APC, by applying Pythagoras theorem

AP2 + PC2 = AC2

⇒ 122 + 52 = AC2

⇒ AC2 = 144 + 25 = 169

⇒ AC = 13 Therefore, distance between their tops = 13m.

13. D andE are points on the sidesCA andCB respectively of a triangleABC right

angled atC. Prove thatAE2 + BD2 = AB2 + DE2.

Solution:

In∆ACE,

AC2 + CE2 = AE2 … (i)

In ∆BCD,

BC2 + CD2 = BD2 … (ii)

Adding (i) and (ii)

AC2 + CE2 + BC2 + CD2 = AE2 + BD2 … (iii)


Page - 44

⇒ CD2 + CE2 + AC2 + BC2 = AE2 + BD2

In ∆CDE

DE2 = CD2 + CE2

In ∆ABC

AB2 = AC2 + CB2

Putting the values in equation (iii)

DE2 + AB2 = AE2 + BD2

14. The perpendicular fromA on side BCof a∆ABC intersectsBC atD such thatDB =3CD(see Fig.). Prove that 2AB2 = 2AC2 + BC2.

Solution:

Given that 3DC = DB

DC =BC

4[DB: DC = 3: 1] … (1)

and

DB =3BC

4… (2)

In ∆ACD

AC2 = AD2 + DC2

AD2 = AC2 − DC2 … (3)

In ∆ABD

AB2 = AD2 + DB2

AD2 = AB2 − DB2 … (4)

From equation (3) and (4)

AC2 − DC2 = AB2 − DB2

Since, given that 3DC = DB


Page - 45

AC2 − (BC

4)

2

= AB2 − (3BC

4)

2

(from(1)and (2))

⇒ AC2 −BC2

16= AB2 −

9BC2

16

⇒ 16AC2 − BC2 = 16AB2 − 9BC2

⇒ 16AB2 − 16AC2 = 8BC2

⇒ 2AB2 = 2AC2 + BC2

15. In an equilateral triangleABC, Dis a point on sideBC such thatBD =1

3BC. Prove

that 9AD2 = 7AB2.

Solution:

Let side of equilateral triangle be a, and AEbe the altitude of∆ABC

So, BE = EC =BC

2=

a

2

And, AE =a√3

2

Given that BD =1

3BC =

a

3

So, DE = BE − BD =a

2−

a

3=

a

6

Now, in ∆ADEby applying Pythagoras theorem

AD2 = AE2 + DE2

⇒ AD2 = (a√3

2)

2

+ (a

6)

2

= (3a2

4) + (

a2

36) =

28a2

36

or, 9AD2 = 7AB2

16. In an equilateral triangle, prove that three times the square of one side is equal to

four times the square of one of its altitudes.

Solution:


Page - 46

Let side of equilateral triangle be 𝑎. And AE be the altitude of ∆ABC

So, BE = EC =BC

2=

𝑎

2

And, AE =𝑎√3

2

Now in ∆ABEby applying Pythagoras theorem

AB2 = AE2 + BE2

⇒ 𝑎2 = AE2 + (𝑎

2)

2

⇒ AE2 = 𝑎2 −𝑎2

4

⇒ AE2 =3𝑎2

4

⇒ 4AE2 = 3𝑎2

or, 4AE2 = 3 × square of one side

17. Tick the correct answer and justify: In∆ABC, AB = 6√3cm, AC = 12cm andBC =6cm. The angleB is:

(A) 120o

(B) 60o

(C) 90o

(D) 45o

Solution: (C)


Page - 47

Given that AB = 6√3cm, AC = 12cm and BC = 6cm

We may observe that

AB2 = 108AC2 = 144

And BC2 = 36

AB2 + BC2 = AC2

Thus the given ∆ABC is satisfying Pythagoras theorem Therefore triangle is a

right angled triangle right angled at B

Therefore, ∠B = 90o.

EXERCISE 6.6 (Optional)*

1. In Fig.,PS is the bisector of∠QPR of ∆PQR. Prove that

QS

SR=

PQ

PR

Solution:

Let us draw a line segment RT parallel to SP which intersects extended line

segment QP at point T.

Given that PS is angle bisector of ∠QPR

∠QPS = ∠SPR (1)

∠SPR = ∠PRT (As PS || TR, alternate interior angles) (2)

∠QPS = ∠QTR (As PS || TR, corresponding angles) (3)


Page - 48

Using these equations we may find

∠PRT = ∠QTR from (2) and (3)

So, PT = PR (Since ΔPTR is isosceles triangle)

Now in ΔQPS andΔQTR

∠QSP = ∠QRT (As PS || TR)

∠QPS = ∠QTR (As PS || TR)

∠Q is common

ΔQPS~ΔQTR (by AAA property)

So, QR

QS=

QT

QP

⇒ QR

QS− 1 =

QT

QP− 1

⇒𝑆𝑅

𝑄𝑆=

PT

QP

⇒QS

SR=

QP

PT

⇒QS

SR=

PQ

PR

2. In Fig.,D is a point on hypotenuseAC of ∆ABC, such thatBD ⊥ AC, DM ⊥ BC and

DN ⊥ AB. Prove that:

(i) DM2 = DN ∙ MC

(ii) DN2 = DM ∙ AN

Solution:

(i) Let us join DB.


Page - 49

DN||CB

DM||AB

So, DN = MB

DM = NB

Then ∠CDB = ∠ADB = 90°

∠2 + ∠3 = 90o … (1)

In ΔCDM

∠1 + ∠2 + ∠DMC = 180°

∠1 + ∠2 = 90° … (2)

In ΔDMB

∠3 + ∠DMB + ∠4 = 180°

∠3 + ∠4 = 90° … (3)


∠1 = ∠3


∠2 = ∠4

ΔBDM~ΔDCM

BM

DM=

DM

MC

⇒DN

DM=

DM

MC

⇒ DM2 = DN × MC

(ii) Similarly in ΔDBN


Page - 50

∠4 + ∠3 = 90o …(4)

In ∆DAN

∠5 + ∠6 = 90o…(5)

In ΔDAB

∠4 + ∠5 = 90o …(6)


∠3 = ∠5


∠4 = ∠6

ΔDNA~ΔBND

AN

DN=

DN

NB

⇒ DN2 = AN × NB

⇒ DN2 = AN × DM

(as NB = DM)

3. In Fig., ABC is a triangle in which ∠ABC > 90o and AD ⊥ CB produced. Prove

that

AC2 = AB2 + BC2 + 2BC ∙ BD.

Solution:

In ΔADB, applying Pythagoras theorem

AB2 = AD2 + DB2 … (1)

In ΔACD, applying Pythagoras theorem

AC2 = AD2 + DC2

⇒ AC2 = AD2 + (DB + BC)2

⇒ AC2 = AD2 + DB2 + BC2 + 2DB × BC

Now using equation (1)


Page - 51

AC2 = AB2 + BC2 + 2BC. BD

4. In Fig., ABC is a triangle in which ∠ABC < 90o and AD ⊥ BC. Prove that AC2 =AB2 + BC2 − 2BC. BD.

Solution:

In ΔADB, applying Pythagoras theorem

AD2 + DB2 = AB2

⇒ AD2 = AB2 − DB2 … (1)

In ΔADC applying Pythagoras theorem

D2 + DC2 = AC2 (2)

Now using equation (1)

AB2 − BD2 + DC2 = AC2

⇒ AB2 − BD2 + (BC − BD)2 = AC2

⇒ AC2 = AB2 − BD2 + BC2 + BD2 − 2BC. BD

= AB2 + BC2 − 2BC. BD

5. In Fig., ADis a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) AC2 = AD2 + BC ∙ DM + (BC

2)

2

(ii) AB2 = AD2 − BC ∙ DM + (BC

2)

2

(iv) AC2 + AB2 = 2AD2 +1

2BC2

Solution:

(i) In ΔAMD


Page - 52

AM2 + MD2 = AD2 … (1)

In ΔAMC

AM2 + MC2 = AC2 … (2)

⇒ AM2 + (MD + DC)2 = AC2

⇒ (AM2 + MD2) + DC2 + 2MD. DC = AC2

Using equation (1) we may get

AD2 + DC2 + 2MD. DC = AC2

Now using the result, DC =BC

2

AD2 + (BC

2)

2

+ 2MD. (BC

2) = AC2

⇒ AD2 + (BC

2)

2

+ MD × BC = AC2

(ii) In ΔABM, applying Pythagoras theorem

AB2 = AM2 + MB2

= (AD2 − DM2) + MB2

= (AD2 − DM2) + (BD − MD)2

= AD2 − DM2 + BD2 + MD2 − 2BD. MD

= AD2 + BD2 − 2BD. MD

= AD2 + (BC

2)

2

− 2 (BC

2) × MD

= AD2 + (BC

2)

2

− BC × MD

(iii) In ΔAMB

AM2 + MB2 = AB2 … (1)

In ΔAMC

AM2 + MC2 = AC2 … (2)

Adding equation (1) and (2)

2AM2 + MB2 + MC2 = AB2 + AC2

⇒ 2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2


Page - 53

⇒ 2AM2 + BD2 + DM2 − 2BD. DM + MD2 + DC2 + 2MD. DC= AB2 + AC2

⇒ 2AM2 + 2MD2 + BD2 + DC2 + 2MD(−BD + DC) = AB2 + AC2

⇒ 2(AM2 + MD2) + (BC

2)

2

+ (BC

2)

2

+ 2MD (−BC

2+

BC

2) = AB2 + AC2

⇒ 2AD2 +BC2

2= AB2 + AC2

6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the

sum of the squares of its sides.

Solution:

Let ABCD be a parallelogram

Let us draw perpendicular DE on extended side BA and AF on side DC.

In ΔDEA

DE2 + EA2 = DA2 … (i)

In ΔDEB

DE2 + EB2 = DB2

⇒ DE2 + (EA + AB)2 = DB2

⇒ (DE2 + EA2) + AB2 + 2EA. AB = DB2

⇒ DA2 + AB2 + 2EA. AB = DB2 … (ii)

In ΔADF

AD2 = AF2 + FD2

In ΔAFC


Page - 54

AC2 = AF2 + FC2

= AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC ∙ FD

= (AF2 + FD2) + DC2 − 2DC. FD

⇒ AC2 = AD2 + DC2 − 2DC ∙ FD … (iii)

Since ABCD is a parallelogram

AB = CD (iii)

And BC = AD (iv)

In ΔDEA and ΔADF

∠DEA = ∠AFD

∠EAD = ∠FDA(EA ∥ DF)

∠EDA = ∠FAD(AF ∥ ED)

AD is common in both triangles.

Since respective angles are same and respective sides are same

ΔDEA ≅ ΔAFD

SOEA = DF

Adding equation (ii) and (iii)

⇒ DA2 + AB2 + 2EA. AB + AD2 + DC2 − 2DC. FD = DB2 + AC2

⇒ DA2 + AB2 + AD2 + DC2 + 2EA. AB − 2DC. FD = DB2 + AC2

⇒ BC2 + AB2 + AD2 + DC2 + 2EA. AB − 2AB. EA = DB2 + AC2

⇒ AB2 + BC2 + CD2 + DA2 = AC2 + BD2

7. In Fig., two chordsAB andCD intersect each other at the point P. Prove that:

(i) ∆APC~∆DPB

(ii) AP ∙ PB = CP ∙ DP

Solution:


Page - 55

Let us join CB

(i) In ΔAPC and ΔDPB

∠APC = ∠DPB {Vertically opposite angles}

∠CAP = ∠BDP {Angles in same segment for chord CB}

ΔAPC~ΔDPB {ByAA similarly criterion}

(ii) We know that corresponding sides of similar triangles are proportional

∴AP

DP=

PC

PB=

CA

BD

⇒AP

DP=

PC

PB

∴ AP. PB = PC ∙ DP

8. In Fig., two chordsABandCD of a circle intersect each other at the pointP (when

produced) outside the circle. Prove that

(i) ∆PAC~∆PDB (ii) PA ∙ PB = PC ∙ PD

Solution:

(i) In ΔPAC and ΔPDB

∠P = ∠P(Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is equal to

opposite interior angle)


Page - 56

∠PCA = ∠PBD (remaining angles)

ΔPAC ∼ ΔPDB (by AAA)

(ii) We know that corresponding sides of similar triangles are proportional.

PA

PD=

AC

DB=

PC

PB

⇒PA

PD=

PC

PB

∴ PA. PB = PC. PD

9. In Fig.,D is a point on sideBC of ∆ABC such that BD

CD=

AB

AC. Prove thatAD is the

bisector of ∠BAC.

Solution:

In ΔDBA and ΔDCA

BD

CD=

AB

AC (Given)

AD = AD(Common)

So, ΔDBA ∼ ΔDCA(By SSS)

Now, corresponding angles of similar triangle will be equal.

∠BAD = ∠CAD

AD is angle bisector of ∠BAC

10. Nazima is fly fishing in a stream. The tip of her fishing rod is1.8m above the

surface of the water and the fly at the end of the string rests on the water3.6m

away and2.4mfrom a point directly under the tip of the rod. Assuming that her

string (from the tip of her rod to the fly) is taut, how much string does she have

out (see Fig.)? If she pulls in the string at the rate of5cm per second, what will be

the horizontal distance of the fly from her after12 seconds?


Page - 57

A

Solution:

Let AB be the height of tip of fishing rod from water surface. Let BC Be the

horizontal distance of fly from the tip of fishing rod.

Then, AC is the length of string.

AC Can be found by applying Pythagoras theorem in ΔABC

AC2 = AB2 + BC2

AC2 = (1.8)2 + (2.4)2

AC2 = 3.24 + 5.76

AC2 = 9.00

Thus, length of string out is3m.

Now, she pulls string at rate of 5cm per second.

So, String Pulled in 12Second = 12 × 5 = 60cm = 0.6m


Page - 58

Let after 12second, fly be at point D.

Length of string out after12second is AD

AD = AC − String pulled by Nazima in 12second

= 3.00 − 0.6

= 2.4

In ΔADB

AB2 + BD2 = AD2

⇒ (1.8)2 + BD2 = (2.4)2

⇒ BD2 = 5.76 − 3.24 = 2.52

⇒ BD = 1.587

Horizontal distance of fly = BD + 1.2

= 1.587 + 1.2

= 2.787

= 2.79m

cbse ncert solutions for class 10 mathematics chapter 6 ... · class- x-cbse-mathematics triangles...

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