Lesson: Triangles

Exercise 6.1 (12)

Choose the correct answer from the given four options:

Question: 1

In the adjoining figure, and AD BC. Then,

  BAC 90





BD.CD BC



AB.AC BC



BD.CD AD



AB.AC AD

Solution:

(c)

In and ,

ADC

ADB

    BDA ADC 90

 

     B DAC 90 C

[by AA similarity criteria]

 ADB|| CDA

Since, the triangles are similar the ratio of their

corresponding sides will be equal.

 

AD AB DB

CD CA DA

That is,



AD DB

CD DA



AD BD.DC

Hence, proved.

Question: 2

The lengths of the diagonals of a rhombus are 16 cm and 12

cm. Then, the length of the side of the rhombus is

9 cm

10 cm

8 cm

20 cm

Solution:

(b)

Let, be the side of the rhombus.

The diagonals of the rhombus are and .



d 12 cm



d 16 cm

Diagonals of rhombus intersect each other at . Therefore,

90

is a right-angled triangle.

OPQ

Applying Pythagoras theorem,

 

2 2 2

PQ OQ OP

   

 

   

   

2 2

1 2

d d

2 2

   

 

   

   

2 2

12 16

2 2

   

 

2 2

6 8

 36 64

 100

10 cm

Hence, the side of the rhombus is .

10 cm

Question: 3

If and is not similar to , then

 ABC|| EDF

ABC

DEF

which of the following is not true?

BC.EF AC.FD

AB.EF AC.DE

BC.DE AB.EF

BC.DE AB.FD

Solution:

(c)

Given: Similar triangles have similar

 ABC|| EDF

corresponding sides.

So,

 ABC|| EDF

 

AB AC BC

ED EF DF

By the equalities,



AB AC

ED EF

AB.EF AC.DE



AC BC

EF DF

AC.FD BC.EF



AB BC

ED DF

BC.DE AB.FD

BC.DE AB.EF



Hence, option C is incorrect.

Question: 4

If in two triangles ABC and PQR, , then

 

AB BC CA

QR PR PQ

 PQR|| CAB

 PQR|| ABC

 CBA|| PQR

 BCA|| PQR

Solution:

(a)

Compare the corresponding vertex of the two triangles.

Vertex A corresponds to vertex Q, vertex P corresponds to

vertex C and vertex R corresponds to vertex B.

Therefore, .

 PQR|| CAB

Hence, option A is correct.

Question: 5

In the adjoining figure, two line segments AC and BD

intersect each other at the point P such that PA = 6 cm, PB =

3 cm, PC = 2.5 cm, PD = 5 cm, = and =

APB

50

CDP

. Then, is equal

30

PBA

50

30

60

100

Solution:

(d)

From and ,

APB

PDC

 

PA PB 6

PD PC 5

Also, [vertically opposite

    APB DPC 50

angles]

[By SAS similarity criterion]

 APB|| DPC

Since, angles corresponding to similar triangles are equal,

  PBA PCD

  PDC 30

Sum of angles of a triangle is

180

PDC,

      PCD DPC PDC 180

      PCD 50 30 180

    PCD 180 80

  PCD 100

So,

    PCD PBA 100

Hence, option D is correct.

Question: 6

If in two triangles DEF and PQR, and

  D Q

  R E

then which of the following is not true?



EF DF

PR PQ



DE EF

PQ RP



DE DF

QR PQ



EF DE

RP QR

Solution:

(b)

In and ,

DEF

PQR

  

 

D Q

R E

[AA similarity rule]

 DEF|| PQR

So,

 

DE DF EF

QR PQ RP

Which implies , and .



EF DF

RP PQ



DE DF

QR PQ



DE DF

QR PQ

Thus, option B is incorrect.

Question: 7

In triangles ABC and DEF, , and AB = 3

  B E

  F C

DE. Then, the two triangles are

a. congruent but not similar

b. similar but not congruent

c. neither congruent nor similar

d. congruent as well as similar

Solution:

(b)

Given: AB = 3DE and

     B Eand C F

[AA similarity criterion]

 ABC|| EDF

But AB = 3DE, which implies that the sides of both the

triangles are not equal.

Hence, they are not congruent.

Thus, both are similar but not congruent.

Question: 8

It is given that ,with . Then,

 ABC|| PQR



BC 1

QR 3

is equal to

 



ar PRQ

ar BCA

a. 9

b. 3

Solution:

(a)

Given:



BC 1

QR 3

Ratio of areas of similar triangle,

 







ar PRQ

ar BCA BC

That is,

 







ar PRQ

ar BCA BC

 



 

 



Hence, option A is correct.

Question: 9

It is given that , , AB = 5

 ABC|| DFE

  A 30

  C 50

cm, AC = 8 cm and DF= 7.5 cm. Then, the following is true:

DE 12 cm,

  F 50

DE 12 cm,

  F 100

EF 12 cm,

  D 100

EF 12 cm,

  D 30

Solution:

(b)

Given: , AB = 5 cm, AC = 8 cm and

  A 30

  C 50

DF= 7.5 cm

Since, ,

 ABC|| DEF

 

AB AC BC

DF DE FE

AB = 5 cm, AC = 8 cm and DF = 7.5 cm.



AB AC

DF DE





AC DF





8 7.5

12 cm

    

A D 30

C E 50

      F 180 50 30

 100

Thus,

    B F 100

Hence, option B is correct.

Question: 10

If in triangles ABC and DEF, , then they will be



AB BC

DE FD

similar, when

  B E

  A D

  B D

  A F

Solution:

(c)

In and ,

ABC

EFD



AB BC

DE FD

AB and BC formed an angle B.

DE and FD formed an angle D.

  B D

Both the triangles will be similar by SAS similarity rule.

Hence, option C is correct.

Question: 11

If , , AB = 18 cm and BC = 15

  ABC QRP

 



ar ABC

ar PQR 4

cm, then PR is equal to

a. 10 cm

b. 12 cm

c. cm

d. 8 cm

Solution:

(a)

Given: AB = 18 cm, BC = 15 cm and

 



ar ABC

ar PQR 4

Since, ,

 ABC|| QRP

 

 

2 2

ar ABC

BC AB

ar PQR RP QR

 

2 2

9 15 18

4 RP QR



9 15

4 RP





15 4

 



15 15 4

 100

10 cm

Hence, PR is equals to .

10 cm

Question: 12

If S is a point on side PQ of a such that

PQR

PS = QS = RS, then



PR.QR RS

 

2 2 2

QS RS QR

 

2 2 2

PR QR PQ

 

2 2 2

PS RS PR

Solution:

(c)

In ,

PQR

PS = QS = RS

In , PS = RS,

PSR

Since, angle opposite to equal sides are equal.

  P 1

Similarly, In , QS = RS,

SQR

  Q 2

Now, in ,

PQR

      P Q R 180

 

        1 2 1 2 180

    1 2 90

  PRQ 90

Thus, is a right-angled triangle.

PQR

By applying Pythagoras theorem,

 

2 2 2

PQ PR RQ

Thus, option C is correct.

Exercise 6.2 (12)

Question: 1

Is the triangle with sides 25 cm, 5 cm and 24 cm a right

triangle? Give reasons for your answer.

Solution:

A triangle is a right-angled triangle if it follows Pythagoras

theorem.

Let, AB = 24 cm, BC = 5 cm and CA = 25 cm be the sides of

the triangle.

By applying Pythagoras theorem,

 

2 2 2

AC AB BC

 

2 2 2

25 5 24

 625 25 576

625 601

Since, Pythagoras theorem is not followed, the given triangle

is not a right-angled triangle.

Question: 2

It is given that . Is it true to say that

 DEF|| RPQ

  D R

and ? Why?

  F P

Solution:

Since, , all the corresponding angles of both

 DEF|| RPQ

triangles are equal.

  D R

  E P

  F Q

Hence, is true but

  D R

  F P

Question: 3

A and B are respectively the points on the sides PQ and PR

of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR =

6 cm and PB = 4 cm. Is ?

AB||QR

Give reasons for your answer.

Solution:

According to the converse of basic proportionality theorem,

AB || QR if AB divides PQ and PR in the same ratio.





PA PA

AQ PQ AP





12.5 5



7.5





5 10



PB 4

BR 6



Since, ,

 

PA PB 2

AQ BR 3

AB||QR

Question: 4

In the adjoining figure, BD and CE intersect each other at

the point P. Is ? Why?

 PBC|| PDE

Solution:

In , [Vertically opposite

 PBC and PDE

  BPC DPE

angles]



BP 5

PD 10



PC 6

PE 12



So,



BP PC

PD PE

Thus, [By SAS similarity rule]

 PBC|| PDE

Hence, the given statement is correct.

Question: 5

In triangles PQR and MST, , ,

  P 55

  Q 25

  M 100

and . Is ? Why?

  S 25

 PQR|| TSM

Solution:

In , and

PQR

  P 55

= 25 Q

+ + = 180   P R Q

55 + + 25 = 180   R

= 180 55 25     R

= 100 R

Similarly, in ,

TSM

  M 100

  S 25

      T S M 180

      T 180 100 25

  T 55

Since,

  P S

  Q T

is not similar to .

PQR

TSM

Question: 6

Is the following statement true? Why?

“Two quadrilaterals are similar, if their corresponding angles

are equal”.

Solution:

False.

Two quadrilaterals will be similar if their angles as well as

ratio of their corresponding sides will be equal.

Question: 7

Two sides and the perimeter of one triangle are respectively

three times the corresponding sides and the perimeter of the

other triangle. Are the two triangles similar? Why?

Solution:

Let the two triangles be and .

ABC

PQR

In ,

ABC

Let the sides be AB, BC and CA.

Then the perimeter of ,

ABC

  

p AB BC CA

Similarly, in ,

PQR

Let the sides be PQ, QR and RP.

And the perimeter will be given by

  

p PQ QR RP

 PQ 3 AB

 QR 3 BC

 

2 1

p 3 p

   

     PQ QR RP 3 AB BC CA

   

       3 AB 3 BC RP 3 AB+3 BC+3 CA

 RP 3 CA



AB BC CA

= = 3

PQ QR RP

Hence, [By SSS similarity criterion]

 ABC|| PQR

Question: 8

If in two right triangles, one of the acute angles of one

triangle is equal to an acute angle of the other triangle, can

you say that the two triangles will be similar? Why?

Solution:

and are right angled triangles.

ABC

PQR

    B Q 90

  C R

Thus, and the given statement is true.

 ABC|| PQR

Question: 9

The ratio of the corresponding altitudes of two similar

triangles is . Is it correct to say that ratio of their areas is

? Why?

Solution:



PS 3

JM 5

Ratio of areas of the two similar triangles PQR and

JKL can be given by,

 



ar PQR

ar JKL JM

 



 

 

 

9 6

25 5

Hence, the above statement is not correct.

Question: 10

D is a point on side QR of such that

PQR

PD QR. Will it be correct to say that ?



 PQD RPD:

Why?

Solution:

In and ,

PQD

PDR

[ ]

    PDR PDQ 90

PD QR

[PD does not bisect ]

  1 2

P

[PQ QR]

  Q R



Ratio of sides are also not equal.

Hence, is not similar to and the above

PQD

PDR

statement is false.

Question: 11

In the adjoining figure, if , then is it true that

  D C

? Why?

 ADE|| ACB

Solution:

In and ,

ADE

ABC

  D C

  A A

Hence, [By AA similarity criterion]

 ADE ABC:

Hence, the above statement is true.

Question: 12

Is it true to say that if in two triangles, an angle of one

triangle is equal to an angle of another triangle and two sides

of one triangle are proportional to the two sides of the other

triangle, then the triangles are similar? Give reasons for your

answer.

Solution:

The ratio of two sides of the triangles are proportional to

each other but the angle corresponding to those sides is not

included.

Hence, the given statement is false.

Exercise 6.3 (15)

Question: 1

In a PQR, and M is a point on side





2 2 2

PR – PQ QR

PR such that QM PR. Prove that .



 

QM PM MR

Solution

Given: A PQR such that .





2 2 2

PR – PQ QR

And M is a point on side PR such that QM PR.



To Prove:

 

QM PM MR

Proof:

In ,

PQR



2 2 2

PR – PQ QR

 

2 2 2

PR PQ QR

By the converse of Pythagoras theorem,

  PQR 90

In triangles QMP and QMR,

QM PR

Thus,

    1 2 90

We know that the sum of all the interior angles in a

triangle is .

180

Thus,

      2 3 R 180

      90 3 R 180

    3 90 R ......(1)

Similarly,

    P 90 R ......(2)

From equations (1) and (2),

  3 P

Thus,

 QMP|| QMR



PM QM

QM MR

 

QM PM MR

Hence proved.

Question: 2

Find the value of for which DE||AB in the given

figure.

Solution

Given: A triangle ABC in which DE||AB.

From the figure,

 AD 3x 19

 DC x 3

 BE 3x 4

EC x

In , DE||AB.

ABC

Thus,



AD BE

DC EC

Put the values of and .

AD,DC,BE

 





3x 19 3x 4

x 3 x

    

   x 3x 19 x 3 3x 4

    

2 2

3x 19x 3x 4x 9x 12

 

2 2

3x – 3x 19x–13x 12

6x 12

x 2

Thus, the value of is 2.

Question: 3

In the adjoining figure, if and

  1 2

, then prove that .

  NSQ MTR

 PTS|| PRQ

Solution

Given: A in which points S and T are on PQ

PQR

and PR respectively such that

  1 2

Also,

  NSQ MTR

To Prove:

 PTS|| PRQ

Proof:

  NSQ MTR

Thus, … (1)

 

SQ TR By CPCT

  1 2

We know that the sides opposite to equal angles in a

triangle are equal.

Thus, … (2)

PT PS

From equations (1) and (2),



PT PS

TR SQ

Thus, by the converse of BPT,

ST||QR

In triangles PTS and PRQ,

ST||QR

  1 3

 

Corresponding angles

  2 4

 

Corresponding angles

Thus,

 PTS|| PRQ

Hence proved.

Question: 4

Diagonals of a trapezium PQRS intersect each other at

the point O, PQ RS and . Find the ratio of

PQ 3RS

the areas of triangles POQ and ROS.

Solution

Given: A trapezium PQRS in which diagonals

intersect at the point O.

Also, PQ||RS and .

PQ 3RS

In triangles POQ and ROS,

PQ||RS

  1 3

 

Aletrnate interior angles

  2 4

 

Aletrnate interior angles

Thus,

 POQ|| ROS

 

By AA similarity criterion

By area theorem,

 



 



 



 

ar POQ

ar ROS RS

Given that

PQ 3RS

 



 



 



 

ar POQ

3RS

ar ROS RS



Thus, the ratio of the areas of triangles POQ and ROS

is .

9 :1

Question: 5

In the adjoining figure, if AB||DC and AC and PQ

intersect each other at the point O, prove that

OA.CQ OC.AP

Solution

Given: A quadrilateral ABCD such that AB DC and

AC and PQ intersect each other at the point O.

Fig. Exm_6.3_5 (Sol)

To prove:

OA.CQ OC.AP

Proof: In triangles OPA and OQC,

  1 2

 

Aletrnate interior angles

  3 4

 

Aletrnate interior angles

Thus,

 

 OPA|| OQC By AA similarity criterion

 

OQ OC CQ

OP OA AP

OA.CQ OC.AP

Hence proved.

Question: 6

Find the altitude of an equilateral triangle of side 8

cm.

Solution

Let PQR be an equilateral triangle of side 8 cm.

PS is an altitude.

PS QR

Thus,

    1 2 90

In triangles PSQ and PSR,

 

PQ PR Equilateral triangle property

    1 2 90

 

PS PS Common

Thus,

 

  PSQ PSR By RHS congruence criterion

QS SR

 

CPTC

Thus,

    

QR PQ 8

QS SR 4 cm

2 2 2

By Pythagoras theorem in ,

PQS

 

2 2 2

PS QS PQ

   

 

2 2

PS 4 8

 

PS 64 16



PS 48

PS 4 3

Thus, the altitude of the equilateral triangle is

4 3 cm

Question: 7

If , AB = 4 cm, DE = 6 cm, EF = 9 cm

 ABC|| DEF

and FD = 12 cm, find the perimeter of ABC.



Solution

Given: Two triangles ABC and DEF such that

 ABC|| DEF

Also, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12

Given:

 ABC|| DEF

Thus,

 

AB AC BC

DE DF EF

 

4 AC BC

6 12 9

Take first and second part.



4 AC

6 12

  

AC 12 8

Take first and third part.



4 BC

6 9

  

BC 9 6

   The perimeter of ABC AB BC AC

  4 6 8

18

Thus, the perimeter of ABC is 18 cm.



Question: 8

In the adjoining figure, if DE||BC, find the ratio of

and .

 

ar ADE

 

ar DECB

Solution

Given: A triangle ABC such that DE||BC.

Also, DE = 6 cm and BC = 12 cm.

In triangles ADE and ABC,

DE BC:

  1 2

 

Corresponding angles

  3 4

 

Corresponding angles

Thus,

 ADE|| ABC

 

AA similarity criterion

We know that in two similar triangles, the ratio of the

areas is equal to the squares of the ratio of their

corresponding sides.

Thus,

 



 



 



 

ar ABC

ar ADE DE

   

 

 

 



 



 

ar DECB ar ADE

ar ADE DE

 

 

 

 



 

ar DECB

ar ADE 6

 

 



ar DECB

2 1

ar ADE

 





ar DECB

ar ADE 1

 





ar ADE

ar DECB 3:

Thus, the ratio of and is .

 

ar ADE

 

ar DECB

1:3

Question: 9

ABCD is a trapezium in which AB||DC and P and Q

are points on AD and BC, respectively such that

PQ||DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,

find AD.

Solution

Given: A trapezium ABCD in which AB||DC and P

and Q are points on AD and BC such that PQ||DC.

Also, PD = 18 cm, BQ = 35 cm and QC = 15 cm.

In trapezium ABCD,

AB|| CD

PQ||CD

Thus,

AB||CD||PQ ......(1)

In , from the equation (1).

BCD

OQ CD:

Thus,



BO BQ

OD QC

 

By BPT

......(2)

Similarly, in ,

DAB

PO AB:



BO AP

OD PD

 

By BPT

......(3)

From equations (2) and (3).



AP BQ

PD QC



AP 35

18 15

 

AP 18

AP 42

 AD AP PD

 AD 42 18

AD 60

Thus,

AD 60 cm

Question: 10

Corresponding sides of two similar triangles are in the

ratio of . If the area of the smaller triangle is

2 :3

, find the area of the larger triangle.

48 cm

Solution

Let LMN and PQR be two similar triangles with sides

in the ratio of .

2 :3

Let the area of smaller be .

LMN

48 cm

Given that

 LMN PQR:

By Area theorem,

 



 



 



 

ar LMN

ar PQR PQ

 

 



 



 

48 2

ar PQR 3

 





48 4

ar PQR 9

 



 

48 9

ar PQR

 

 ar PQR 108

Thus, the area of the larger triangle is

PQR

108 cm

Question: 11

In a triangle PQR, N is a point on PR such that QN



PR. If , prove that .



PN.NR QN

  PQR 90

Solution

Given: A triangle PQR with a point N on PR such that

QN PR and





PN.NR QN

To prove:

  PQR 90

Proof:

In triangles QNP and QNR,

QN PR

    1 2 90



PN.NR QN

Thus,



QN NP

NR QN

Thus,

 QNP QNR:

 

By SAS similarity criterion

   P RQN x ......(1)

    1 2 90

   PQN R y ......(2)

In ,

PQR

      P PQR R 180

 

Angle sum property

From equations (1) and (2),

    

  

x x y y 180

2x 2y 180

x y 90

Thus,

  PQR 90

Hence proved.

Question: 12

Areas of two similar triangles are 36 and 100 .

If the length of a side of the larger triangle is 20 cm,

find the length of the corresponding side of the smaller

triangle.

Solution

Let LMN and PQR be two similar triangles with areas

36 and 100 respectively.

 

 

ar LMN 36 cm

 

 

ar PQR 100 cm

PQ 20 cm

Given that,

 LMN PQR:

 



 



 



 

ar LMN

ar PQR PQ

 



 

 

36 LM

100 20



LM 6

20 10





6 20

LM 12

Thus, the length of the corresponding side of the

smaller triangle is 12 cm.

Question: 13

In the adjoining figure, if , AC = 8 cm

  ACB CDA

and AD = 3 cm, find BD.

Solution

In triangles ACD and ACB,

  ACB CDA

 

  A A Common

 ACD ACB:

 

AA similarity criterion

Thus,

 

AC DC AD

AB BC AC

 

8 DC 3

AB BC 8

Take first and third part.



8 3

AB 8





8 8



 BD AB AD

 



18.33

Thus,

BD 18.33 cm

Question: 14

A 15 metres high tower casts a shadow 24 metres long

at a certain time and at the same time, a telephone pole

casts a shadow 16 metres long. Find the height of the

telephone pole.

Solution

Let AB be a tower of height 15 m and CB be its

shadow.

Let PQ be a pole of height m and RQ be its shadow.

AB = 15 m

CB = 24 m

PQ x m

RQ = 16 m

In triangles ABC and PQR,

    B Q 90

  C R

 

Angular elevation of sun

Thus,

 ABC PQR:

 

AB AC CB

PQ PR RQ

Take first and third part.



AB CB

PQ RQ



15 24

x 16





15 16

x 10

Hence, the height of the telephone pole.is 10 m.

Question: 15

Foot of a 10 m long ladder leaning against a vertical

wall is 6 m away from the base of the wall. Find the

height of the point on the wall where the top of the

ladder reaches.

Solution

Let RP be a ladder leaning against vertical wall PQ.

RP = 10 m

QR = 6 m

Let

PQ x m

By Pythagoras theorem,

     

 

2 2 2

RP PQ QR

     

 

2 2 2

10 x 6

 

 

x 100 36

 



x 64

x 8

Thus, the height of the point on the wall where the top

of the ladder reaches is 8 m.

Exercise 6.4 (18)

Question: 1

In the adjoining figure, if , AB = 6 cm, BP =

  A C

15 cm, AP = 12 cm and CP = 4 cm, then find the

lengths of PD and CD.

Solution

Given: , AB = 6 cm, BP = 15 cm, AP = 12

  A C

cm and CP = 4 cm

In triangles ABP and CDP,

  A C

  1 2

 

Vertically opposite angles

Thus,

 

 ABP CDP AA similarity criterion:

Thus,

 

AB AP BP

CD CP PD

 

6 12 15

CD 4 PD

Take first and second part.



6 12

CD 4





6 4

CD 2

Take second and third part.



12 15

4 PD





4 15

PD 5

Thus, PD = 5 cm and CD = 2 cm

Question: 2

It is given that such that AB = 5 cm,

 ABC EDF:

AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the

lengths of the remaining sides of the triangles.

Solution

Given:

 ABC EDF:

 

AB AC BC

ED EF DF

 

5 7 BC

12 EF 15

Take first and second part.



5 7

12 EF





12 7



EF 16.8

Take first and third part.



5 BC

12 15





15 5



BC 6.25

Hence, the lengths of the remaining sides EF and BC

are 16.8 cm and 6.25 cm respectively.

Question: 3

Prove that if a line is drawn parallel to one side of a

triangle to intersect the other two sides, then the two

sides are divided in the same ratio.

Solution

Let PQR be a triangle and the line ST is drawn parallel

to the side QR that intersects the other two sides.

To prove:



PS PT

SQ TR

Construction: Draw TU PQ and SV PR. Join SR



and QT.

Proof:

 

 



 



 

PS TU

ar PST

......(1)

ar SQT SQ

SQ TU

 

 



 



 

PT SV

ar PTS

......(2)

ar TRS TR

TR SV

From the figure, SQT and TRS are situated on



same base between same parallel lines.

Thus,

   

  ar SQT ar TRS ......(3)

From equations (2) and (3).

 







ar PTS

......(3)

ar SQT TR

From equations (1) and (4).



PS PT

SQ TR

Hence proved.

Question: 4

In the adjoining figure, if PQRS is a parallelogram and

AB PS, then prove that OC SR.

Solution

Given: A parallelogram PQRS inside and AB

ABC

PS.

To Prove: OC SR

Proof:

In triangles OAB and OPS,

AB PS:

  1 2

 

Corresponding angles

  3 4

 

Corresponding angles

Thus,

 OAB OPS:

 

By AA similarity criterion

 

OP OS PS

......(1)

OA OB AB

… (2)

PS||QR

 

Parallelogram property

So,

QR AB ......(3):

In triangles CQR and CAB,

QR AB:

  CAB 5

 

Corresponding angles

  CBA 5

 

Corresponding angles

Thus,

 CQR CAB:

 

By AA similarity criterion

 

CQ CR QR

CA CB AB

From equation (2),

PS QR:

Thus,

 

PS CR CQ

......(4)

AB CB CA

From equations (1) and (4).



CR OS

CB OB

Thus, the ratios of two sides of BOC are equal.



By the converse of BPT,

OC SR:

Hence proved.

Question: 5

A 5 m long ladder is placed leaning towards a vertical

wall such that it reaches the wall at a point 4 m high. If

the foot of the ladder is moved 1.6 m towards the wall,

then find the distance by which the top of the ladder

would slide upwards on the wall.

Solution

Let AB be a ladder leaning towards vertical wall DC.

The ladder is moved 1.6 m towards the wall.

 AB DE 5m

AC = 4 m

BE = 1.6 m

In triangle ABC, by the Pythagoras theorem,

     

 

2 2 2

AB AC BC

     

 

2 2 2

5 4 BC

 

 

BC 25 16

 



BC 9

BC 3

Now,

 EC BC BE

 EC 3 1.6

EC 1.4

In triangle DEC, by the Pythagoras theorem,

     

 

2 2 2

DE DC EC

     

 

2 2 2

5 DC 1.4

 

 

DC 25 1.96

 



DC 23.04

DC 4.8

Now,

 DA DC AC

 DA 4.8 4

DA 0.8

Thus, the distance by which the top of the ladder

would slide upwards on the wall is 0.8 m.

Question: 6

For going to a city B from city A, there is a route via

city C such that AC CB, km and



AC 2x

km. It is proposed to construct a 26 km

 

 CB 2 x 7

highway which directly connects the two cities A and

B. Find how much distance will be saved in reaching

city B from city A after the construction of the

highway.

Solution

Given: Two cities A and B and a route between them

via city C.

AC CB

AC 2x

 

 CB 2 x 7

AB 26

By Pythagoras theorem,

 

2 2 2

AC BC AB

   

 



 





2x 2 x 7 26

   

 



 





2x 2 x 7 26

   

2 2

4 x x 49 14x[ ] 676

  

676

2x 14x 49

  

2x 14x 49– 169 0

 

2x 14x– 120 0

 

x 7x– 60 0

Split the middle term to find factors.

 

x 12x – 5x– 60 0

   

  x x 12 – 5 x 12 0

  

 x 12 x – 5 0

That is,

 

x 12 0

 

x– 5 0

 x 12

x 5

Distance cannot be negative,

Thus,

x 5



 



 

 

The saved distance after the

AC BC – AB

construction of the highway

  2x 2x 14 – 26

 4x–12

Put the value of in the above expression.

  The saved distance 4 5 12

 20 12

 8

Thus, the distance that will be saved in reaching city B

from city A after the construction of the highway is 8

km.

Question: 7

A flag pole 18 m high casts a shadow 9.6 m long. Find

the distance of the top of the pole from the far end of

the shadow.

Solution

Let AB be a flag pole and CB be its shadow.

AB = 18 m

CB = 9.6 m

By Pythagoras theorem,

     

 

2 2 2

AC AB BC

     

 

2 2 2

AC 18 9.6

 

 

AC 324 92.16

 



41AC 6.16

AC 20.4

Hence, the distance of the top of the pole from the far

end of the shadow.is 20.4 m.

Question: 8

A street light bulb is fixed on a pole 6 m above the

level of the street. If a woman of height 1.5 m casts a

shadow of 3 m, find how far she is away from the base

of the pole.

Solution

Let the bulb is fixed at point A on the pole AC. DE be

the woman and BE be her shadow.

AC = 6 m

DE = 1.5 m

BE = 3 m

In triangles ABC and DBE.

    1 2 90

  B B

 

Common

Thus,

 ABC DBE:

 

By AA similarity criterion

 

AC AB BC

DE AD BE

 

6 AB BC

1.5 AD 3

Take first and third part.



6 BC

1.5 3





6 3

1.5

BC 12

Now,

 EC BC BE

 EC 12 3

EC 9

Thus, the woman is 9 m away from the base of the

pole.

Question: 9

In the given figure, ABC is a triangle right angled at B

and . If AD = 4 cm, and CD = 5 cm, find BD

BD AC

and AB.

Solution

According to the question,

In ,

ABC

  ABC 90

Also,

BD AC

So,

 

BD AD CD

Now, put the values of AD and CD in the above

equation.

 

BD 4 5



BD 20

BD 20

BD 2 5 cm

Now, in ,

BDA

It is given that

BD AC

So,

  BDA 90

Now, by Pythagoras theorem,

 

2 2 2

AB AD BD

Put the values of AD and BD in the above equation.

 

2 2 2

AB AD BD

 

 

AB 4 2 5

 

AB 16 20



AB 36



2 2

AB 6

AB 6 cm

Thus, the lengths of BD and AB are

2 5 cm and 6 cm

respectively.

Question: 10

In the given figure, PQR is a right triangle right angled

at Q and . If PQ = 6 cm and PS = 4 cm, find

QS PR

QS, RS and QR.

Solution

According to the question,

In ,

PQR

  PQR 90

Also,

QS PR

So,

 

QS PS RS ......(1)

Now, in , apply the Pythagoras theorem.

PSQ

 

2 2 2

PQ QS PS

Put the values of PS and PQ in the above equation.

 

2 2 2

6 QS 4

 

QS 36 16



QS 20

QS 2 5 cm

Now, substitute the value of QS and PS in equation

 

 

2 5 4 RS



RS 5 cm

Now, in ,

QSR

  QSR 90

 

QS PR

Apply, Pythagoras theorem.

 

2 2 2

QR QS RS

Now, substitute the value of QS and RS in the above

equation.

 

 

QR 2 5 5

 

QR 20 25



QR 45

QR 45

Thus, the lengths of QS, RS and QR are

respectively.

2 5 cm, 5 cm and 3 5 cm

Question: 11

In , such that D lies on QR. If PQ

PQR

PD QR

 a

, PR , QD and DR , prove that

 b

 c

 d

     

    a b a b c d c d

Solution

According to the question,

Given: In ,

PQR

PD QR

So,

    PDQ PDR 90

It is also given that, PQ , PR , QD and DR

 a

 b

 c

 d

To prove:

     

    a b a b c d c d

Proof: In , right angled at D,

PDQ

Apply, Pythagoras theorem.

 

2 2 2

PD QD PQ

Substitute the values of QD and PQ in the above

equation.

 

2 2 2

PD c a

 

2 2 2

PD a c ......(1)

Now, in , right angled at D,

PDR

Apply, Pythagoras theorem.

 

2 2 2

PD DR PR

Substitute the values of DR and PR in the above

equation.

 

2 2 2

PD d b

 

2 2 2

PD b d ......(2)

Now, from equations and .

 

  

2 2 2 2

a c b d

  

2 2 2 2

a b c d

     

    a b a b c d c d

Hence proved.

Question: 12

In a quadrilateral ABCD, . Prove that

    A D 90

[Hint: Produce AB and DC to

  

2 2 2 2

AC BD AD BC

meet at E.]

Solution

Given: ABCD is a quadrilateral having

    A D 90

To prove:

  

2 2 2 2

AC BD AD BC

Proof: First, produce the sides AB and DC to meet

each other at point E.

Now, in , it is given that,

EAD

    EAD EDA 90

So, (By interior angles property of triangle)

  E 90

Now, apply Pythagoras theorem in and .

EAD

EBC

So,

 

2 2 2

AD AE DE ......(1)

And,

 

2 2 2

BC BE CE ......(2)

Add equations .

   

1 and 2

    

2 2 2 2 2 2

AD BC AE DE BE CE ......(3)

Similarly, apply Pythagoras theorem in

 ECAand EBD

So,

 

2 2 2

AC AE CE ......(4)

And,

 

2 2 2

BD BE DE ......(5)

Add equations .

   

4 and 5

    

2 2 2 2 2 2

AC BD AE CE BE DE ......(6)

Now, from equations and .

 

  

2 2 2 2

AC BD AD BC

Hence proved.

Question: 13

In the given figure, and line segments AB, CD

l m:

and EF are concurrent at point P. Prove that

 

AE AC CE

BF BD FD

Solution

Given: Two lines parallel to each other.

l and m

Also, line segments AB, CD and EF all meet at point

To prove:

 

AE AC CE

BF BD FD

Proof: In ,

 AEP and BFP

l m:

… (Alternate interior angles)

  PAE PBF

Also, … (Alternate interior angles)

  AEP BFP

So, … (By similarity criterion)

 AEP|| BFP

Thus,

 

AE AP EP

......(1)

BF BP PF

Now, in and ,

CEP

DFP

l|| m

… (Alternate interior angles)

  PCE PDF

Also, … (Alternate interior angles)

  CEP DFP

So, … (By similarity criterion)

 CEP|| DFP

Thus,

 

CE CP EP

......(2)

DF DP PF

Now, in ,

 ACP and BPD

l|| m

… (Alternate interior angles)

  PAC PBD

Also, … (Alternate interior angles)

  ACP PDB

So, … (By similarity criterion)

 ACP BPD:

Thus,

 

AC AP CP

......(3)

BD BP DP

Now, from equations and ,

 

1 ,

 

    

AE AP EP CE CP AC

BF BP PF DF DP BD

So,

 

AE AC CE

BF BD FD

Hence proved.

Question: 14

In the given figure, PA, QB, RC and SD are all

perpendiculars to a line , AB = 6 cm, BC = 9 cm, CD

= 12 cm and SP = 36 cm. Find PQ, QR and RS.

Solution

According to the question,

   PA l, QB l,RC l and SD l

Produce the line and SP, so that they meet at point T.

Now, in ,

TDS

PA QB RC SD: : :

So,

PQ :QR:RS AB:BC : CD

It is given that, AB = 6 cm, BC = 9 cm and CD = 12

So,

PQ :QR:RS 6 : 9 :12

Let us assume, PQ , QR and RS

 6x

 9x

12x

Now, as SP

 36

So,

  PQ QR RS 36

Now, substitute the values of PQ, QR and RS in the

above equation.

  6x 9x 12x 36

27x 36



Now, find PQ, QR and RS.

 6x

 

 8 cm

 9x

 

12 cm

12x

 

16 cm

Hence, the lengths of PQ, QR and RS are 8 cm, 12 cm

and 16 cm respectively.

Question: 15

O is the point of intersection of the diagonals AC and

BD of a trapezium ABCD with . Through O,

AB DC:

a line segment PQ is drawn parallel to AB meeting

AD in P and BC in Q. Prove that .

PO QO

Solution

Given: A trapezium ABCD with .

AB DC:

Also,

PQ AB:

To prove:

PO QO

Proof: In ,

ADB

PO AB:

So,



AP BO

......(1)

PD OD

Now, in , (As, )

BDC

OQ CD:

AB DC:

So,



BO BQ

......(2)

OD QC

From equations and ,

 



AP BQ

PD QC

Add 1 on both sides of the above equation.

  

AP BQ

1 1

PD QC

Now, solve the above equation.

 



AP PD BQ QC

PD QC



AD BC

PD QC

The above equation can also be written as,



PD QC

......(3)

AD BC

Now, in ,

 DOP and DBA

It is given that,

AB PO:

Therefore, (corresponding angles)

  DPO DAB

And, (corresponding angles)

  DOP DBA

So, (By similarity criterion)

 DOP DBA:



PO DP

......(4)

AB DA

Now, in

 COQ and CAB

AB OQ:

Therefore, (corresponding angles)

  COQ CAB

And, (corresponding angles)

  CQO CBA

So, (By similarity criterion)

 COQ CAB:



OQ QC

......(5)

AB BC

Now, from equations ,

     

3 , 4 and 5

  

PD QC PO OQ

AD BC AB AB

So,



PO OQ

AB AB

Thus,

PO QO

Hence proved.

Question: 16

In the given figure, line segment DF intersect the side

AC of a triangle ABC at the point E such that E is the

mid–point of CA and . Prove that

  AEF AFE

[Hint: Take point G on AB such that



BD BF

CD CE

]

CG||DF

Solution

Given: In , AE = CE = AF

ABC

To prove:



BD BF

CD CE

Proof: Draw a line CG parallel to EF.

In ,

ACG

CG EF:

And, E is the mid–point of CA.

So, F will be the mid–point of GA.

Therefore, AF = FG

But it is given that, AE = CE = AF

So, FG = FA = EA = EC

......(1)

Now, in ,

 BCG and BDF

CG EF:

So,



BC BG

CD GF

Add 1 on both sides of the above equation.

  

BC BG

1 1

CD GF

Now, solve the above equation.



BC+CD BG+GF

CD GF



BD BF

CD GF

Now, from equation ,

 

FG = CE

So,



BD BF

CD CE

Hence proved.

Question: 17

Prove that the area of the semicircle drawn on the

hypotenuse of a right angled triangle is equal to the

sum of the areas of the semicircles drawn on the other

two sides of the triangle.

Solution

Given: is right angled at B.

PQR

The semicircles are drawn on sides QR,

1 2 3

S , S and S

PQ and PR of .

PQR

To prove: Area of semicircles Area of

 

 

1 2

S S

semi–circle

Proof: Firstly, let us assume the radii of hemispheres

be .

1 2 3

S , S and S

1 2 3

r ,r and r

   PQR, Q 90

Now, apply Pythagoras theorem.

 

2 2 2

PQ QR PR

Now,

   

 

 

2 2

2 1 3

2r 2r 2r

 

 

2 2 2

1 2 3

4 r r 4r

 

 

2 2 2

1 2 3

r r r

Now, multiply on both the sides of the above

equation.

 

 

2 2 2

1 2 3

1 1

π r r πr

2 2

 

2 2 2

1 2 3

1 1 1

πr πr πr

2 2 2

So, area of semicircles Area of semi–circle

 

 

1 2

S S

Hence proved.

Question: 18

Prove that the area of the equilateral triangle drawn on

the hypotenuse of a right angled triangle is equal to the

sum of the areas of the equilateral triangles drawn on

the other two sides of the triangle.

Solution

Given: A right angled

PQR

Let PQ , QR , PR and

 a

 b

 c

  Q 90

Also, equilateral triangles are drawn with sides PQ,

QR and PR

To prove: Area of equilateral triangle with side

c

Sum of the areas of equilateral triangles with sides

a and b

Proof: In

   PQR, Q 90

Now, apply Pythagoras theorem.

So,

 

2 2 2

PQ QR PR

Substitute the values of PQ, QR and PR in the above

equation.

 

2 2 2

a b c

Multiply on both sides of the above equation.

 

2 2 2

3 3 3

a b c

4 4 4

Or,

 

2 2 2

3 3 3

c a b

4 4 4

So, area of equilateral triangle with side Sum of

c

the areas of equilateral triangles with sides

a and b

Hence proved.