Lesson: Triangles

Exercise 6.1 (3)

Question: 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 are

similar, if (a) their corresponding angles are

____and (b) their corresponding sides are

______. (equal, proportional)

Solution

(i) Similar

(ii) Similar

(iii) Equilateral

(iv)

(a) Equal

(b) Proportional

Question: 2

Give two different examples of pair of

(i) similar figures

(ii) non-similar figures

Solution

(i) Two equilateral triangles having sides 1 cm and

2 cm.

Fig. Exc_6.1_2 [i(a)]

Two squares having sides 1 cm and 2 cm.

Fig. Exc_6.1_2 [i(b)]

(ii) Trapezium and square

Fig. Exc_6.1_2 [ii(a)]

Triangle and parallelogram

Fig. Exc_6.1_2 [ii(b)]

Question: 3

State whether the following quadrilaterals are similar

or not:

Fig. Exc_6.1_3

Solution

The corresponding sides of quadrilateral PQRS and

ABCD are proportional, but their corresponding

angles are not equal.

Hence, quadrilateral PQRS and ABCD are not

similar.

Exercise 6.2 (10)

Question: 1

In the given figure, DE

BC. Find EC in (i) and AD in

(ii).

Fig. Exc_6.2_1

Solution

(i) Let

EC a cm

Given: .

DE||BC

By basic proportionality theorem,



AD AE

DB EC



1.5 1

3 a





3 1

1.5

a 2

Hence,

EC 2cm

(ii) Let

AD a cm

Given: .

DE||BC

By basic proportionality theorem,



AD AE

DB EC



a 1.8

7.2 5.4





1.8 7.2

5.4

a 2.4

Hence,

AD 2.4cm

Question: 2

E and 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 3.9 cm, EQ 3 cm, PF 3.6 cm and FR



2.4 cm

(ii) PE 4 cm, QE 4.5 cm, PF 8 cm and RF 9



(iii) PQ 1.28 cm, PR 2.56 cm, PE 0.18 cm and



PF 0.36 cm



Solution

(i) Given:

PE 3.9 cm, EQ 3 cm, PF 3.6 cm and FR



2.4 cm

Fig. Exc_6.2_2 (i)



PE 3.9

EQ 3

1.3



PF 3.6

FR 2.4

1.5

Thus,



PE PF

EQ FR

Hence, EF is not parallel to QR.

(ii) Given:

PE 4 cm, QE 4.5 cm, PF 8 cm and RF 9



Fig. Exc_6.2_2 (ii)



PE 4

EQ 4.5



PF 8

FR 9

Thus,



PE PF

EQ FR

Hence, .

EF||QR

(iii) Given:

PQ 1.28 cm, PR 2.56 cm, PE 0.18 cm and



PF 0.36 cm



Fig. Exc_6.2_2 (iii)



PE 0.18

PQ 1.28



PF 0.36

PR 2.56



Thus,



PE PF

PQ PR

Hence, .

EF||QR

Question: 3

In the given figure, if LM

CB and LN

CD, prove

that .



AM AN

AB AD

Fig. Exc_6.2_3

Solution

Given: LM || CB.

By basic proportionality theorem,



AM AL

......(1)

AB AC

Also, given: LN || CD.

By basic proportionality theorem,



AN AL

......(2)

AD AC

From equations (1) and (2).



AM AN

AB AD

Hence proved.

Question: 4

In the given figure, DE || AC and DF

AE. Prove that



BF BE

FE EC

Fig. Exc_6.2_4

Solution

Given: DE || AC.

By basic proportionality theorem,



BD BE

......(1)

DA EC

Also, given: DF || AE.

By basic proportionality theorem,



BD BF

......(2)

DA FE

From equations (1) and (2).



BF BE

FE EC

Hence proved.

Question: 5

In the given figure, DE || OQ and DF || OR. Show that

EF || QR.

Fig. Exc_6.2_5

Solution

Given: DE || OQ.

By basic proportionality theorem,



PE PD

......(1)

EQ DO

Also, given: DF || OR.

By basic proportionality theorem,



PF PD

......(2)

FR DO

From equations (1) and (2).



PE PF

EQ FR

By converse of basic proportionality theorem,

QR.

Hence proved.

Question: 6

In the given figure, A, B and C are points on OP, OQ

and OR respectively such that AB || PQ and AC || PR.

Show that BC || QR.

Fig. Exc_6.2_6

Solution

Given: AB || PQ.

By basic proportionality theorem,



OA OB

......(1)

AP BQ

Also, given: AC || PR.

By basic proportionality theorem,



OA OC

......(2)

AP CR

From equations (1) and (2).



OB OC

BQ CR

By converse of basic proportionality theorem,

BC || QR.

Hence proved.

Question: 7

Using Theorem 6.1, prove that a line drawn through

the mid-point of one side of a triangle parallel to

another side bisects the third side.

Solution

Let LMN be a triangle and the line AB is drawn

through the mid-point A of LM.

Also, .

AB||MN

Fig. Exc_6.2_7

By basic proportionality theorem,



LB LA

BN AM



 

LA AM

LB BN

Thus, B is the mid-point of LN.

Hence, AB bisects LN at point B.

Question: 8

Using Theorem 6.2, prove that the line joining the

mid-points of any two sides of a triangle is parallel to

the third side.

Solution

Let LMN be a triangle and the line AB is drawn

through the mid-point A of LM to the mid-point B of

LN.

Thus, and

LA AM

LB BN

Fig. Exc_6.2_8

From the above results,

and



LA 1

AM 1



LB 1

BN 1

Thus,



LA LB

AM BN

By the converse of basic proportionality theorem,

AB||MN

Hence proved.

Question: 9

ABCD is a trapezium in which AB

DC and its

diagonals intersect each other at the point O. Show

that .



AO CO

BO DO

Solution

Given: A trapezium ABCD in which AB

DC and

diagonals of the trapezium intersect each other at

the point O.

Fig. Exc_6.2_9

Draw through O.

EF||CD

In , .

ADC

EO||CD

By the basic proportionality theorem,

)



AE AO

......(1)

ED CO

In , .

ABD

OE||AB

By the basic proportionality theorem,



AE BO

......(2)

ED DO

From equations (1) and (2).



AO BO

CO DO



AO CO

BO DO

Hence proved.

Question: 10

The diagonals of a quadrilateral ABCD intersect each

other at the point O such that . Show that



AO CO

BO DO

ABCD is a trapezium.

Solution

Given: A quadrilateral ABCD in which diagonals

intersect each other at the point O such that



AO CO

BO DO

Fig. Exc_6.2_10

Draw .

OE||AB

In ABD, OE || AB.



By the basic proportionality theorem,



AE BO

......(1)

ED DO

Given:



AO CO

BO DO



AO BO

......(2)

CO DO

From equations (1) and (2).



AE AO

ED CO

By the converse of basic proportionality theorem,

EO||DC

AB||EO||DC

AB||CD

Hence, ABCD is a trapezium.

Exercise 6.3 (16)

Question: 1

State which pairs of triangles in the given figure 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)

Fig. Exc_6.3_1 (i)

(ii)

Fig. Exc_6.3_1 (ii)

(iii)

Fig. Exc_6.3_1 (iii)

(iv)

Fig. Exc_6.3_1 (iv)

(v)

Fig. Exc_6.3_1 (v)

(vi)

Fig. Exc_6.3_1 (vi)

Solution

(i) From the figure,

    A P 60

    B Q 80

    C R 40

Thus,

 

  By AAA similarity crABC|| itePQR rion

 

AB BC CA

QR RP PQ

(ii) From the figure,

 

AB 2 1

QR 4 2

 

CA 3 1

PQ 6 2

 

BC 2.5 1

RP 5 2

Thus,

 

AB CA BC

QR PQ RP

Hence,

 

  By SSS similarity crABC|| itePQR rion

(iii) From the figure,

The corresponding sides of the triangles are not

proportional.

Hence, the given triangles are not similar.

(iv) From the figure,

The corresponding sides of the triangles are not

proportional.

Hence, the given triangles are not similar.

(v) From the figure,

The corresponding sides of the triangles are not

proportional.

Hence, the given triangles are not similar.

(vi) We know that the sum of the interior angles of

a triangle is .

180

Thus, in ,

DEF

      D E F 180

      70 80 F 180

  F 30

Similarly, in ,

PQR

      P Q R 180

      P 80 30 180

  P 70

In and ,

DEF

PQR

  D P

  E Q

  F R

Hence,

 

  By AAA similarity crDEF|| itePQR rion

Question: 2

In the given figure, , and

 ODC|| OBA

  BOC 125

. Find , and .

  CDO 70

DOC

DCO

OAB

Fig. Exc_6.3_2

Solution

From the figure,

The line DOB is straight.

Thus,

    DOC COB 180

      DOC 180 125 55

We know that the sum of the interior angles of a

triangle is .

180º

In ,

DOC

      DCO CDO DOC 180

      DCO 70 55 180

  DCO 55

Given: .

 ODC|| OBA

Thus,

 

  OAB DCO Corresponding angles

Hence, .

  OAB 55

Question: 3

Diagonals AC and BD of a trapezium ABCD with

intersect each other at the point O. Using a

AB||DC

similarity criterion for two triangles, show that



OA OB

OC OD

Solution

From the figure,

  CDO ABO

 

Alternate interior angles

  DCO BAO

 

Alternate interior angles

  DOC BOA

 

Vertically opposite angles

Fig. Exc_6.3_3

Thus,

 

 DOC|| BOA By AAA similarity criterion

Thus,



DO OC

BO OA



OA OB

OC OD

Hence proved.

Question: 4

In the given figure, and . Show



QR QT

QS PR

  1 2

that .

 PQS TQR:

Fig. Exc_6.3_4

Solution

Given:

  1 2

Thus,

 .....PQ PR .(1)

Also, given:



QR QT

QS PR

From equation (1),



QR QT

......(2)

QS QP

In and ,

PQS

TQR



QR QT

QS QP

  So, Q Q

Thus,

 

 PQS|| TQR By SASsimilarity criterion

Hence proved.

Question: 5

S and T are points on sides PR and QR of such

PQR

that . Show that .

  P RTS

 RPQ|| RTS

Solution

From the figure,

In and ,

RPQ

RST

  RTS QPS

  R R

 

Common angle

Fig. Exc_6.3_5

Thus,

 

 RPQ|| RTS By AA similarity criterion

Hence proved.

Question: 6

In the given figure, if , show that

  ABE ACD

 ADE|| ABC

Fig. Exc_6.3_6

Solution

Given:

  ABE ACD

 

AB AC By CPCT ......(1)

 

 By CPCT .....AD AE .(2)

Divide equation (2) by (1).



AD AE

AB AC

In and ,

ADE

ABC

  A A

 

Common angle



AD AE

AB AC

Thus,

 

 ADE|| ABC By SAS similarity criterion

Hence proved.

Question: 7

In the given figure, altitudes AD and CE of

ABC

intersect each other at the point P.

Fig. Exc_6.3_7

Show that:

(i)

 AEP|| CDP

(ii)

 ABD|| CBE

(iii)

 AEP|| ADB

(iv)

 PDC|| BEC

Solution

(i)

 AEP|| CDP

In and ,

AEP

CDP

  AEP CDP

 

 Each angleisof 90

  APE CPD

 

Vertically opposite angles

Hence,

 

 AEP|| CDP AA similarity criterion

(ii)

 ABD|| CBE

In and ,

ABD

CBE

  ADB CEB

 

Each angleisof 90

  ABD CBE

 

Common

Hence,

 

 ABD|| CBE By AA similarity criterion

(iii)

 AEP|| ADB

In and ,

AEP

ADB

  AEP ADB

 

Each angleisof 90

  PAE DAB

 

Common

Hence,

 

 AEP|| ADB By AA similarity criterion

(iv)

 PDC|| BEC

In and ,

PDC

BEC

  PDC BEC

 

Each angleisof 90

  PCD BCE

 

Common angle

Hence,

 

 PDC|| BEC By AA similarity criterion

Question: 8

E is a point on the side AD produced of a

parallelogram ABCD and BE intersects CD at F.

Show that .

 ABE|| CFB

Solution

From the figure,

In and ,

ABE

CFB

Fig. Exc_6.3_8

  A C

 

Opposite angles of a parallelogram

Thus,

  AEB CBF

 

Alternate interior angles

 ABE|| CFB

 

By AA similarity criterion

Question: 9

In the given figure, ABC and AMP are two right

triangles, right angled at B and M respectively.

Fig. Exc_6.3_9

Prove that:

(i)

 ABC|| AMP

(ii)



CA BC

PA MP

Solution

(i) To Prove:

 ABC|| AMP

In and ,

ABC

AMP

  ABC AMP

 

Each angleisof 90

  A A

 

Common

Thus,

 

 ABC|| AMP By AA similarity criterion

(ii) To Prove:



CA BC

PA MP

We have proved that .

 ABC|| AMP

We know that the corresponding sides of similar

triangles are proportional.

Thus, in and ,

ABC

AMP



CA BC

PA MP

Hence, proved.

Question: 10

CD and GH are respectively the bisectors of

ACB

and such that D and H lie on sides AB and FE

EGF

of ABC and EFG respectively. If ,



 ABC FEG:

show that:

(i)



CD AC

GH FG

(ii)

 DCB|| HGE

(iii)

 DCA|| HGF

Solution

Fig. Exc_6.3_10

Given: .

 ABC|| FEG

Thus,

  A F

  B E

  ACB FGE

Now,

 

  ACD FGH Angle bisector

 

  DCB HGE Angle bisector

(i) To Prove:



CD AC

GH FG

In and ,

ACD

FGH

  A F

  ACD FGH

 

 ACD|| FGH By AA similarity criterion

Thus,



CD AC

GH FG

(ii) To Prove:

 DCB|| HGE

In and ,

DCB

HGE

  DCB HGE

  B E

Thus,

 

 DCB|| HGE By AA similarity criterion

(iii) To Prove:

 DCA|| HGF

In and ,

DCA

HGF

  ACD FGH

  A F

Thus,

 

 DCA|| HGF By AA similarity criterion

Question: 11

In the given figure, E is a point on side CB produced

of an isosceles triangle ABC with AB AC. If AD





BC and EF AC, prove that .



 ABD|| ECF

Fig. Exc_6.3_11

Solution

Given: An isosceles triangle ABC.

AB AC



Thus, .

  ABD ECF

In and ,

ABD

ECF

 

   ADB EFC Each angleisof 90

  BAD CEF

Thus,

 

 ABD|| ECF By AA similarity criterion

Question: 12

Sides AB and BC and median AD of a triangle ABC

are respectively proportional to sides PQ and QR and

median PM of . Show that

PQR

 ABC|| PQR

Fig. Exc_6.3_12

Solution

Given:

 

AB BC AD

PQ QR PM

 

AB AD

PQ PM

We know that the median of a triangle divides the

opposite side.

Thus, and



Thus,

 

AB BD AD

PQ QM PM

In and ,

ABD

PQM

 

AB BD AD

PQ QM PM

Thus,

 

 ABD|| PQM By SSS similarity criterion

 

  ABD PQM Corresponding angles

In and ,

ABC

 PQR

 

  ABD PQM Corresponding angles



AB BC

PQ QR

Thus,

 

 ABC|| PQR By SAS similarity criterion

Hence proved.

Question: 13

D is a point on the side BC of a triangle ABC such

that . Show that .

  ADC BAC



CA CB.CD

Solution

Given:

  ADC BAC

Fig. Exc_6.3_13

In and ,

ADC

BAC

  ADC BAC

 

  ACD BCA Common angle

Thus,

 

 ADC|| BAC By AA similarity criterion

In similar triangles, corresponding sides are in

proportion.

Thus,



CA CD

CB CA

 

CA CB CD

Hence proved.

Question: 14

Sides AB and AC and median AD of a triangle ABC

are respectively proportional to sides PQ and PR and

median PM of another triangle PQR. Show that

 ABC|| PQR

Solution

Given:

 

AB AC AD

PQ PR PM

Fig. Exc_6.3_14 (i)

Expand AD and PM to the points E and L such that

AD DE and PM ML



Fig. Exc_6.3_14 (ii)

Join B to E, C to E, Q to L, and R to L.

Since median of a triangle divides the opposite side.

So,BD DC

QM MR

 

AD DE By construction

 

PM ML By construction

Diagonals AE and BC of quadrilateral ABEC, bisect

each other at point D.

Thus, ABEC is a parallelogram.

 

AC BE Opposite sides of a parallelogram

 

AB EC Opposite sides of a parallelogram

Similarly,

Quadrilateral PQLR is a parallelogram.

Thus,



PR QL

PQ LR

Given:

 

AB AC AD

PQ PR PM

 

AB AC 2AD

PQ PR 2PM

From the figure,

and

2AD AE

2PM PL

Thus,

 

AB AC AE

PQ PR PL

Thus,

 

 ABE|| PQL By SSS similarity criterion

Similarly,

 AEC|| PLR

 

  BAE QPL Corresponding angles ......(1)

 

  CAE RPL Corresponding angles ......(2)

Add equations (1) and (2).

      BAE CAE QPL RPL

  CAB RPQ ......(3)

In ABC and PQR,





AB AC

PQ PR

  CAB RPQ

Thus,

 

 ABC|| PQR By SAS similarity criterion

Hence proved.

Question: 15

A vertical pole of length 6 m casts a shadow 4 m long

on the ground and at the same time a tower casts a

shadow 28 m long. Find the height of the tower.

Solution

Let LM be a tower and PQ be a pole of length 6 m.

Let MN and QR be the shadow of LM and PQ

respectively.

MN 28m

PQ 6m

QR 4m

Fig. Exc_6.3_15

The rays of the sun falls on the tower and the pole at

the same time and at the same angle.

So, from the figure

  QPR MLN

  QRP MNL

(Tower and pole are standing

  PQR LMN

vertically on the ground)

Thus,

 

 || By AAA similarity critPQR LMN erion



LM MN

PQ QR



LM 28

6 4

 

LM 6

LM 42

Hence, the height of the tower is 42 m.

Question: 16

If AD and PM are medians of triangles ABC and

PQR, respectively where , prove that

 ABC|| PQR



AB AD

PQ PM

Solution

Given: .

 ABC|| PQR

Since, the corresponding sides of similar triangles

remain in proportion.

 

AB AC BC

PQ PR QR

  A P

  B Q

  C R

Fig. Exc_6.3_16

Also, .

 

AB AC

PQ PR

We know that medians divide their opposite sides

equally.

Thus, and



Thus,

 

AB AC BD

PQ PR QM

In ABD and PQM,





  

AB BD

B Q

Thus,

 

ABD|| PQM By SAS similarity criterion

Thus,

 

AB BD AD

PQ QM PM

Hence proved.

Exercise 6.4 (9)

Question: 1

Let and their areas be, respectively, 64

 ABC|| DEF

and 121 . If EF 15.4 cm, find BC.



Solution

Given: .

 ABC|| DEF

Thus,

 



     

  

     



     

2 2 2

ar ABC

AB BC AC

ar DEF DE EF DF

Also, given:

EF 15.4 cm



 

 

ar ABC 64cm

 

 

ar DEF 121cm

 



 



 



 

ar ABC

ar DEF EF

 



121

15.4



BC 8

15.4 11

 

BC 15.4

BC 11.2

Hence, cm

BC 11.2

Question: 2

Diagonals of a trapezium ABCD with AB DC

intersect each other at the point O. If AB 2CD, find



the ratio of the areas of triangles AOB and COD.

Solution

Given: A trapezium ABCD with AB DC.

Fig. Exc_6.4_2

In and ,

AOB

COD

 

  AOB COD Vertically opposite angles

 

  OAB OCD Alternate interior angles

 

  OBA ODC Alternate interior angles

Thus,

 

 AOB|| COD By AAA similarity criterion

Thus,

 



 



 



 

ar AOB

ar COD CD

Given: AB 2CD



Thus,

 



 



 



 

ar AOB

2CD

ar COD CD

 



 



 



 

ar AOB

ar COD 1

 







ar AOB

ar COD 1

Hence, the ratio of the areas of triangles AOB and

COD is .

4:1

Question: 3

In the given figure, ABC and DBC are two triangles

on the same base BC. If AD intersects BC at O, show

that .

 



ar ABC

ar DBC DO

Fig. Exc_6.4_3 (Ques.)

Solution

Given: Two triangles ABC and DBC on the same

base BC and AD intersects BC at O.

Draw two perpendiculars AP and DM on BC.

Fig. Exc_6.4_3 (Sol)

  Area of a triangle

base height

 

 

 

 

BC AP

ar ABC

ar DBC DM

BC DM

......(1)

In and ,

APO

DMO

 

   APO DMO Each angleisof 90

 

  AOP DOM Vertically opposite angles

Thus,

 

 APO DMO By AA similarity criterion:

So,

 ..

AP AO

...

M DO

.(2)

From (1) and (2),

 



ar ABC

ar DBC DO

Hence proved.

Question: 4

If the areas of two similar triangles are equal, prove

that they are congruent.

Solution

Let ABC and XYZ be two similar triangles.

 

     

  

     

     

2 2 2

ar ABC

AB BC AC

ar PQR XY Y

....

Z XZ

..(1)

Given:

   

ar ABC ar XYZ

 



ar ABC

1 .....

ar XYZ

.(2)

From equations (1) and (2).

     

  

     

     

2 2 2

AB BC AC

XY YZ XZ

Thus,

AB XY

BC YZ

AC XZ

Thus,

 

  ABC XYZ By SSScongruencecriterion

Hence proved.

Question: 5

D, E and F are respectively the mid-points of sides

AB, BC and CA of ABC. Find the ratio of the areas



of DEF and ABC.



Solution

Given: A triangle ABC in which D and E are the mid-

points of AB and BC.

Fig. Exc_6.4_5

and

DE||AC



In and ,

BED

BCA

 

  BED BCA Corresponding angles

 

  BDE BAC Corresponding angles

 

  EBD CBA Common angles

Thus,

 

 BED BCA By AAA similarity crite|| rion

 



 



 



 

ar BED

ar BCA AC

Put in above expression.



 

 

 





 



 

 

ar BED

ar BCA AC

 







ar BED

ar BCA 4

   

   

ar BED ar BC ....A

..(1)

Similarly,

   

   

ar CFE ar CB ....A

..(2)

   

   

ar ADF ar AB ....C

..(3)

         

        

 

 

ar DEF ar ABC ar BED ar CFE ar ADF

From equations (1), (2) and (3).

     

    

ar DEF ar ABC ar ABC

   

  

ar DEF ar ABC

 







ar DEF

ar ABC 4

Question: 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

Let LMN and ABC be two similar triangles with

medians LO and AD respectively.

Fig. Exc_6.4_6

Since ,

 ||LMN ABC

 

LN NM LM

AB BC AC

 

LN L

.(1)

Since LO and AD are medians,

and



Thus,

 

LN NO LM

AB BD AC

From the figure,

  L A

  N B

  M C

In and ,

LNO

ABD

  N B



LN NO

AB BD

 

 || By SAS similarity critLNO ABD erion

Thus,

 

LN NO LO

....

B BD AD

.(2)

 



     

  

     



     

2 2 2

ar LMN

LN NM LM

ar ABC AB BC AC

From equations (1) and (2).

  

LN NM LM LO

AB BC AC AD

Thus,

 



 



 



 

ar LMN

ar ABC AD

Hence proved.

Question: 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 LMNO be a square of side .

The length of diagonal of LMNO

 2k

Fig. Exc_6.4_7

LMP is an equilateral triangle described on one of



the sides of the square and OMQ is an equilateral



triangle described on one of the diagonals of the

square.

Length of the side of LMP



 k

Length of the side of OMQ



 2k

All equilateral triangles have sides of the same length

and their angles as .

60

Thus, equilateral triangles are similar to each other.

Now, find the ratio between the areas of LMP and



OMQ



 

 

 

 

 



k 1

LMP

OMQ 2

Thus, the ration between the areas of LMP and



OMQ is .

1:2

Question: 8

Tick the correct answer and justify:

ABC and BDE are two equilateral triangles such that

D is the mid-point of BC. Ratio of the areas of

triangles ABC and BDE is

(A)

2:1

(B)

1:2

(C)

4:1

(D)

1:4

Solution

such that D is the mid-point of BC.

Fig. Exc_6.4_8

The sides of equilateral triangles are of same length

and all the angles are of .

60

Thus, equilateral triangles are similar to each other.

Thus, the ratio of the areas of equilateral triangles is

equal to the square of the ratio of their sides.

Let us consider the side of ABC be .



Thus, the side of BDE =



 

 

 



 

 



 

 

ar ABC

x 4

ar BDE 1

Hence, the ratio of the areas of triangles ABC and

BDE is 4 : 1.

Question: 9

Tick the correct answer and justify:

Sides of two similar triangles are in the ratio .

4:9

Areas of these triangles are in the ratio

(A)

2:3

(B)

4:9

(C)

81:16

(D)

16:81

Solution

(D)

Given: Sides of two similar triangles in the ratio

4:9

In similar triangles, the ratio of the areas is

equal to the square of the ratio of the

corresponding sides.

The ratio of the areas of given triangles

 



 

 



Thus, the ratio is 16 : 81.

Exercise 6.5 (17)

Question: 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) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Solution

(i) Given: 7 cm, 24 cm, 25 cm

Let us consider the lengths of the sides ,

a 7

, and

b 24

c 25

   

  

2 2

a b 7 24

  

2 2

a b 49 576

 

2 2

.....a b 625 .(1)

Now

 



c 25



c 625

From equations (1) and (2),

 

2 2 2

a b c

This satisfies the Pythagoras theorem.

Thus, the given triangle is a right triangle with

hypotenuse 25 cm.

(ii) Given: 3 cm, 8 cm, 6 cm

Let us consider the lengths of the sides ,

a 3

, and

b 6

c 8

   

  

2 2

a b 3 6

  

2 2

a b 9 36

 

2 2

.....a b 45 .(1)

Now

 



c 8



c 64

From equations (1) and (2),

 

2 2 2

a b c

This does not satisfy the Pythagoras theorem.

Thus, the given triangle is not a right triangle.

(iii) Given: 50 cm, 80 cm, 100 cm

Let us consider the lengths of the sides ,

a 50

, and

b 80

c 100

   

  

2 2

a b 50 80

  

2 2

a b 2500 6400

 

2 2

a b 8900 ......(2)

Now

 



c 100



c 10000

From equations (1) and (2),

 

2 2 2

a b c

This does not satisfy the Pythagoras theorem.

Thus, the given triangle is not a right triangle.

(iv) Given: 13 cm, 12 cm, 5 cm

Let us consider the lengths of the sides ,

a 5

, and

b 12

c 13

   

  

2 2

a b 5 12

  

2 2

a b 25 144

 

2 2

.....a b 169 .(1)

Now

 



c 13



c 169

From equations (1) and (2),

 

2 2 2

a b c

This satisfies the Pythagoras theorem.

Thus, the given triangle is a right triangle.

Question: 2

PQR is a triangle right angled at P and M is a point

on QR such that PM QR. Show that





PM QM.MR

Solution

Given: A triangle PQR right angled at P and M is a

point on QR such that PM QR.



Fig. Exc_6.5_2

To prove:



PM QM.MR

Proof: It is given that PM QR.



Thus, .

 PQM|| PRM



PM MR

QM PM



PM QM.MR

Hence proved.

Question: 3

In the given figure, ABD is a triangle right angled at

A and AC BD.



Fig. Exc_6.5_3

Show that

(i)



AB BC.BD

(ii)



AC BC.DC

(iii)



AD BD.CD

Solution

Given: A triangle ABD, right angled at A and AC



BD.

(i) To Prove:



AB BC.BD

It is given that AC BD.



Thus,.

 ||ABC ABD

So,



AB BC

BD AB



AB BC.BD

Hence proved.

(ii) To Prove:



AC BC.DC

From equation (1),

 ||ABC ADC



AC DC

BC AC



AC BC.DC

Hence proved.

(iii) To Prove:



AD BD.CD

From equation (2),

 ||ADC ABD



AD BD

CD AD



AD BD.CD

Hence proved.

Question: 4

ABC is an isosceles triangle right angled at C. Prove

that .



2 2

AB 2AC

Solution

Given: An isosceles triangle ABC, right angled at C

and

AC BC

Fig. Exc_6.5_4

By the Pythagoras theorem,

     

 

2 2 2

AB AC BC

     

 

  

2 2 2

AB AC AC AC BC

   



2 2

AB 2 AC

Hence proved.

Question: 5

ABC is an isosceles triangle with . If

AC BC

, prove that ABC is a right triangle.



2 2

AB 2AC

Solution

Given: An isosceles triangle ABC such that

AC BC

and .



2 2

AB 2AC

Fig. Exc_6.5_5



2 2

AB 2AC

 

2 2 2

AB AC AC

 

 

2 2 2

AB AC B ACC BC

Thus, by Pythagoras theorem, ABC is a triangle,

right angled at C.

Hence proved.

Question: 6

ABC is an equilateral triangle of side . Find each

of its altitudes.

Solution

Given: An equilateral triangle ABC with side .

Thus,

  AB BC CA 2a ......(1)

Draw AD BC such that D be the mid-point of BC.



Fig. Exc_6.5_6



 .....BD a .(2)

In right triangle ABD, by the Pythagoras theorem,

 

2 2 2

AB AD BD

From equations (1) and (2).

 

 

2 2

2a AD a



2 2

AD 3a

AD 3a

Hence, each of the altitudes of the triangle is .

Question: 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

Let PQRS be a rhombus with diagonals PS and QR

that intersect each other at O.

Fig. Exc_6.5_7

We know that, the diagonals of a rhombus, bisect

each other at right angles.

Thus,

       POQ QOS SOR ROP 90

and

PO SO

QO OR

Now, by Pythagoras theorem in right triangle POQ,

     

 

2 2 2

PQ PO QO

 

   

 

   

   

2 2

PS QR

2 2

     

 

2 2 2

4 PQ PS QR ......(1)

Similarly,

     

 

2 2 2

4 QS PS QR ......(2)

     

 

2 2 2

4 SR PS QR ......(3)

     

 

2 2 2

4 RP PS QR ......(4)

Add equations (1), (2), (3), and (4).

           

   

    

   

2 2 2 2 2 2

4 PQ QS SR RP 4 PS QR

           

    

2 2 2 2 2 2

PQ QS SR RP PS QR

Hence proved.

Question: 8

In the given figure, O is a point in the interior of a

triangle ABC, OD BC, OE AC and OF AB.



Fig. Exc_6.5_8 (Ques.)

Show that

(i)

    

2 2 2 2 2 2 2 2 2

OA OB OC – OD – OE – OF AF BD CE

(ii)

    

2 2 2 2 2 2

AF BD CE AE CD BF

Solution

Given: A triangle ABC such that OD BC, OE AC



and OF AB.



Fig. Exc_6.5_8 (Sol.)

Join AO, BO and CO.

(i)

    

2 2 2 2 2 2 2 2 2

OA OB OC – OD – OE – OF AF BD CE

By Pythagoras theorem in right triangles OFA,

ODB and OEC.

 

2 2 2

....OA AF OF ..(1)

 

2 2 2

....OB BD OD ..(2)

 

2 2 2

....OC CE OE ..(3)

Add equations (1), (2), and (3),

       

2 2 2 2 2 2 2 2 2

OA OB OC AF OF BD OD CE OE

    

2 2 2 2 2 2 2 2 2

OA OB OC – OD – OE – OF AF BD CE

Hence proved.

(ii)

    

2 2 2 2 2 2

AF BD CE AE CD BF

By Pythagoras theorem in right triangles ODB

and ODC.

 

2 2 2

....OB BD OD ..(1)

 

2 2 2

....OC OD CD ..(2)

Subtract equation (2) from (1).

  

2 2 2 2

OB OC BD C .....D .(3)

Similarly,

  

2 2 2 2

OC OA CE A .....E .(4)

  

2 2 2 2

OA OB AF B .....F .(5)

Add equations (3), (4), and (5),

          

2 2 2 2 2 2 2 2 2 2 2 2

OB OC OC OA OA OB BD CD CE AE AF BF

     

2 2 2 2 2 2

BD CD CE AE AF BF 0

    

2 2 2 2 2 2

AF BD CE AE CD BF

Hence proved.

Question: 9

A ladder 10 m long reaches a window 8 m above the

ground. Find the distance of the foot of the ladder

from base of the wall.

Solution

Let LM be a ladder that reaches the window and MN

be the wall.

and

LM 10m

MN 8m

Fig. Exc_6.5_9

From the figure,

LMN is a right triangle, right angled at N.

Now, by Pythagoras theorem in triangle LMN,

   

 

2 2

10 LN 8

 

LN 100 64



LN 36

LN 6m

Hence, the distance of the foot of the ladder from

base of the wall is 6 m.

Question: 10

A guy wire attached to a vertical pole of height 18 m

is 24 m 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

Let LM be a guy wire and MN be a vertical pole.

Let the wire is fixed to a stake at L to keep the wire

taut.

 LM 24m andMN 18m

Fig. Exc_6.5_10

Now, by Pythagoras theorem in right triangle LMN.

 

2 2 2

LM LN MN

   

 

2 2

24 LN 18

 

LN 576 324



LN 252

LN 6 7

Thus, the stake should be driven at a distance of

6 7

m from the base of the pole so that the wire will be

taut.

Question: 11

An aeroplane leaves an airport and flies due north at

a speed of 1000 km per hour. At the same time,

another aeroplane leaves the same airport and flies

due west at a speed of 1200 km per hour. How far

apart will be the two planes after hours?

Solution

Let the first aeroplane leaves the airport from M at a

speed of 1000 km per hour and goes upto N towards

north.

Fig. Exc_6.5_11

We know that

 Distance speed time

Thus,

 

MN 1000 1

 

MN 1000

MN 1500km

Let the second aeroplane leaves the airport from M

at a speed of 1200 km per hour and goes upto L

towards north.

 

LM 1200 1

 

LM 1200

LM 1800km

Now, by Pythagoras theorem in triangle LMN,

 

2 2 2

LN LM MN

   

 

2 2

LN 1800 1500

 

LN 3240000 2250000



LN 5490000

LN 300 61

Hence, the two planes will be at a distance of

300 61

km after hours.

Question: 12

Two poles of heights 6 m and 11 m stand on a plane

ground. If the distance between the feet of the poles

is 12 m, find the distance between their tops.

Solution

Let LM and NO be the two poles at a distance of 12

LM 6m

NO 11m

MO 12m

Draw and join LN.

LP NO

Fig. Exc_6.5_12

From the figure,

 LP MO 12m

 LM PO 6m

 



NP NO PO

NP 11 6

NP 5m

Now, by Pythagoras theorem in .

LNP

 

2 2 2

LN LP NP

   

 

2 2

LN 12 5

 

LN 144 25



LN 169

LN 13

Hence, the tops of the two poles are 13 m apart.

Question: 13

D and E are points on the sides CA and CB

respectively of a triangle ABC right angled at C.

Prove that .

  

2 2 2 2

AE BD AB DE

Solution

Given: A triangle ABC right angled at C with points

D and E on the sides CA and CB respectively.

Fig. Exc_6.5_13

By Pythagoras theorem in triangles ACE and DCB.

 

2 2 2

....AE AC CE ..(1)

 

2 2 2

....BD DC BC ..(1)

Add equations (1) and (2).

    

2 2 2 2 2 2

AE BD AC CE DC BC

    

2 2 2 2 2 2

AE BD AC BC D ....C CE ..(1)

Again by Pythagoras theorem,

and

 

2 2 2

AC BC AB

 

2 2 2

DC CE DE

Put the above result in equation (3).

So,

  

2 2 2 2

AE BD AB DE

Hence proved.

Question: 14

The perpendicular from A on side BC of a ABC



intersects BC at D such that DB 3 CD (see the



given figure).

Fig. Exc_6.5_14

Prove that .

 

2 2 2

2AB 2AC BC

Solution

Given: A ABC, in which perpendicular from A on



BC intersects BC at D.

And,

 .....DB 3CD .(1)

From the figure,

 BC DB CD

From equation (1),

 BC 3CD CD

BC 4CD

 .....

.(2)

From equation (1),

DB 3CD



3BC

DB ......(3)

By Pythagoras theorem in ,

ABD

 

2 2 2

....AB AD DB ..(4)

Similarly, in ,

ACD

 

2 2 2

....AC AD CD ..(5)

Subtract equation (5) from (4).

    

2 2 2 2 2 2

AB AC AD DB AD CD

  

2 2 2 2

AB AC DB CD

From equations (2) and (3).

   

  

   

   

2 2

3 1

AB AC BC BC

4 4

 

  

 

 

2 2 2

9 1

AB AC BC

16 16

 

2 2 2

AB AC BC

 

2 2 2

2AB 2AC BC

 

2 2 2

2AB 2AC BC

Hence proved.

Question: 15

In an equilateral triangle ABC, D is a point on side

BC such that . Prove that .



BD BC



2 2

9AD 7AB

Solution

Given: An equilateral triangle ABC with a point D on

side BC and



BD BC

Join AD and draw .

AE BC

Fig. Exc_6.5_15

In triangles AEB and AEC,

 

 Equilateral trianglepropAB AC erty

 

   AEB AEC Each angleisof 90

 

AE AE Common

Thus,

 

  By SASsimilarity criAEB A teE nC rio:

So,

BE EC

Thus,

and



BD ,



2BC

DC ,

  ...

BE EC

...(1)

We know that in an equilateral triangle, all angles

are of .

60

Thus, .

  C 60

So, is an acute angle triangle.

ADC

   

2 2 2

AD AC DC 2DC EC

From equation (1).

 

    

 

 

2 2

2 2 1

AD AC BC 2 BC BC

3 3 2

  

2 2 2 2

4 2

AD AC BC BC

9 3

 

    

2 2 2 2

4 2

AD AB AB AB AB BC AC

9 3





2 2

AD AB



2 2

9AD 7AB

Hence proved.

Question: 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

Let PQR be an equilateral triangle such that,

PS QR

Fig. Exc_6.5_16

In triangles PQS and PSR,

 

PQ PR Given

 

    Q R 60 Given

 

   PSQ PSR Each angleisof 90

Thus,

 

   By RHS criterion of congrPSQ PSR uence

Thus,

QS SR

 

QS SR

By Pythagoras theorem in ,

PQS

 

2 2 2

PQ PS QS

 

 

 

 

2 2

PQ PS

 

 

  

 

 

2 2

PQ PS QR PQ



 

2 2

PQ PS

 

2 2 2

4PQ PQ 4PS



2 2

3PQ 4PS

Hence proved.

Question: 17

Tick the correct answer and justify: In ΔABC,

cm, AC 12 cm and BC 6 cm. The angle

AB 6 3



B is:

(A)

120

(B)

60

(C)

90

(D)

45

Solution

(C)

Given: A triangle ABC, in which, cm, AC

AB 6 3



12 cm and BC 6 cm



Fig. Exc_6.5_17

 

  

2 2

AB BC 6 3 6

 

  

2 2

AB BC 6 3 6

 

2 2

AB BC 144 ......(1)

 



AC 12



.....AC 144 .(2)

From equations (1) and (2).

 

2 2 2

AB BC AC

Thus, by Pythagoras theorem, triangle ABC is a right

triangle.

Thus,

  B 90

Exercise 6.6 (10)(Optional)

Question: 1

In the given figure, PS is the bisector of of

QPR



PQR. Prove that .



QS PQ

SR PR

Fig. Exc_6.6_1 (Ques.)

Solution

Given: A triangle PQR with PS as the internal

bisector of .

QPR

  QPS SPR

To prove:



QS PQ

SR PR

Construction:

Extend QP and draw RT parallel to SP, such that it

intersects the extended part of QP at T.

Fig. Exc_6.6_1 (Sol.)

Proof:

… (1)

 

  SPR PRT Alternateangles

… (2)

 

  QPS PTR Corresponding angles

Given that, … (3)

  QPS SPR

From equations (1), (2), and (3),

  PRT PTR

We know that, the opposite sides of equal angles are

equal.

Thus, … (4)

PT PR

In ,

QRT

 

RT SP By constru|| ction

Thus, by Thales theorem,



QS PQ

SR PT

From equation (4),



QS PQ

SR PR

Hence proved.

Question: 2

In the given figure, D is a point on hypotenuse AC of

ABC, DM BC and DN AB.





Fig. Exc_6.6_2 (Ques.)

Prove that:

(i)



DM DN.MC

(ii)



DN DM.AN

Solution

Given: A triangle ABC in which, D is a point on

hypotenuse.

DM BC and DN AB.



Fig. Exc_6.6_2 (Sol.)

AB BC

DM BC



Thus, .

AB DM:

Similarly,

BC AB

DN AB

Thus,

BC DN:

So, the quadrilateral BMDN forms a rectangle.

Thus,

BM ND

(i) To prove:



DM DN.MC

We know that the sum of all the interior angles

in a triangle is .

180

Thus, in ,

BMD

      1 BMD 2 180

      1 90 2 180

      1 2 180 90

… (1)

    1 2 90

Similarly, in ,

DMC

… (2)

    3 4 90

From the figure,

BD AC

Thus, … (3)

    2 3 90

From equations (1) and (3).

      1 2 2 3

  1 3

From equations (2) and (3).

      3 4 2 3

  4 2

In triangles BMD and DMC,

and

  1 3

  4 2

Thus, .

 BMD DMC:



BM MD

DM MC

 

 

DN DM

BM DN

DM MC





DM DN.MC

Hence proved.

(ii) To prove:



DN DM.AN

Similarly, as in first part,

 BND DNA:

Thus,



BN DN

DN NA

 

 

DM DN

BN DM

DN AN





DN DM.AN

Hence proved.

Question: 3

In the given figure, ABC is a triangle in which

and AD CB produced.

  ABC 90



Fig. Exc_6.6_3

Prove that .

  

2 2 2

AC AB BC 2BC.BD

Solution

Given: A triangle ABC such that and

  ABC 90

AD CB produced.



To prove:

  

2 2 2

AC AB BC 2BC.BD

Proof: By Pythagoras theorem in ,

ADB

… (1)

 

2 2 2

AB AD BD

Now by Pythagoras theorem in ,

ADC

 

2 2 2

AC AD DC

 

 

    

2 2

AC AD BD BC DC BD BC

    

2 2 2 2

AC AD BD BC 2BD BC

From equation (1),

   

2 2 2

AC AB BC 2BD BC

  

2 2 2

AC AB BC 2BC.BD

Hence proved.

Question: 4

In the given figure, ABC is a triangle in which

and AD BC.

  ABC 90



Fig. Exc_6.6_4

Prove that .

  

2 2 2

AC AB BC 2BC.BD

Solution

Given: A triangle ABC such that and

  ABC 90

AD CB.



To prove:

  

2 2 2

AC AB BC 2BC.BD

Proof: By Pythagoras theorem in ,

ADB

… (1)

 

2 2 2

AB AD BD

Now, by Pythagoras theorem in ,

ADC

 

2 2 2

AC AD DC

 

 

    

2 2

AC AD BC BD DC BC BD

    

2 2 2 2

AC AD BD BC 2BD BC

From equation (1),

   

2 2 2

AC AB BC 2BD BC

  

2 2 2

AC AB BC 2BC.BD

Hence proved.

Question: 5

In the given figure, AD is a median of a triangle ABC

and AM BC.



Fig. Exc_6.6_5

Prove that:

(i)

 

  

 

 

2 2

AC AD BC.DM

(ii)

 

  

 

 

2 2

AB AD BC.DM

(iii)

  

2 2 2

AC AB 2AD

Solution

Given: A triangle ABC with median AD and AM



BC.

  AMD 90

AD is a median on BC.

Thus, and



Thus, and .

  ADM 90

  ADC 90

(i) Since ,

  ADC 90

In ,

ADC

  

2 2 2

AC AD DC 2DC.DM

     

   

   

 

     

2 2

BC BC BC

AC AD 2 .DM DC

2 2 2



 

  

 

 

2 2

AC AD BC.DM

 

  

 

 

2 2

AC AD BC.DM

Hence proved.

(ii) Since ,

  ADM 90

  

2 2 2

AB AD BD 2BD.DM

     

   

   

 

     

2 2

BC BC BC

AB AD 2 .DM BD

2 2 2



 

  

 

 

2 2

AB AD BC.DM

 

  

 

 

2 2

AB AD BC.DM

Hence proved.

(iii) From above two proofs,

… (1)

 

  

 

 

2 2

AC AD BC.DM

… (2)

 

  

 

 

2 2

AB AD BC.DM

Add equations (1) and (2).

   

      

   

   

2 2

2 2 2 2

BC BC

AC AB AD BC.DM AD BC.DM

2 2

 

  

 

 

2 2 2

AC AB 2AD 2

  

2 2 2

AC AB 2AD

Hence proved.

Question: 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 PQRS be a parallelogram with diagonals PR and

QS.

Fig. Exc_6.6_6

PS is a median of .

PQR

Thus, .

  

2 2 2

PQ PR 2PS

The medians of triangles PQR and PSR are QO and

SO because diagonals in a parallelogram bisect each

other

Thus,

… (1)

  

2 2 2 2

PQ QR 2QO PR

… (2)

  

2 2 2 2

PS RS 2SO PR

Add equations (1) and (2).

      

2 2 2 2 2 2 2 2

1 1

PQ QR PS RS 2QO PR 2SO PR

2 2

 

     

2 2 2 2 2 2 2

PQ QR PS RS 2 QO SO PR

 

     

       

 

   

 

 

     

 

2 2

2 2 2 2 2

QS QS QS

PQ QR PS RS 2 PR QO SO

2 2 2



 

    

 

 

2 2 2 2 2

2QS

PQ QR PS RS 2 PR

    

2 2 2 2 2 2

PQ QR PS RS QS PR

Hence proved.

Question: 7

In the given figure, two chords AB and CD intersect

each other at the point P.

Fig. Exc_6.6_7

Prove that:

(i)

 APC|| DPB

(ii)

AP.PB CP.DP

Solution

(i) To prove:

 APC|| DPB

In triangles APC and DPB,

 

  APC DPB Vertically oppositeangles

 

  CAP BDP Anglesin samesegment

Thus,

 

 APC|| DPB By AA similarity criterion

Hence proved.

(ii) To prove:

AP.PB CP.DP

From the first part,

 APC|| DPB

Thus,



AP CP

DP PB

AP.PB CP.DP

Hence proved.

Question: 8

In the given figure, two chords AB and CD of a circle

intersect each other at the point P (when produced)

outside the circle.

Fig. Exc_6.6_8

Prove that

(i)

 PAC PDB:

(ii)

PA.PB PC.PD

Solution

(i)

 PAC|| PDB

In triangles PAC and PDB,

 

  APC DPB Common

… (1)

    BAC 180 PAC

      PDB CDB 180 BAC

From equation (1),

 

        PDB CDB 180 180 PAC

  PDB PAC

Thus,

 

 PAC|| PDB By AA similarity criterion

Hence proved.

(ii)

PA.PB PC.PD

From the first part,

 PAC|| PDB

Thus,



PA PC

PD PB

PA.PB PC.PD

Hence proved.

Question: 9

In the given figure, D is a point on side BC of ABC



such that



BD AB

CD AC

Fig. Exc_6.6_9 (Ques.)

Prove that AD is the bisector of .

BAC

Solution

Given: A ABC in which D is a point on side BC



such that .



BD AB

CD AC

To prove: AD is the bisector of .

BAC

Construction: Extend BA to a point E such that,

and join CE.

AE AC

Fig. Exc_6.6_9 (Sol.)

Proof: In ,

AEC

AE AC

We know that the opposite angles of equal sides in a

triangle are equal.

Thus, … (1)

  AEC ACE

Given that,



BD AB

CD AC

 

 

BD AB

AC AE

CD AE



By the converse of basic proportionality theorem,

DA CE:

… (2)

 

  BAD AEC Corresponding angles

… (3)

 

  DAC ACE Alternateangles

From equations (2) and (3),

  BAD DAC

 

  AEC ACE

Thus, AD is the bisector of .

BAC

Hence proved.

Question: 10

Nazima is fly fishing in a stream. The tip of her

fishing rod is 1.8 m above the surface of the water

and the fly at the end of the string rests on the water

3.6 m away and 2.4 m from 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 given figure)? If she pulls in the

string at the rate of 5 cm per second, what will be the

horizontal distance of the fly from her after 12

seconds?

Fig. Exc_6.6_10 (Ques.)

Solution

Let R be the tip of the fishing rod and the fly at the

end of the string rests on point P.

Fig. Exc_6.6_10 (Sol.) (i)

From the figure,

PQ 2.4m

QR 1.8m

By Pythagoras theorem in ,

PQR

     

 

2 2 2

PR PQ QR

     

 

2 2 2

PR 2.4 1.8

 

 

PR 5.76 3.24

 



PR 9

PR 3

Thus, length of the string out of the water is 3 m.

Length of the string that is pulled

in 12 seconds at the rate of 5 cm / sec

5 12

 

 

 

 

60 cm

 

0.60 m 1 m 100 cm 

The remaining part of the string left out 3 0.6 

2.4 m

Now, find the horizontal distance of the fly from

Nazima after 12 seconds.

AR be the remaining part of the string after 12

seconds.

Fig. Exc_6.6_10 (Sol.) (ii)

By Pythagoras theorem in triangle AQR,

     

2 2 2

AR AQ QR 

     

2 2 2

AR AQ QR 

 

AQ 5.76 3.24 

 

AQ 2.52

AQ 1.59

Thus,

The horizontal distance of the fly

1.59 1.2

from Nazima after 12 seconds

 

 

 

 

2.79 m

Hence, the horizontal distance of the fly from

Nazima is 2.79 m.