NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 6
Chapter Name : Triangles

Exercise 6.1

Q1 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)

Answer. (i) Similar
(ii) Similar
(iii) Equilateral
(iv) (a) Equal
(b) Proportional

Page : 122 , Block Name : Exercise 6.1

Q2 Give two different examples of pair of
(i) similar gures.
(ii) non-similar gures.

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

Two squares with sides 1 cm and 2 cm.

(ii) Trapezium and square

Page : 122 , Block Name : Exercise 6.1

Q3 State whether the following quadrilaterals are similar or not:

Answer. Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional,
i.e 1:2, but their corresponding angles are not equal.

Page : 122 , Block Name : Exercise 6.1

Exercise 6.2

Q1 In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer. (i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
AD AE
=
DB EC

1.5 1
=
3 x

3×1
x =
1.5

x = 2

∴ EC = 2cm

(ii)

Let AD = x cm
It is given that DE || BC
By using basic proportionality theorem, we obtain

AD AE
=
DB EC

x 1.8
=
7.2 5.4

1.8×7.2
x =
5.4

x = 2.4

∴ AD = 2.4cm

Page : 128 , Block Name : Exercise 6.2

Q2 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 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Answer. (i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
PE 3.9
= = 1.3
EQ 3

PF 3.6
= = 1.5
FR 2.4

Hence, PE

EQ
≠
PF

FR

Therefore, EF is not parallel to QR.
(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
PE 4 8
= =
EQ 4.5 9

PF 8
=
FR 9

Hence
PE PF
=
EQ FR

T heref ore, EF isparalleltoQR.

(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
PE 0.18 18 9
= = =
PQ 1.28 128 64

PF 0.36 9
= =
PR 2.56 64
PE PF
Hence, =
PQ PR

Therefore, EF is parallel to QR.

Page : 128 , Block Name : Exercise 6.2

Q3 In Figure, if LM || CB and LN || CD, prove that
AM AN
=
AB AD

In the given gure, LM || CB
By using basic proportionality theorem, we obtain
..(i)
AM AL
=
AB AC

Similarly, LN || CD
AN AL
∴ =
AD AC

(ii)
From (i) and (ii), we obtain
AM AN
=
AB AD

Page : 128 , Block Name : Exercise 6.2

Q4 In Figure, DE || AC and DF || AE. Prove that
BF BE
=
FE EC

In △ABC, DE∥AC
(Basic Proportionality Theorem) (i)
BD BE
∴ =
DA EC

In △BAE, DF ∥AE
∴
BD
=
BF
( Basic Proportionality Theorem) (ii)
DA FE

From (i) and (ii), we obtain
BE BF
=
EC FE

Page : 128 , Block Name : Exercise 6.2

Q5 In Figure, DE || OQ and DF || OR. Show that EF || QR.

In △POQ, DE∥OQ
∴
PE
=
PD
( Basic proportionality theorem ) (i)
EQ DO

In △POR, DF∥OR
(Basic proportionality theorem)
PF PD
∴ =
FR DO

(ii)
From (i) and (ii), we obtain
PE PF
=
EQ FR

∴ EF∥OR (Converse of basic proportionality theorem)

Page : 129 , Block Name : Exercise 6.2

Q6 In Figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.

In △POQ, AB∥PQ
(Basic proportionality theorem) (i)
OA OB
∴ =
AP BQ

In △POR, AC∥PR
∴
OA
=
OC

CR
(By basic proportionality theorem) (ii)
AP

From (i) and (ii), we obtain.
OB OC
=
BQ CR

∴ BC∥QR (By the converse of basic proportionality theorem)

Page : 129 , Block Name : Exercise 6.2

Q7 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. (Recall that you have proved it in Class IX).

Consider the given gure in which PQ is a line segment drawn through the mid-point P of line AB,
such that PQ || BC
By using basic proportionality theorem, we obtain
AQ AP
=
QC PB

AQ 1
= (P is the mid-point of AB. ∴ AP = PB)
QC 1

⇒ AQ = QC

Or, Q is the mid-point of AC.

Page : 129 , Block Name : Exercise 6.2

Q8 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. (Recall that you have done it in Class IX).

Consider the given gure in which PQ is a line segment joining the mid-points P and Q of line AB and
AC respectively.
I.e., AP = PB and AQ = QC
It can be observed that
AP 1
=
PB 1

and
AQ 1
=
QC 1

AP AQ
∴ =
PB QC

Hence, by using basic proportionality theorem, we obtain
PQ∥BC

Page : 129 , Block Name : Exercise 6.2

Q9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
that
AO CO
=
BO DO

Draw a line EF through point O, such that EF || CD
In △ADC, EO∥CD
By using basic proportionality theorem, we obtain
AE

ED
= ...(1)
AO

OC

In △ABD, OE ∥AB
So, by using basic proportionality theorem, we obtain
ED OD
=
AE BO

... (2)
AE BO
⇒ =
ED OD

From equations (1) and (2), we obtain
AO BO
=
OC OD

AO OC
⇒ =
BO OD

Page : 129 , Block Name : Exercise 6.2

Q10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that .
AO CO
=
BO DO

Show that ABCD is a trapezium.

Answer. Let us consider the following gure for the given question.

Draw a line OE || AB

In △ABD, OE∥AB
By using basic proportionality theorem, we obtain
AE

ED
=
BO

OD
(1)
However, it is given that
(2)
AO OB
=
OC OD

From equations (1) and (2), we obtain
AE AO
=
ED OC

⇒ EO ∥ DC [By the converse of basic proportionality theorem]

⇒ AB∥OE∥DC

⇒ AB∥CD

∴ ABCD is a trapezium.

Page : 129 , Block Name : Exercise 6.2

Exercise 6.3

Q1 State which pairs of triangles in 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 :

Answer. (i) ∠A = ∠P = 60 ∘

∘
∠B = ∠Q = 80
∘
∠C = ∠R = 40

Therefore, △ABC − △PQR[By AAA similarity criterion ]
AB BC CA
= =
QR RP PQ

(ii) ∴ △ABC − ΔQRP [ By SSS similarity criterion ]

(iii) The given triangles are not similar as the corresponding sides are not proportional.

(iv) The given triangles are not similar as the corresponding sides are not proportional.

(v) THe given triangles are not similar as the corresponding sides are not proportional.

(vi) In ΔDEF
∘
∠D + ∠E + ∠F = 180

( Sum of the measures of the angles of a triangle is 180 . )∘

∘ ∘ ∘
70 + 80 + ∠F = 180
∘
∠F = 30

Similarly, In △PQR
∘
∠P + ∠Q + ∠R = 180

( Sum of the measures of the angles of a triangle is 180 . )
∘

∘ ∘ ∘
∠P + 80 + 30 = 180
∘
∠P = 70

In △DEF and ΔP QR,
∘
∠D = ∠P (Each70 )
∘
∠E = ∠Q (Each80 )
∘
∠F = ∠R ( Each 30 )

∴ △DEF ∼ △PQR[By AAA similarity criterion ]

Page : 138 , Block Name : Exercise 6.3

Q2 In Figure, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer. DOB is a straight line.
∘
∴ ∠DOC + ∠COB = 180
∘ ∘
⇒ ∠DOC = 180 − 125
∘
= 55

In ΔDOC
∘
∠DCO + ∠CDO + ∠DOC = 180

( Sum of the measures of the angles of a triangle is 180 . )
∘

∘ ∘ ∘
⇒ ∠DCO + 70 + 55 = 180
∘
⇒ ∠DCO = 55

It is given that △ODC − △OBA
∴ ∠OAB = ∠OCD[ Corresponding angles are equal in similar triangles.]
∘
⇒ ∠OAB = 55

Page : 139 , Block Name : Exercise 6.3

Q3 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O.
Using a similarity criterion for two triangles, show that
OA OB
=
OC OD

In △DOC and △BOA,
∠CDO = ∠ABO[ Alternate interior angles as AB∥CD]

∠DCO = ∠BAO[ Alternate interior angles as AB∥CD]

∠DOC = ∠BOA[ Vertically opposite angles ]

∴ △DOC − △BOA [AAA similarity criterion]
[Corresponding sides are proportional]
DO OC
∴ =
BO OA

OA OB
⇒ =
OC OD

Page : 139 , Block Name : Exercise 6.3

Q4 In Figure, and ∠1 = ∠2. Show that ΔPQS − ΔTQR.
QR QT
=
QS PR

In ΔP QR, ∠P QR = ∠P RQ
∴ PQ = PR(i)

Given,
QR QT
=
OS PR

Using (i), we obtain
(ii)
QR QT
=
QS QP

In △PQS and ΔTQR

QR QT
= [Using(ii)]
QS QP

∠Q = ∠Q

∴ △PQS ∼ △TQR [SAS similarity criterion]

Page : 140 , Block Name : Exercise 6.3

Q5 S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆ RPQ ~ ∆ RTS.

In △RPQ and ΔRST,
∠RTS = ∠QPS( Given )

∠R = ∠R( Common angle )

∴ △RP Q − ΔRT S(By AA similarity criterion)

Page : 140 , Block Name : Exercise 6.3

Q6 In Figure, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.

Answer. It is given that △ABE ≅ △ACD
∴ AB = AC[ByCP CT ](1)

And, AD = AE[BY CP CT ](2)

In △ADE and △ABC
AD AE
= [ Dividing equation (2)by(1)]
AB AC

∠A = ∠A[ Common angle ]

∴ △ADE − △ABC[By SAS similarity criterion]

Page : 140 , Block Name : Exercise 6.3

Q7 In Figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:

(i) ∆AEP ~ ∆ CDP
(ii) ∆ABD ~ ∆ CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆ PDC ~ ∆ BEC

Answer. (i)

In △AEP and △CDP .
∘
∠AEP = ∠CDP (Each90 )

∠APE = ∠CPD( Vertically opposite angles )

Hence, by using AA similarity criterion,
△AEP ∼ ΔCDP

(ii)

In △ABD and ΔCBE
∘
∠ADB = ∠CEB (Each90 )

∠ABD = ∠CBE( Common )

Hence, by using AA similarity criterion,

△ABD − ΔCBE

(iii)

In △AEP and △ADB ,
∘
∠AEP = ∠ADB (Each90 )

∠P AE = ∠DAB( Common )

Hence, y using AA similarity criterion,
△AEP ∼ △ADB

(iv)
In ΔPDC and ΔBEC
∘
∠PDC = ∠BEC ( Each 90 )

∠PCD = ∠BCE( Common angle )

Hence, by using AA similarity criterion,
△P DC − △BEC

Page : 140 , Block Name : Exercise 6.3

Q8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show
that ∆ ABE ~ ∆ CFB.

In △ABE and ΔCFB,
∠A = ∠C( Opposite angles of a parallelogram)

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

∴ △ABE − ΔCFB(By AA similarity criterion)

Page : 140 , Block Name : Exercise 6.3

Q9 In Figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ΔABC ∼ ΔAMP

(ii) CA

PA
=
BC

MP

Answer. In △ABC and △AM P ,
∘
∠ABC = ∠AM P (Each90 )

∠A = ∠A( Common )

∴ △ABC − △AMP(By AA similarity criterion)

⇒
CA

PA
=
BC

MP
( Corresponding sides of similar triangles are proportional)

Page : 140 , Block Name : Exercise 6.3

Q10 CD and GH are respectively the bisectors of ∠ACB and ∠ EGF such that D and H lie on sides AB
and FE of ∆ ABC and ∆ EFG respectively. If ∆ABC ~ ∆ FEG, show that:
(i) CD AC
=
GH FG

(ii) ΔDCB − ΔHGE
(iii) ΔDCA − ΔHGF

It is given that △ABC − ΔFEG
∴ ∠A = ∠F , ∠B = ∠E, and ∠ACB = ∠F GE

∠ACB = ∠F GE

∴ ∠ACD = ∠F GH ( Angle bisector )

And, ∠DCB = ∠HGE (Angle bisector)

In △ACD and ΔFGH ,

∠A = ∠F( Proved above)

∠ACD = ∠FGH( Proved above )

∴ △ACD ∼ ΔF GH (By AA similarity criterion)
CD AC
⇒ =
GH FG

In △DCB and ΔH GE,
∠DCB = ∠HGE( Proved above )

∠B = ∠E( Proved above )

△△DCB ∼ ΔH GE(By AA similarity criterion)

In △DCA and ΔH GF

∠ACD = ∠F GH ( Proved above )

∠A = ∠F ( Proved above )

∴ △DCA ∼ ΔHGF( By AA similarity criterion)

Page : 140 , Block Name : Exercise 6.3

Q11 In 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.

Answer. It is given that ABC is an isosceles triangle.
∴ AB = AC

⇒ ∠ABD = ∠ECF

In △ABD and ΔECF
∘
∠ADB = ∠EF C (Each90 )

∠BAD = ∠CEF ( Proved above )

∴ △ABD ∼ ΔECF(By using AA similarity criterion )

Page : 141 , Block Name : Exercise 6.3

Q12 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and
QR and median PM of ∆ PQR (see Figure). Show that ∆ ABC ~ ∆ PQR.

Answer. Median divides the opposite side.
BC QR
∴ BD = and QM =
2 2

Given that,

AB BC AD
= =
PQ QR PM

1
BC
AB 2 AD
⇒ = =
1
PQ QR PM
2

AB BD AD
⇒ = =
PQ QM PM

In △ABD and ΔP QM
AB

PQ
= =
BD

QM
(Proved above)
AD

PM

△△ABD ∼ △PQM(By SSS similarity criterion )

⇒ ∠ABD = ∠P QM ( Corresponding angles of similar triangles)

In △ABC and ΔP QR ,

∠ABD = ∠P QM (Proved above)
AB BC
=
PQ QR

∴ △ABC ∼ ΔP QR(By SAS similarity criterion)

Page : 141 , Block Name : Exercise 6.3

Q13 D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA = CB.CD.
2

In △ADC and ΔBAC ,
∠ADC = ∠BAC( Given )

∠ACD = ∠BCA( Common angle )

∴ △ADC ∼ ΔBAC(By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
CA CD
∴ =
CB CA
2
⇒ CA = CB × CD

Page : 141 , Block Name : Exercise 6.3

Q14 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.

Given that,
AB AC AD
= =
PQ PR PM

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then,
join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML(By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
AB AC AD
= =
PQ PR PM

AB BE 2AD
⇒ = =
PQ QL 2P M

AB BE AE
⇒ = =
PQ QL PL

∴ △ABE ∼ △PQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QP L … (1)

Similarly, it can be proved that △AEC ∼ △PLR and ∠CAE = ∠RP L … (2)
Adding equation (1) and (2), we obtain
∠BAE + ∠CAE = ∠QP L + ∠RP L

⇒ ∠CAB = ∠RP Q … (3)

In △ABC and ΔP QR ′

AB

PQ
= (Given)
AC

PR

∠CAB = ∠RP Q[U sing equation (3)]

∴ △ABC ∼ ΔPQR(By SAS similarity criterion)

Page : 141 , Block Name : Exercise 6.3

Q15 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.

Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, ∠DCF = ∠BAE
And, ∠DFC = ∠BEA
∠CDF = ∠ABE( Tower and pole are vertical to the ground)

∴ △ABE ∼ △CDF (AAA similarity criterion)
AB BE
⇒ =
CD DF

AB 28
⇒ =
6m 4

⇒ AB = 42m

Therefore, the height of the tower will be 42 metres.

Page : 141 , Block Name : Exercise 6.3

Q16 If AD and PM are medians of triangles ABC and PQR, respectively where ∆ ABC ~ ∆ PQR, prove
that AB

PQ
=
AD

PM

It is given that △ABC ∼ △PQR
We know that the corresponding sides of similar triangles are in proportion.

∴
AB

PQ
=
AC

PR
=
BC

QR
…(1)
Since AD and PM are medians, they will divide their opposite sides.
…(3)
BC QR
BD = and QM =
2 2

From equations (1) and (3), we obtain
AB

PQ
= …(4)
BD

QM

In △ABD and ΔPQM
∠B = ∠Q[U sin g equation (2)]
AB

PQ
=
BD

QM
[Using equation (4)]
∴ △ABD ∼ ΔPQM (By SAS similarity criterion)
AB BD AD
⇒ = =
PQ QM PM

Page : 141 , Block Name : Exercise 6.3

Exercise 6.4

Q1 Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64cm 2
and 121cm
2
. If EF = 15.4 cm, nd BC.

Answer. It is given that △ABC − ΔDEF .
2 2 2
ar(ΔABC)
AB BC AC
∴ = ( ) = ( ) = ( )
ar(ΔDEF) DE EF DF

Given that,
EF = 15.4cm ,
2
ar(ΔABC) = 64cm
2
ar(ΔDEF) = 121cm
2
ar(ABC) BC
∴ = ( )
ar(DEF) EF

2 2
64cm BC
⇒ ( ) = 2
2
121cm (15.4cm)

BC 8
⇒ = ( ) cm
15.4 11

8×15.4
⇒ BC = ( ) cm = (8 × 1.4)cm = 11.2cm
11

Page : 143 , Block Name : Exercise 6.4

Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD,
nd the ratio of the areas of triangles AOB and COD.

Since AB || CD,
∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC( Alternate interior angles)

In △AOB and ΔCOD,
∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD( Alternate interior angles )

∠OBA = ∠ODC( Alternate interior angles )

∴ △AOB ∼ ΔCOD(By AAA similarity criterion )
2
ar(ΔAOB)
AB
∴ = ( )
ar(ΔCOD) CD

Since AB = 2CD
2
ar(ΔAOB) 2CD 4
∴ = ( ) = = 4 : 1
ar(ΔCOD) CD 1

Page : 143 , Block Name : Exercise 6.4

Q3 In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show
that
ar(ABC) AO
=
ar(DBC) DO

Answer. Let us draw two perpendicular AP and DM on line BC.

We know that area of a triangle = 1/2 x Base x Height
1
ar(ΔABC) BC×AP
2 AP
∴ = =
1
ar(ΔDBC) BC×DM DM
2

In △APO and △DMO

∘
∠AP O = ∠DM O (Each = 90 )

∠AOP = ∠DOM( Vertically opposite angles)

∴ △APO ∼ △DMO(By AA similarity criterion)
AP AO
∴ =
DM DO
ar(ΔABC) AO
⇒ =
ar(ΔDBC) DO

Page : 144 , Block Name : Exercise 6.4

Q4 If the areas of two similar triangles are equal, prove that they are congruent.

Answer. Let us assume two similar triangles are △ABC ∼ ΔPQR
2 2 2
ar(ΔABC)
= (
AB

PQ
) = (
BC

QR
) = (
AC

PR
) (1)
ar(ΔPQR)

Given that, ar (ΔABC) = ar (ΔPQR)
ar(ΔABC)
⇒ = 1
ar(ΔPQR)

Putting this value in equation (1), we obtain
2 2 2
AB BC AC
I = ( ) = ( ) = ( )
PQ QR PR

⇒ AB = PQ, BC = QR, and AC = PR

∴ ΔABC ≅ ΔPQR ( By SSS congruence criterion )

Page : 144 , Block Name : Exercise 6.4

Q5 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.

1
∴ DE ∥
∥AC and DE = AC
2

In △BED and ΔBCA,
∠BED = ∠BCA (Corresponding angles)

∠BDE = ∠BAC (Corresponding angles)

∠EBD = ∠CBA ( Common angles)

∴ △BED ∼ ΔBCA (AAA similarity criterion)
ar(ΔBED)
1
⇒ =
ar(ΔBCA) 4

1
⇒ ar(ΔBED) = ar(ΔBCA)
4

Similarly, ar(ΔCFE) = 1

4
ar(CBA) and ar (ΔADF) =
1

4
ar(ΔABC)

Also (ΔDEF) = ar(ΔABC) − [ar(ΔBED) + ar(ΔCFE) + ar(ΔADF)]
3 1
⇒ ar(ΔDEF) = ar(ΔABC) − ar(ΔABC) = ar(ΔABC)
4 4
ar(ΔDEF) 1
⇒ =
ar(ΔABC) 4

Page : 144 , Block Name : Exercise 6.4

Q6 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their
corresponding medians.

Let us assume two similar triangles as △ABC ∼ ΔPQR. Let AD and PS be the medians of these
triangles.
∵ △ABC ∼ ΔP QR

∴
AB

PQ
=
BC

QR
=
AC

PR
..(1)
∠A = ∠P , ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,
BC
∴ BD = DC =
2

And, QS = SR =
QR

2

Equation (1) becomes
AB

PQ
=
BD
=
QS
…(3)AC

PR

In △ABD and ΔP QS,
∠B = ∠Q[ Using equation (2)]

And, AB

PQ
=
BD

QS
[U sing equation (3)]

∴ △ABD ∼ ΔPQS (SAS similarity criterion)
Therefore, it can be said that
AB

PQ
=
BD
=
QS
…(4)
AD

PS

2 2 2
ar(ΔABC) AC
AB BC
= ( ) = ( ) = ( )
ar(ΔPQR) PQ QR PR

From equations (1) and (4), we may nd that
AB BC AC AD
= = =
PQ QR PR PS

And hence,
2
ar(ΔABC) AD
= ( )
ar(ΔPQR) PS

Page : 144 , Block Name : Exercise 6.4

Q7 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.

Let ABCD be a square of side a.
Therefore, its diagonal = √2a
Two desired equilateral triangles are formed as △ABE and △DBF
Two desired equilateral triangles are formed as △ABE and ΔDBF.
Side of an equilateral triangle, △ABE, described on one of its sides = a
Side of an equilateral triangle, △DBF , described on one of its diagonals = √2a
We know that equilateral triangles have all its angles as 60 and all its sides of the same length.
∘

Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of
these triangles will be equal to the square of the ratio between the sides of these triangles.
2
Area of ΔABE a 1
= ( ) =
Area of ΔDBF √2a 2

Page : 144 , Block Name : Exercise 6.4

Q8 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

We know that equilateral triangles have all its angles as 60 and all its sides of the same length.
∘

Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of
these triangles will be equal to the square of the ratio between the sides of these triangles.
Let side of ΔABC = x
Therefore, side of ΔBDE = x

2
2
area(ΔABC) x 4
∴ = ( x ) =
area(ΔBDE) 1
2

Hence, the correct answer is (C).

Page : 144 , Block Name : Exercise 6.4

Q9 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

Answer. If two triangles are similar to each other, then the ratio of the areas of these triangles will be
equal to the square of the ratio of the corresponding sides of these triangles.
It is given that the sides are in the ratio 4:9.
2

Therefore, ratio between areas of these triangles = ( 4

9
) =
16

81

Hence, the correct answer is (D).

Page : 144 , Block Name : Exercise 6.4

Exercise 6.5

Q1 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

Answer. (i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths
of these sides, we will obtain 49, 576, and 625.
49 + 576 = 625
Or, 7 + 24 = 25
2 2 2

The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest sides of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will obtain 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 3 + 6 ≠ 8
2 2 2

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of
the third side.
Hence, it is not a right triangle
(iii) Given that sides are 50 cm, 80 cm. And 100 cm.
Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50 + 80 ≠ 100
2 2 2

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of
the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will obtain 169, 144, and 24.
Clearly, 144 + 25 = 169.
Or, 12 + 5 = 13
2 2 2

The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 13 cm.

Page : 150 , Block Name : Exercise 6.5

Q2 PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM = 2

QM . MR

Let ∠MPR = x
In ΔMPR
∘ ∘
∠M RP = 180 − 90 − x
∘
∠M RP = 90 − x

Similarly, in ΔMPQ
∘
∠MPQ = 90 − ∠MPR
∘
= 90 − x
∘ ∘ ∘
∠MQP = 180 − 90 − (90 − x)

∠MOP = x

In △QMP and ΔPMR,
∠MPQ = ∠MRP

∠PMQ = ∠RMP

∠MQP = ∠MPR

∴ △QMP − APMR ( By AAA similarity criterion )
QM MP
⇒ =
PM MR

2
⇒ PM = QM × MR

Page : 150 , Block Name : Exercise 6.5

Q3 In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB = BC. BD
2

(ii) AC = BC ⋅ DC
2

(iii) AD = BD. CD
2

Answer. (i) In △ADB and △CAB,
∘
∠DAB = ∠ACB ( Each 90 )

∠ABD = ∠CBA ( Common angle )

∴ △ADB − △CAB (AA similarity criterion)
AB BD
⇒ =
CB AB
2
⇒ AB = CB × BD

(ii) Let ∠CAB = x
In ΔCBA,
∘ ∘
∠CBA = 180 − 90 − x
∘
∠CBA = 90 − x

Similarly, In △CAD,
∘
∠CAD = 90 − ∠CAB
∘
= 90 − x
∘ ∘ ∘
∠CDA = 180 − 90 − (90 − x)

∠CDA = x

In △CBA and ΔCAD
∠CBA = ∠CAD

∠CAB = ∠CDA
∘
∠ACB = ∠DCA (Each 90 )

∴ ΔCBA ∼ ΔCAD ( By AAA rule )
AC BC
⇒ =
DC AC
2
⇒ AC = DC × BC

(iii) In △DCA and ΔDAB,
∘
∠DCA = ∠DAB (Each90 )

∠CDA = ∠ADB( Common angle )

∴ ΔDCA ∼ ΔDAB (AA similarity criterion)
DC DA
⇒ =
DA DB
2
⇒ AD = BD × CD

Page : 150 , Block Name : Exercise 6.5

Q4 ABC is an isosceles triangle right angled at C. Prove that AB 2
= 2AC
2
.

Given that △ABC is an isosceles triangle.
∴ AC = CB

Applying Pythagoras theorem in △ABC (i.e., right-angled at point C), we obtain
2 2 2
AC + CB = AB
2 2 2
⇒ AC + AC = AB (AC = CB)
2 2
⇒ 2AC = AB

Page : 150 , Block Name : Exercise 6.5

Q5 ABC is an isosceles triangle with AC = BC. If AB 2
= 2AC
2
. prove that ABC is a right triangle.

Given that,
2 2
AB = 2AC
2 2 2
⇒ AB = AC + AC
2 2 2
⇒ AB = AC + BC (As AC = BC)

The triangle is satisfying the pythagoras theorem.
Therefore, the given triangle is a right-angled triangle.

Page : 150 , Block Name : Exercise 6.5

Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Let AD be the altitude in the given equilateral triangle, △ABC.
We know that altitude bisects the opposite side.
∴ BD = DC = a

In △ADB
Applying pythagoras theorem, we obtain
2 2 2
AD + DB = AB
2 2 2
⇒ AD + a = (2a)

2 2 2
⇒ AD + a = 4a

2 2
⇒ AD = 3a

⇒ AD = a√3

In an equilateral triangle, all the altitudes are equal in length.
Therefore, the length of each altitude will be √3a.

Page : 150 , Block Name : Exercise 6.5

Q7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of
its diagonals.

In △AOB, ΔBOC, ΔCOD, △AOD
Applying Pythagoras theorem, we obtain
AB = AO + OB …(1)
2 2 2

BC = BO + OC ….(2)
2 2 2

CD = CO + OD …(3)
2 2 2

AD = AO + OD …(4)
2 2 2

Adding all these equations, we obtain
2 2 2 2 2 2 2 2
AB + BC + CD + AD = 2 (AO + OB + OC + OD )

2 2 2 2
AC BD AC BD
= 2 (( ) + ( ) + ( ) + ( ) )
2 2 2 2

( (Diagonals bisect each other) \)
2 2
(AC) (BD)
= 2( + )
2 2

2 2
= (AC) + (BD)

Page : 150 , Block Name : Exercise 6.5

Q8 In Figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥AC and OF ⊥AB. Show
that

(i) OA + OB + OC − OD − OE − OF = AF
2 2 2 2 2 2 2 2
+ BD + CE
2

(ii) AF + BD + CE = AE + CD + BF
2 2 2 2 2 2

(i) Applying Pythagoras theorem in △AOF, we obtain
2 2 2
OA = OF + AF

Similarly, in △BOD,
2 2 2
OB = OD + BD

Similarly, in ΔCOE,
2 2 2
OC = OE + EC

Adding these equations,
2 2 2 2 2 2 2 2 2
OA + OB + OC = OF + AF + OD + BD + OE + EC

2 2 2 2 2 2 2 2 2
OA + OB + OC − OD − OE − OF = AF + BD + EC

(ii) From the above result,
2 2 2 2 2 2 2 2 2
AF + BD + EC = (OA − OE ) + (OC − OD ) + (OB − OF )

2 2 2 2 2 2
∴ AF + BD + EC = AE + CD + BF

Page : 151 , Block Name : Exercise 6.5

Q9 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.

Let OA be the wall and AB be the ladder.
Therefore, by Pythagoras theorem,
2 2 2
AB = OA + BO

2 2 2
(10m) = (8m) + OB

2 2 2
100m = 64m + OB
2 2
OB = 36m

OB = 6m

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

Page : 151 , Block Name : Exercise 6.5

Q10 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?

Let OB be the pole and AB be the wire.
By Pythagoras theorem,
2 2 2
AB = OB + OA
2 2 2
(24m) = (18m) + OA
2 2 2
OA = (576 − 324)m = 252m

OA = √252m = √6 × 6 × 7m = 6√7m

Therefore, the distance from the base is 6√7m.

Page : 151 , Block Name : Exercise 6.5

Q11 An aeroplane leaves an airport and ies due north at a speed of 1000 km per hour. At the same
time, another aeroplane leaves the same airport and ies due west at a speed of 1200 km per hour.
How far apart will be the two planes after 1 hours? 1

2

Distance travelled by the plane ying towards north in 1
1
hrs
2
1
= 1, 000 × 1 = 1, 500km
2

Similarly, distance travelled vu the plane ying towards west in 1 1

2
hrs
1
= 1, 200 × 1 = 1, 800km
2

Let these distances be represented by OA and OB respectively.
Applying Pythagoras as theorem,
Distance between these planes after 1 hrs, AB = √OA + OB
1

2
2 2

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

= (√5490000)km = (√9 × 610000)km = 300√61km

Therefore, the distance between these planes will be 300√61km after 1 1

2
hrs

Page : 151 , Block Name : Exercise 6.5

Q12 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, nd the distance between their tops.

Let CD and AB be the poles of height 11 m and 6 m.
Therefore, Cp = 11 - 6 = 5 m
From the gure, it can be observed that AP = 12m
Applying Pythagoras theorem for △APC, we obtain
2 2 2
AP + PC = AC

2 2 2
(12m) + (5m) = AC

2 2 2
AC = (144 + 25)m = 169m

AC = 13m

Therefore, the distance between their tops is 13 m.

Page : 151 , Block Name : Exercise 6.5

Q13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove
that AE + BD = AB + DE .
2 2 2 2

Applying Pythagoras theorem in △ACE, we obtain
AC + CE = AE …(1)
2 2 2

Applying Pythagoras theorem in △BCD , we obtain
BC + CD = BD ..(2)
2 2 2

Using equation (1) and equation (2), we obtain
AC + CE + BC + CD = AE + BD …(3)
2 2 2 2 2 2

Applying Pythagoras theorem in ΔCDE, we obtain
2 2 2
AB = AC + CB

Putting the values in equation (3), we obtain
2 2 2 2
DE + AB = AE + BD

Page : 151 , Block Name : Exercise 6.5

Q14 The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see
Figure). Prove that 2AB = 2AC + BC . 2 2 2

Answer. Applying Pythagoras theorem for △ACD, we obtain
2 2 2
AC = AD + DC

AD
2
= AC …(1)
2
− DC
2

Applying Pythagoras theorem in △ABD, we obtain
2 2 2
AB = AD + DB

AD
2
= AB
2
…(2)
− DB
2

From equation (1) and equation (2), we obtain
… (3)
2 2 2 2
AC − DC = AB − DB

It is given that 3DC = DB
BC 3BC
∴ DC = and DB =
4 4

Putting these values in equation (3), we obtain,
2 2
2 BC 2 3BC
AC − ( ) = AB − ( )
4 4

2 2
2 BC 2 9BC
AC − = AB −
16 16

2 2 2 2
16AC − BC = 16AB − 9BC

2 2 2
16AB − 16AC = 8BC

2 2 2
2AB = 2AC + BC

Page : 151 , Block Name : Exercise 6.5

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that
9AD = 7AB .
2 2

Let the side of the equilateral triangle be a, and AE be the altitude of △ABC
BC a
∴ BE = EC = =
2 2

And, AE =
a√3

2

Given that, BD = 1/3 BC
a
∴ BD =
3
a a a
DE = BE − BD = − =
2 3 6

Applying Pythagoras theorem in △ADE, we obtain
2 2 2
AD = AE + DE
2 2
2 a√3 a
AD = ( ) + ( )
2 6

2 2
3a a
= ( ) + ( )
4 36

2
28a
=
36

7 2
= AB
9
2 2
⇒ 9AD = 7AB

Page : 151 , Block Name : Exercise 6.5

Q16 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.

Let the side of the equilateral triangle be a, and AE be the altitude of △ABC
BC a
∴ BE = EC = =
2 2

Applying Pythagoras theorem in △ABE, we obtain
2 2 2
AB = AE + BE
2
2 2 a
a = AE + ( )
2

2
2 2 a
AE = a −
4
2
2 3a
AE =
4
2 2
4AE = 3a

⇒ 4 × ( Square of altitude ) = 3× (Square of one side)

Page : 151 , Block Name : Exercise 6.5

Q17 Tick the correct answer and justify : In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The
angle B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°

Given that, AB = = 6√3cm, AC = 12 cm, and BC = 6 cm
It can be observed that
2
AB = 108

2
AC = 144
2
And, BC = 36

2 2 2
AB + BC = AC

The given triangle, △ABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∘
∴ ∠B = 90

Hence, the correct answer is (C)

Page : 151 , Block Name : Exercise 6.5

Exercise 6.6

Q1 In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that
QS PQ
=
SR PR

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T,
Given that, PS is the angle bisector of ∠QPR.
∠QPS = ∠SPR … (1)

By construction,
∠SP R = ∠P RT (ASP S∥T R) … (2)

∠QP S = ∠QT R(ASP S||T R) … (3)

Using these equations, we obtain
∠PRT = ∠QTR

∴ PT = PR

By construction,
PS || TR
By using basic proportionality theorem for ΔQTR,
QS QP
=
SR PT

(PT =TR)
QS PQ
⇒ =
SR PR

Page : 152 , Block Name : Exercise 6.6

Q2 In Figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB.
Prove that :
(i) DM = DN ⋅ MC
2

(ii) DN = DM. AN
2

Answer. (i) Let us join DB.

We have, DN || CB, DM || AB, and ∠B = 90 ∘

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
∘
∴ ∠CDB = 90
∘
⇒ ∠2 + ∠3 = 90 … (1)

In △CDM,
∘
∠1 + ∠2 + ∠DMC = 180
∘
⇒ ∠1 + ∠2 = 90 … (2)

In △DMB,
∘
∠3 + ∠DM B + ∠4 = 180
∘
⇒ ∠3 + ∠4 = 90 … (3)

From equation (1) and (2), we obtain
∠2 = ∠4

In △DCM and △BDM,
∠1 = ∠3 (Proved above)

∠2 = ∠4( Proved above )

∴ △DCM ∼ △BDM( AA similarity criterion)
BM DM
⇒ =
DM MC
DN DM
⇒ = (BM = DN)
DM MC
2
⇒ DM = DN × MC

(ii) In right triangle DBN,
∘
∠5 + ∠7 = 90 … (4)

I nrighttriangleDAN ,
∘
∠6 + ∠8 = 90 … (5)

D is the foot of the perpendicular drawn from B to AC.
∘
∴ ∠ADB = 90
∘
⇒ ∠5 + ∠6 = 90 … (6)

From equation (5) and (6), we obtain
∠8 = ∠5

In △DNA and △BND,
∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ △DNA ∼ ΔBND (AA similarity criterion)
AN DN
⇒ =
DN NB
2
⇒ DN = AN × NB

2
⇒ DN = AN × DM(ASNB = DM)

Page : 152 , Block Name : Exercise 6.6

Q3 In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
2 2 2
AC = AB + BC + 2BC. BD

Answer. Applying Pythagoras theorem in △ADB, we obtain
2 2 2
AB = AD + DB … (1)

Applying Pythagoras theorem in △ACD, we obtain
2 2 2
AC = AD + DC

2 2 2
AC = AD + (DB + BC)

2 2 2 2
AC = AD + DB + BC + 2DB × BC
2 2 2
AC = AB + BC + 2DB × BC[ Using equation (1)]

Page : 152 , Block Name : Exercise 6.6

Q4 In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
AC = AB + BC − 2BC ⋅ BD.
2 2 2

Answer. Applying Pythagoras theorem in △ADB, we obtain
2 2 2
AD + DB = AB

2 2 2
⇒ AD = AB − DB … (1)

Applying Pythagoras theorem in △ADC, we obtain
2 2 2
AD + DC = AC
2 2 2 2
AB − BD + DC = AC [ Using equation (1)]

2 2 2 2
AB − BD + (BC − BD) = AC

2 2 2 2
AC = AB + (BC − BD) = AC

2 2 2 2 2
AC = AB − BD + BC + BD − 2BC × BD

2 2
= AB + BC − 2BC × BD

Page : 152 , Block Name : Exercise 6.6

Q5 In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
2

(i) AC
2 2 BC
= AD + BC ⋅ DM + ( )
2

2

(ii) AB
2 2 BC
= AD − BC ⋅ DM + ( )
2

(iii) AC
2 2 2 1 2
+ AB = 2AD + BC
2

Answer. (i) Applying Pythagoras theorem in △AMD, we obtain
2 2 2
AM + MD = AD … (1)

Applying Pythagoras theorem in △AMC, we obtain
2 2 2
AM + MC = AC

2 2 2
AM + (MD + DC) = AC

2 2 2 2
(AM + MD ) + DC + 2MD ⋅ DC = AC

2 2 2
AD + DC + 2M D. DC = AC [ Using equation (1)]

Using the result, DC =
BC

2
, , we obtain
2
2 BC BC 2
AD + ( ) + 2MD ⋅ ( ) = AC
2 2

2
2 BC 2
AD + ( ) + MD × BC = AC
2

(ii) Applying Pythagoras theorem in △ABM , we obtain
2 2 2
AB = AM + MB
2 2 2
= (AD − DM ) + MB

2 2 2
= (AD − DM ) + (BD − M D)

2 2 2 2
= AD − DM + BD + MD − 2BD × M D

2 2
= AD + BD − 2BD × M D
2
2 BC BC
= AD + ( ) − 2( ) × MD
2 2

2
2 BC
= AD + ( ) − BC × MD
2

(iii) Applying Pythagoras theorem in △ABM, we obtain
2 2 2
AM + MB = AB … (1)

Applying Pythagoras theorem in △AMC, we obtain
2 2 2
AM + MC = AC … (2)

Adding equations (1) and (2), we obtain
2 2 2 2 2
2AM + MB + MC = AB + AC

2 2 2 2 2
2AM + (BD − DM) + (MD + DC) = AB + AC

2 2 2 2 2 2
2AM + BD + DM − 2BD ⋅ DM + MD + DC + AC

2 2 2 2 2 2
2AM + 2MD + BD + DC + 2MD(−BD + DC) = AB + AC
2 2
2 2 BC BC BC BC 2 2
2 (AM + MD ) + ( ) + ( ) + 2MD (− + ) = AB + AC
2 2 2 2

2
2 BC 2 2
2AD + = AB + AC
2

Page : 152 , Block Name : Exercise 6.6

Q6 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the
squares of its sides.

Let ABCD be a parallelogram.
Let us draw perpendicular DE on extended side AB, and AF on side DC.
Applying Pythagoras theorem in △DEA, we obtain
2 2 2
DE + EA = DA … (i)

Applying Pythagoras theorem in △DEA, we obtain
(i)
2 2 2
DE + EA = DA …

Applying Pythagoras theorem in △DEB, we obtain
2 2 2
DE + EB = DB

2 2 2
DE + (EA + AB) = DB

2 2 2 2
(DE + EA ) + AB + 2EA × AB = DB

2 2 2
DA + AB + 2EA × AB = DB … (ii)

Applying Pythagoras theorem in △ADF, we obtain
2 2 2
AD = AF + FD

Applying Pythagoras theorem in △AFC, we obtain
2 2 2
AC = AF + FC
2 2
= AF + (DC − FD)

2 2 2
= AF + DC + FD − 2DC × FD

2 2 2
= (AF + FD ) + DC − 2DC × FD

2 2 2
AC = AD + DC − 2DC × FD … (ii)

Since ABCD is a parallelogram,
AB = CD … (iv)

And, BC = AD … (v)

In △DEA and △ADF
∘
∠DEA = ∠AF D (Both90 )

∠EAD = ∠ADF (EA∥DF )

AD = AD( Common )

∴ ΔEAD ≅ △F DA (AAS congruence criterion)
⇒ EA = DF … (vi)

Adding equations (i) and (iii), we obtain

2 2 2 2 2 2
DA + AB + 2EA × AB + AD + DC − 2DC × FD = DB + AC

2 2 2 2 2 2
DA + AB + AD + DC + 2EA × AB − 2DC × FD = DB + AC

2 2 2 2 2 2
BC + AB + AD + DC + 2EA × AB − 2AB × EA = DB + AC

[Using equations (iv) and (vi)]
2 2 2 2 2 2
AB + BC + CD + DA = AC + BD

Page : 153 , Block Name : Exercise 6.6

Q7 In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP

(i) In △AP C and ΔDP B,
∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP ( Angles in the same segment for chord CB)

△APC ∼ ΔDPB (By AA similarity criterion)

(ii) We have already proved that
△APC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.
AP PC CA
∴ = =
DP PB BD

AP PC
⇒ =
DP PB

∴ AP. PB = PC. DP

Page : 153 , Block Name : Exercise 6.6

Q8 In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced)
outside the circle. Prove that
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD

Answer. (i) In △PAC and ΔPDB,
∠P = ∠P ( Common )

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

opposite interior angle)

∴ △PAC ∼ ΔPDB

(ii) We know that the corresponding sides of similar triangles are proportional.
PA AC PC
∴ = =
PD DB PB
PA PC
⇒ =
PD PB

∴ PA.PB = PC. PD

Page : 153 , Block Name : Exercise 6.6

Q9 In Figure, D is a point on side BC of ∆ ABC such that BD

CD
=
AB

AC
Prove that AD is the bisector of ∠
BAC.

Answer. Let us extend BA to P such that AP =AC, Join PC.

It is given that,
BD AB
=
CD AC
BD AP
⇒ =
CD AC

By using the converse of basic proportionality theorem, we obtain AD || PC
AD∥PC

⇒ ∠BAD = ∠AP C( Corresponding angles ) … (1)
And ∠DAC = ∠ACP (Alternate interior angles)...(2)

By construction, we have
AP = AC

⇒ ∠AP C = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain
∠BAD = ∠AP C

⇒ AD is the bisector of the angle BAC.

Page : 153 , Block Name : Exercise 6.6

Q10 Nazima is y shing in a stream. The tip of her shing rod is 1.8 m above the surface of the water
and the y 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 y) is taut, how
much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second,
what will be the horizontal distance of the y from her after 12 seconds?

Let AB be the height of the tip of the shing rod from the water surface. Let BC be the horizontal
distance of the y from the tip of the shing rod.
Then, AC is the length of the string.
AC can be found by applying Pythagoras theorem in △ABC
2 2 2
AC = AB + BC
2 2 2
AB = (1.8m) + (2.4m)
2 2
AB = (3.24 + 5.76)m

2 2
AB = 9.00m

⇒ AB = √9m = 3m

Thus, the length of the string out is 3 m.
She pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m.

Page : 153 , Block Name : Exercise 6.6