What is the Value of X Given That Pq Bc

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Mathematics Part II Solutions Solutions for Class 10 Math Chapter 1 Similarity are provided here with simple step-by-step explanations. These solutions for Similarity are extremely popular among Class 10 students for Math Similarity Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 5:

Question 1:

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Answer:

Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.
Given:
BC = 9, AB = 5, PQ = 6 and QR = 10.




A



ABC




A



PQR




=


AB
×
BC


PQ
×
QR



=


5
×
9


6
×
10



=

3
4

Page No 6:

Question 2:

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find



A



ABC




A



ADB




.

Answer:

Given:
BC = 4
AD = 8




A



ABC




A



ADB




=


AB
×
BC


AB
×
AD



=

BC
AD

































BC
=
4

and

AD
=
8



=

4
8

=

1
2

Page No 6:

Question 3:

In adjoining figure, seg PS ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT.

Answer:

Given:
PS ⊥ RQ
QT ⊥ PR
RQ = 6, PS = 6 and PR = 12
With base PR and height QT,

A



PQR


=

1
2

×
PR
×
QT

With base QR and height PS,

A



PQR


=

1
2

×
QR
×
PS




A



PQR




A



PQR




=




1
2


×
PR
×
QT




1
2


×
QR
×
PS




1
=


PR
×
QT


QR
×
PS




PR
×
QT
=
QR
×
PS


QT
=


QR
×
PS

PR


=


6
×
6

12


=
3

Hence, the measure of side QT is 3 units.

Page No 6:

Question 4:

In adjoining figure, AP ⊥ BC, AD || BC, then Find A(∆ABC) : A(∆BCD).

Answer:

Given:
AP ⊥ BC
AD || BC




A



ABC




A



BCD




=


AP
×
BC


AP
×
BC



=

1
1

Hence, the ratio of A(∆ABC) and A(∆BCD) is 1 : 1.

Page No 6:

Question 5:

In adjoining figure PQ ⊥ BC, AD⊥ BC then find following ratios.

(i)



A



PQB




A



PBC




(ii)



A



PBC




A



ABC




(iii)



A



ABC




A



ADC




(iv)



A



ADC




A



PQC




Answer:

(i)



A



PQB




A



PBC




=


PQ
×
BQ


PQ
×
BC



=

BQ
BC

(ii)



A



PBC




A



ABC




=


PQ
×
BC


AD
×
BC



=

PQ
AD

(iii)



A



ABC




A



ADC




=


AD
×
BC


AD
×
DC



=

BC
DC

(iv)



A



ADC




A



PQC




=


AD
×
DC


PQ
×
QC


Page No 13:

Question 1:

Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
(1)


(2)


(3)

Answer:

(1)

In


QMP
,



QM
QP

=


3
.
5

7


=

1
2

In


MRP
,


MR
RP

=


1
.
5

3


=

1
2



QM
QP

=

MR
RP

By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

(2)

In


QMP
,



QM
QP

=

8
10


=

4
5

In


MRP
,


MR
RP

=

6
7



QM
QP



MR
RP

Therefore, ray PM is not the the bisector of ∠QPR.
(3)

In


QMP
,



QM
QP

=


3
.
6

9


=

2
5

In


MRP
,


MR
RP

=

4
10


=

2
5



QM
QP

=

MR
RP

By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

Page No 13:

Question 2:

In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Answer:

Given:
PM = 15,
PQ = 25,
PR = 20 and
NR = 8
Now, PN = PR − NR
= 20 − 8
= 12
Also, MQ = PQ − PM
= 25 − 15
= 10

In


PRQ
,



PR
NR

=

12
8


=

3
2

Also
,


PM
MQ

=

15
10


=

3
2



PR
NR

=

PM
MQ

By converse of basic proportionality theorem, NM is parallel to side RQ or NM || RQ.

Page No 14:

Question 3:

In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then Find QP.

Answer:

In


PNM
,



QM
QP

=

MN
PN

























By

angle

bisector

theorem






2
.
5

QP

=

5
7


QP
=


2
.
5
×
7

5


=
3
.
5

Hence, the measure of QP is 3.5.

Page No 14:

Question 4:

Measures of some angles in the figure are given. Prove that


AP
PB

=

AQ
QC

Answer:

Given:
∠APQ = 60

∠ABC = 60

Since, the corresponding angles ∠APQ and ​∠APC are equal.
Hence, line PQ || BC.

In


ABC
,

PQ

B
C


AP
PB

=

AQ
QC

























By

Basic

proportionality
theorem



Page No 14:

Question 5:

In trapezium ABCD, side AB || side PQ || side ∆C, AP = 15, PD = 12, QC = 14, Find BQ.

Answer:

Construction: Join BD intersecting PQ at X.


In △ABD, PX || AB


DP
PA

=

DX
XB




















.
.
.

1








By

Basic

proportionality

theorem


In △BDC, XQ||DC


DX
XB

=

CQ
QB




















.
.
.

2








By

Basic

proportionality

theorem


From (1) and (2), we get


DP
PA

=

CQ
QB




12
15

=

14
QB



QB
=
17
.
5

Page No 14:

Question 6:

Find QP using given information in the figure.

Answer:

In


PNM
,



QM
QP

=

MN
PN

























By

angle

bisector

theorem





14
QP

=

25
40


QP
=


14
×
40

25


=
22
.
4

Hence, the measure of QP is 22.4.

Page No 14:

Question 7:

In the given figure, if AB || CD || FE then Find
x
and AE.

Answer:

Construction: Join AFintersecting CD at X.


In △ABF, DX || AB


FD
DB

=

FX
XA
















.
.
.

1














By

Basic

proportionality

theorem


In △AEF, XC||FE


FX
XA

=

EC
CA
















.
.
.

2















By

Basic

proportionality

theorem


From (1) and (2), we get


FD
DB

=

EC
CA




4
8

=

x
12



x
=
6

Now, AE = AC + CE
= 12 + 6
= 18

Page No 15:

Question 8:

In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT.

Answer:

In


LNM
,



LT
NT

=

LM
NM

























By

angle

bisector

theorem





LT
8

=

6
10


LT
=


8
×
6

10


=
4
.
8

Hence, the measure of LT is 4.8.

Page No 15:

Answer:

In △ABC, ∠ABD = ∠DBC


AD
DC

=

AB
CB































By

angle

bisector

theorem






x

2


x
+
2


=

x

x
+
5





x
2

+
3
x

10
=

x
2

+
2
x


3
x

2
x
=
10


x
=
10

Page No 15:

Question 10:

In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Answer:

Given:
Seg PQ || seg DE
seg QR || seg EF
In △DXE, PQ || DE


XP

PD


=


XQ

QE




























.
.
.

I




By

basic

proportionality

theorem



In △XEF, QR || EF                ….Given




XQ


QE


=


XR


RF





























.
.
.
.
.

II




By

basic

proportionality

theorem






XP


PD


=


XR


RF




























From


I


and


II

∴ seg PR || seg DF        (Converse of basic proportional theorem)

Page No 15:

Question 11:

In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

Answer:

Given:
ray BD bisects ∠ABC
ray CE bisects ∠ACB.
seg AB ≅ seg AC


In △ABC, ∠ABD = ∠DBC


AD
DC

=

AB
BC




























.
.
.

I




By

angle

bisector

theorem


In △ABC, ∠BCE = ∠ACE


AE
EB

=

AC
BC




























.
.
.

II




By

angle

bisector

theorem


From (I) and (II)


AD
DC

=

AE
EB

















seg

AB



seg

AC


∴ ​ED || BC        (Converse of basic proportional theorem)

Page No 21:

Question 1:

In the given figure, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Answer:

Given:
∠ABC = 75°, ∠EDC = 75°
Now, in △ABC and △EDC
∠ABC = ∠EDC = 75°         (Given)
∠C = ∠C         (Common)
By AA test of similarity
△ABC ∼ △EDC

Page No 21:

Question 2:

Are the triangles in the given figure similar? If yes, by which test ?

Answer:

Given:
PQ = 6
PR = 10
QR = 8
LM = 3
LN = 5
MN = 4
Now,


PQ
LM

=

6
3

=
2
,



QR


MN


=


8


4


=
2
,



RP


NL


=


10


5


=
2



PQ
LM

=


QR


MN


=


RP


NL


By SSS test of similarity
△PQR ∼ △LMN

Page No 21:

Question 3:

As shown in the given figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time ?

Answer:

Given:
PR = 4
RL = 6
AC = 8
In △PLR and △ABC
∠PRL = ∠ACB         (Vertically opposite angles)
∠LPR = ∠BAC         (Angles made by sunlight on top are congruent)
By AA test of similarity
△PLR ∼ △ABC



PR
AC

=


LR


BC


























Corresponding

sides

are

proportional





4
8

=

6
x



x
=
12

Hence, the length of shadow of bigger pole due to sunlight is 12 m.

Page No 21:

Question 4:

In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC.

Answer:

Given:
AP ⊥ BC
BQ ⊥ AC
To prove: ∆CPA ~ ∆CQB
Proof: In ∆CPA and ∆CQB
∠CPA = ∠CQB = 90         (Given)
∠C = ∠C                             (Common)
By AA test of similarity
∆CPA ~ ∆CQB
Hence proved.

Now
,


AP
BQ

=


AC


BC


























Corresponding

sides

are

proportional




AC
=

AP
BQ

×
BC

=

7
8

×
12

=
10
.
5

Page No 22:

Question 5:

Given :
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Answer:

Given:
side PQ || side SR
AR = 5AP,
AS = 5AQ
To prove: SR = 5PQ
Proof: In ∆APQ and ∆ARS
∠PAQ = ∠RAS          (Vertically Opposite angles)
∠PQA = ∠RSA          (Alternate angles, side PQ || side SR and QS is a transversal line)
By AA test of similarity
∆APQ ~ ∆ARS


PQ
SR

=


AP


AR


























Corresponding

sides

are

proportional





PQ
SR

=


1


5






















AR
=
5
AP




SR
=
5
PQ


Hence proved.

Page No 22:

Question 6:

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

Answer:

Given:
side AB || side DC
AB = 20,
DC = 6,
OB = 15
In △COD and △AOB
∠COD = ∠AOB         (Vertically opposite angles)
∠CDO= ∠ABO         (Alternate angles, CD ||BA and BD is a transversal line)
By AA test of similarity
△COD ∼ △AOB



CD
AB

=


OD


OB


























Corresponding

sides

are

proportional





6
20

=

OD
15



OD
=
4
.
5

Page No 22:

Question 7:

◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.

Answer:

Given: ◻ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In ∆BET and ∆CED
∠BET = ∠CED         (Vertically opposite angles)
∠BTE = ∠CDE         (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
∆BET ∼ ∆CED



BE
CE

=


ET


ED


























Corresponding

sides

are

proportional




BE
×
ED
=
CE
×
ET

Hence proved.

Page No 22:

Question 8:

In the given figure, seg AC and seg BD intersect each other in point P and


AP
CP

=


BP


DP


. Prove that, ∆ABP ~ ∆CDP

Answer:

Given:


AP
CP

=


BP


DP


To prove: ∆ABP ~ ∆CDP
Proof: In ∆ABP and ∆DCP


AP
CP

=


BP


DP


       (Given)
∠P = ∠P                   (Common)
By SAS test of similarity


AP
CP

=


BP


DP


Page No 22:

Question 9:

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2
= CB × CD

Answer:

Given:  ∠BAC = ∠ADC
To prove: CA2 = CB × CD
Proof: In ∆ABC and ∆DAC
∠BAC = ∠ADC       (Given)
∠C = ∠C                   (Common)
By AA test of similarity
∆ABC ∼ ∆DAC



BC
AC

=

AC
DC

































Corresponding

sides

are

proportional





AC
2

=
BC
×
DC

Hence proved.

Page No 25:

Question 1:

The ratio of corresponding sides of similar triangles is 3 : 5; then Find the ratio of their areas.

Page No 25:

Question 2:

If ∆ABC ~ ∆PQR and AB : PQ = 2 : 3, then fill in the blanks.



A



ABC




A



PQR




=


AB
2











=


2
2


3
2


=




















Answer:

Given:
∆ABC ~ ∆PQR
AB : PQ = 2 : 3
According to theorem of areas of similar triangles “When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides”.




A



ABC




A



PQR




=


AB
2



PQ
2



=


2
2


3
2


=






4









9





Page No 25:

Question 3:

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.



A



ABC




A



.

.

.

.




=

80
125















AB
PQ

=
























Answer:

Given:
∆ABC ~ ∆PQR
A (∆ABC) = 80
A (∆PQR) = 125
According to theorem of areas of similar triangles “When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides”.




A



ABC




A



PQR




=


AB
2


PQ
2





80
125

=


AB
2


PQ
2





16
25

=


AB
2


PQ
2





4
2


5
2


=


AB
2


PQ
2





AB
PQ

=


4


5


Therefore,



A



ABC




A



PQR




=

80
125


and


AB
PQ

=


4


5


Page No 25:

Question 4:

∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN.

Answer:

Given:
∆LMN ~ ∆PQR
9 × A (∆PQR ) = 16 × A (∆LMN)
Consider, 9 × A (∆PQR ) = 16 × A (∆LMN)



A



LMN




A



PQR




=

9
16





MN
2


QR
2


=


3
2


4
2





MN
QR

=

3
4


MN
=

3
4

×
QR


MN
=

3
4

×
20





















QR
=
20




MN
=
15

Page No 25:

Question 5:

Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then Find corresponding side of the bigger triangle.

Answer:

According to theorem of areas of similar triangles “When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides”.




Area

of

bigger

triangle


Area

of

smaller

triangle


=

225
81







Side

of

bigger

triangle


2




Side

of

smaller

triangle


2


=


15
2


9
2






Side

of

bigger

triangle


Side

of

smaller

triangle


=

15
9


Side

of

bigger

triangle
=

15
9

×
Side

of

smaller

triangle


Side

of

bigger

triangle
=

15
9

×
12

=
20

Hence, the corresponding side of the bigger triangle is 20.

Page No 25:

Question 6:

∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.

Answer:

Consider, A(∆ABC) : A (∆DEF) = 1 : 2




A



ABC




A



DEF




=

1
2





AB
2


DE
2


=

1
2




DE
2

=
2

AB
2



DE
2

=
2
×

4
2






















AB
=
4




DE
=

32



DE
=
4

2


Page No 25:

Question 7:

In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.

Answer:

Given:
seg PQ || seg DE
A(∆PQF) = 20 units
PF = 2 DP
Let us assume DP =x

∴ PF = 2x

DF
=
DP
+

PF

=

x

+

2
x

=
3
x

In △FDE and △FPQ
∠FDE = ∠FPQ         (Corresponding angles)
∠FED = ∠FQP         (Corresponding angles)
By AA test of similarity
△FDE ∼ △FPQ




A



FDE




A



FPQ




=




FD
2






FP
2




=




3
x


2




2
x


2


=

9
4


A



FDE


=

9
4

A



FPQ


=

9
4

×

20

=

45


A



DPQE


=
A



FDE



A



FPQ



=

45



20


=

25

Page No 26:

Question 1:

Select the appropriate alternative.
(1) In ∆ABC and ∆PQR, in a one to one correspondence


AB
QR

=

BC
PR

=

CA
PQ

then

(A) ∆PQR ~ ∆ABC
(B) ∆PQR ~ ∆CAB
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~ ∆PQR

(2) If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

(A)


EF
PR

=

DF
PQ

(B)


DE
PQ

=

EF
RP

(C)


DE
QR

=

DF
PQ

(D)


EF
RP

=

DE
QR

(3) In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true ?
(A)The triangles are not congruent and not similar
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.

(4) ∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆DEF) = 1 : 2
If AB = 4 then what is length of DE?
(A)

2

2

(B) 4
(C) 8
(D)

4

2

(5) In the given figure, seg XY || seg BC, then which of the following statements is true?

(A)


AB
AC

=

AX
AY

(B)


AX
XB

=

AY
AC

(C)


AX
YC

=

AY
XB

(D)


AB
YC

=

AC
XB

Answer:

(1)
Given:


AB
QR

=

BC
PR

=

CA
PQ

By SSS test of similarity
∆PQR ~ ∆CAB
Hence, the correct option is (B).

(2)
In ∆DEF and ∆PQR
∠D ≅ ∠Q
∠R ≅ ∠E
By AA test of similarity
∆DEF~ ∆PQR



DE
PQ

=

EF
QR

=

DF
PR

















Corresponding

sides

of

similar

triangles

are

proportional




DE
PQ



EF
RP

Hence, the correct option is (B).

(3)
In ∆ABC and ∆DEF
∠B = ∠E,
∠F = ∠C
By AA test of similarity
∆ABC ~ ∆DEF
Since, there is not any congruency criteria like AA.
Thus, ∆ABC and ∆DEF are not congruent.
Hence, the correct option is (B).

(4)
Given: ∆ABC and ∆DEF are equilateral triangles
Constrcution: Draw a perependicular from vertex A and D on AC and DF in both triangles.


In ∆ABX and ∆DEY
∠B = ∠C = 60             (∆ABC and ∆DEF are equilateral triangles)
∠AXB = ∠DYB           (By construction)
By AA test of similarity
∆ABX ~ ∆DEY



AB
DE

=

AX
DY

















Corresponding

sides

of

similar

triangles

are

proportional




DE
PQ



EF
RP



A



ABC




A



DEF




=

1
2






1
2

×
AB
×
AX



1
2

×
DE
×
DY


=

1
2





AB
2


DE
2


=


1


2


























AB
DE

=

AX
DY







DE
2

=
32


DE
=
4

2

Hence, the correct option is (D).

(5)
Given: seg XY || seg BC
By basic proportionality theorem


AX
BX

=

AY
YC




BX
AX

+
1
=

YC
AY

+
1




BX
+
AX

AX

=


YC
+
AY

AY



AB
AX

=

AC
AY




AB
AC

=

AX
AY

Hence, the correct option is (D).

Page No 27:

Question 2:

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratios.

(1)



A



ABD




A



ADC




(2)



A



ABD




A



ABC




(3)



A



ADC




A



ABC




Page No 27:

Question 3:

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?

Answer:



Area

of


smaller

triangle


Area

of

bigger

triangle


=

2
3







1
2



×

Height

of

smaller

triangle
×
Base

of

smaller

triangle




1
2


×
Height

of

bigger

triangle
×
Base

of

bigger

triangle


=


2


3






6


Base

of

bigger

triangle


=


2


3



Base

of

bigger

triangle
=


3


2


×
6

=
9

Page No 27:

Question 4:

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then



A




ABC




A




DCB




=
?

Answer:

Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8

Now
,



A




ABC




A




DCB




=



1
2

×
AB
×
BC



1
2

×
DC
×
BC



=


6


8



=


3


4



Page No 27:

Question 5:

In the given figure, PM = 10 cm A(∆PQS) = 100 sq.cm A(∆QRS) = 110 sq.cm then Find NR.

Answer:

Given:
PM = 10 cm
A(∆PQS) = 100 sq.cm
A(∆QRS) = 110 sq.cm

Now
,



A




PQS




A




QRS




=

100
110






1
2

×
PM
×
QS



1
2

×
RN
×
QS


=

10
11





10


RN


=

10
11



RN
=
11

cm

Page No 27:

Question 6:

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio



A



MNT




A



QRS




.

Answer:

The areas of two similar triangles are proportional to the squares of their corresponding altitudes.




A



MNT




A



QRS




=



5
9


2


=

25
81

Page No 28:

Question 7:

In the given figure, A – D– C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then Find BE.

Answer:

Given:
AD = 5,
DC = 3,
BC = 6.4
In △ABC,  DE || AB



CD
DA

=

CE
EB


















By

basic

proportionality

theorem






3


5


=


6
.
4

x

x



3
x
=
32

5
x

​​


8
x
=
32


x
=
4

Page No 28:

Question 8:

In the given figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then Find PQ, QR and RS.

Answer:

Given:
AB = 60,
BC = 70,
CD = 80,
PS = 280
Now, AD = AB + BC + CD
= 60 + 70 + 80
= 210
By intercept theorem, we have


PQ
AB

=

QR
BC

=

RS
CD

=

PS
AD





PQ


60


=


QR


70


=


RS


80


=


280


210






PQ


60


=


QR


70


=


RS


80


=


4


3



PQ
=


4


3


×
60
=
80

QR
=


4


3


×
70
=

280
3


RS
=


4


3


×
80
=

320
3


Page No 28:

Question 9:

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Answer:

In △PMQ, ray MX is bisector of △PMQ.




PX


XQ


=


MQ


MP


                       ………. (I) theorem of angle bisector.
In △PMR, ray MY is bisector of△PMR.




PY


YR


=


MR


MP


                       ………. (II) theorem of angle bisector.
But


MP
MQ

=

MP
MR

                      ……… M is the midpoint QR, hence MQ = MR.



PX
XQ

=

PY
YR

∴XY || QR ………. converse of basic proportionality theorem.

Page No 29:

Question 10:

In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find


AX
XY

.

Answer:

In △ABY, ∠YBX = ∠XBA


AX
XY

=

AB
BY






















.
.
.

I





By

angle

bisector

theorem


In △ACY, ∠YCX = ∠XCA


AX
XY

=

AC
CY






















.
.
.

II





By

angle

bisector

theorem


From (I) and (II)


AC
CY

=

AB
BY




AC

BC

BY


=

AB
BY




4

6

BY


=

5
BY


4
BY
=
30

5
BY


9
BY
=
30


BY
=

10
3

From (I), we have


AX
XY

=

5


10
3




=

3
2

Page No 29:

Question 11:

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that


AP
PD

=

PC
BP

Answer:

Given: ▢ABCD is a parallelogram
To prove:


AP
PD

=

PC
BP

Proof: In △APD and △CPB
∠APD = ∠CPB         (Vertically opposite angles)
∠PAD = ∠PCB         (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
△APD ∼ △CPB



AP
PC

=


PD


PB


























Corresponding

sides

are

proportional





AP
PD

=


PC


PB


Hence proved.

Page No 29:

Question 12:

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.

Answer:

Given:
XY || seg AC
2AX = 3BX
XY = 9
Consider, 2AX = 3BX



AX
BX

=


3


2





AX
+
BX

BX

=



3

+

2



2



















.
.
.
.
.
by

componendo


AB
BX

=


5


2





















.
.
.
.

I










BCA
~

BYX









.
.
.

SAS


test

of

similarity



BA
BX

=

AC
XY


















.
.
.
corresponding

sides

of

similar

triangles




5


2


=

AC
9






AC
=

17
.
5




















.
.
.
from


I












Page No 29:

Question 13:

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE2
= BD × EC (Hint : Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)

Answer:

Given: ▢DEFG is a square
To prove: DE2 = BD × EC
Proof: In △GBD and △AGF
∠GDB = ∠GAF =  90°         (Given)
∠AGF = ∠GBF                    (Corresponding, GF || BC and AB is a transversal line)
By AA test of similarity
△GBD ∼ △AGF                         …(1)
In △CFEand △AGF
∠FEC = ∠GAF =  90°         (Given)
∠FCE = ∠AGF                   (Corresponding, GF || BC and AC is a transversal line)
By AA test of similarity
△CFE ∼ △AGF                          …(2)
From (1) and (2), we get
△CFE ∼ △GBD



CE
GD

=


FE


BD


























Corresponding

sides

are

proportional





CE
DE

=


DE


BD
























GD
=
FE
=
DE





DE
2

=
BD
×
CE

Hence proved.

View NCERT Solutions for all chapters of Class 10

What is the Value of X Given That Pq Bc

Sumber: https://www.meritnation.com/maharashtra-class-10/math/mathematics-part-ii-solutions-/similarity/textbook-solutions/91_1_3385_24433

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