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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.

#### Page No 6:

#### Question 2:

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

$\frac{}{}$

#### Answer:

Given:

BC = 4

AD = 8

$=$

#### 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,

With base QR and height PS,

$\mathrm{A}$$\therefore $

$\Rightarrow $

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

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)

$\frac{}{}$(ii)

$\frac{}{}$(iii)

$\frac{}{}$(iv)

$\frac{}{}$#### Answer:

(i)

$\frac{}{}$(ii)

$\frac{}{}$(iii)

$\frac{}{}$(iv)

$\frac{}{}$#### 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)

$\mathrm{In}$$\mathrm{In}$

$\therefore $

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

(2)

$\mathrm{In}$$\mathrm{In}$

$\therefore $

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

(3)

$\mathrm{In}$

$\therefore $

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

$\mathrm{Also}$

$\therefore $

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:

$\mathrm{In}$

Hence, the measure of QP is 3.5.

#### Page No 14:

#### Question 4:

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

$\frac{}{}$

#### Answer:

Given:

∠APQ = 60^{∘}

∠ABC = 60^{∘}

Since, the corresponding angles ∠APQ and ∠APC are equal.

Hence, line PQ || BC.

#### 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

In △BDC, XQ||DC

$\frac{}{}$From (1) and (2), we get

$\frac{}{}$#### Page No 14:

#### Question 6:

Find QP using given information in the figure.

#### Answer:

$\mathrm{In}$

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

In △AEF, XC||FE

$\frac{}{}$From (1) and (2), we get

$\frac{}{}$
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:

$\mathrm{In}$

Hence, the measure of LT is 4.8.

#### Page No 15:

#### Question 9:

In ∆ABC, seg BD bisects ∠ABC. If AB =

*x*, BC =

*x
*+ 5, AD =

*x*

– 2, DC =

*x*

+ 2, then find the value of

*x.*

#### Answer:

In △ABC, ∠ABD = ∠DBC

$\frac{}{}$$\Rightarrow $

#### 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

In △XEF, QR || EF ….Given

$\therefore $$\therefore $

∴ 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

In △ABC, ∠BCE = ∠ACE

$\frac{}{}$From (I) and (II)

$\frac{}{}$∴ 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,

$\therefore $

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

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.

#### 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

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

#### 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

Hence proved.

#### Page No 22:

#### Question 8:

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

$\frac{}{}$. Prove that, ∆ABP ~ ∆CDP

#### Answer:

Given:

$\frac{}{}$
To prove: ∆ABP ~ ∆CDP

Proof: In ∆ABP and ∆DCP

(Given)

∠P = ∠P (Common)

By SAS test of similarity

#### Page No 22:

#### Question 9:

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA^{2}

= CB × CD

#### Answer:

Given: ∠BAC = ∠ADC

To prove: CA^{2} = CB × CD

Proof: In ∆ABC and ∆DAC

∠BAC = ∠ADC (Given)

∠C = ∠C (Common)

By AA test of similarity

∆ABC ∼ ∆DAC

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.

#### 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”.

Therefore, the ratio of the areas of triangles

$=$

#### Page No 25:

#### Question 2:

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

$\frac{}{}$#### 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”.

#### Page No 25:

#### Question 3:

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

$\frac{}{}$#### 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”.

$\Rightarrow $

Therefore,

$\frac{}{}$#### 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)

$\Rightarrow $

#### 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”.

$\therefore $

$\Rightarrow $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

$\Rightarrow $

$\Rightarrow $#### 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 = 2*x*

In △FDE and △FPQ

∠FDE = ∠FPQ (Corresponding angles)

∠FED = ∠FQP (Corresponding angles)

By AA test of similarity

△FDE ∼ △FPQ

$\therefore $

#### Page No 26:

#### Question 1:

Select the appropriate alternative.

(1) In ∆ABC and ∆PQR, in a one to one correspondence

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)

$\frac{}{}$(B)

$\frac{}{}$(C)

$\frac{}{}$(D)

$\frac{}{}$

(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)

(B) 4

(C) 8

(D)

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

(A)

$\frac{}{}$(B)

$\frac{}{}$(C)

$\frac{}{}$(D)

$\frac{}{}$

#### Answer:

(1)

Given:

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

$\therefore $

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

$\therefore $

$\frac{}{}$

Hence, the correct option is (D).

(5)

Given: seg XY || seg BC

By basic proportionality theorem

$\Rightarrow $

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)

$\frac{}{}$(2)

$\frac{}{}$(3)

$\frac{}{}$#### Answer:

Construction: Draw a perpendicular from vertex A to line BC.

(1)

(2)

(3)

#### 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:

$\frac{}{}$

#### Page No 27:

#### Question 4:

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

$\frac{}{}$

#### Answer:

Given:

∠ABC = ∠DCB = 90°

AB = 6

DC = 8

#### 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

#### 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

$\frac{}{}$.

#### Answer:

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

$\therefore $#### 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

$\Rightarrow $#### 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

$\therefore $

#### 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.

$\therefore $ ………. (I) theorem of angle bisector.

In △PMR, ray MY is bisector of△PMR.

………. (II) theorem of angle bisector.

But

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

$\therefore $∴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

$\frac{}{}$.

#### Answer:

In △ABY, ∠YBX = ∠XBA

$\frac{}{}$
In △ACY, ∠YCX = ∠XCA

From (I) and (II)

$\frac{}{}$$\Rightarrow $

From (I), we have

#### 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

$\frac{}{}$

#### Answer:

Given: ▢ABCD is a parallelogram

To prove:

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

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

$\frac{}{}$

#### 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 DE^{2}

= BD × EC (Hint : Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)

#### Answer:

Given: ▢DEFG is a square

To prove: DE^{2} = 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

Hence proved.

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### What is the Value of X Given That Pq Bc

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