Extra practice Questions and Answers on Class 10 Mathematic -2 Lesson 1.Similarity

 

Chapter 1: Similarity 

Q.1 (A)MCQ Solutions

[1 mark each]

If ΔABC ~ ΔPQR and AB : PQ = 3 : 4, then A(ΔABC) : A(ΔPQR) = ?

(A) 9 : 25
(B) 9 : 16
(C) 16 : 9
(D) 25 : 9

Solution:

Since the two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Area ratio = (Corresponding side ratio)²

Given:

AB : PQ = 3 : 4

Therefore,

A(ΔABC) : A(ΔPQR) = 3² : 4²

= 9 : 16

Answer: (B) 9 : 16

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Q.2

Which of the following is not a test of similarity?

(A) AAA
(B) SAS
(C) SAA
(D) SSS

Solution:

The standard tests of similarity of triangles are:

• AAA – Angle-Angle-Angle
• SAS – Side-Angle-Side
• SSS – Side-Side-Side

Among the given options, SAA is not stated as a separate standard test of similarity.

Answer: (C) SAA

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Q.3

If ΔXYZ ~ ΔPQR and A(ΔXYZ) = 25 cm², A(ΔPQR) = 4 cm², then XY : PQ = ?

(A) 4 : 25
(B) 2 : 5
(C) 5 : 2
(D) 25 : 4

Solution:

For two similar triangles,

Area ratio = (Ratio of corresponding sides)²

Therefore,

A(ΔXYZ) / A(ΔPQR) = (XY / PQ)²

Substitute the given values:

25 / 4 = (XY / PQ)²

Taking the square root on both sides:

XY / PQ = √(25 / 4)

XY / PQ = 5 / 2

Therefore,

XY : PQ = 5 : 2

Answer: (C) 5 : 2

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Q.4

Ratio of areas of two similar triangles is 9 : 25. _____ is the ratio of their corresponding sides.

(A) 3 : 4
(B) 3 : 5
(C) 5 : 3
(D) 25 : 81

Solution:

For similar triangles, the ratio of areas is equal to the square of the ratio of corresponding sides.

Given:

Area ratio = 9 : 25

Therefore, the ratio of corresponding sides is obtained by taking the square root:

√9 : √25

= 3 : 5

Answer: (B) 3 : 5

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Q.5

ΔABC ~ ΔDEF, if ∠A = 45° and ∠E = 35°, then ∠B = ?

(A) 45°

(B) 35°

(C) 25°
(D) 40°

Solution:

Given:

ΔABC ~ ΔDEF

The order of the letters tells us which vertices correspond to each other.

A ↔ D
B ↔ E
C ↔ F

Therefore, ∠B corresponds to ∠E.

Corresponding angles of similar triangles are equal.

∠B = ∠E

Given:

∠E = 35°

Hence,

∠B = 35°

Answer: (B) 35°

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Q.6

In the figure, segment DE ∥ segment BC. Identify the correct statement.

(A) AD/DB = AE/EC
(B) AD/DB = AB/AC
(C) AD/DB = AC/EC
(D) AD/DB = AE/EC

Solution:

Given that DE ∥ BC.

Here, D lies on AB and E lies on AC.

We use the Basic Proportionality Theorem (BPT).

The Basic Proportionality Theorem states:

If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same proportion.

Therefore,

AD/DB = AE/EC

Thus, the required relationship is:

AD : DB = AE : EC

Answer: (A) AD/DB = AE/EC

Note: In the uploaded image, options (A) and (D) appear to be the same. If they are printed exactly as shown, both options contain the same correct statement.

Q.7

If ΔXYZ ~ ΔPQR, then XY/PQ = YZ/QR = ?

(A) XZ/PR
(B) XZ/PQ
(C) XZ/QR
(D) YZ/PQ

Solution:

Given:

ΔXYZ ~ ΔPQR

From the order of the vertices:

X ↔ P,   Y ↔ Q,   Z ↔ R

Therefore, the corresponding sides are:

XY ↔ PQ
YZ ↔ QR
XZ ↔ PR

Hence,

XY/PQ = YZ/QR = XZ/PR

Answer: (A) XZ/PR

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Q.8

If ΔABC ~ ΔLMN and ∠A = 60°, then ∠L = ?

(A) 45°
(B) 60°
(C) 25°
(D) 40°

Solution:

Given:

ΔABC ~ ΔLMN

From the order of the triangles:

A ↔ L

Therefore, ∠A and ∠L are corresponding angles.

Corresponding angles of similar triangles are equal.

∠A = ∠L

Given:

∠A = 60°

Therefore,

∠L = 60°

Answer: (B) 60°

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Q.9

ΔDEF and ΔXYZ, DE/XY = EF/YZ and ∠E = ∠Y. Which test gives similarity between ΔDEF & ΔXYZ?

(A) AAA
(B) SAS
(C) SAA
(D) SSS

Solution:

We are given:

DE/XY = EF/YZ

Thus, two pairs of corresponding sides are proportional.

Also,

∠E = ∠Y

These angles are the included angles between the corresponding sides.

Therefore, two corresponding sides are proportional and the included angles are equal.

This satisfies the SAS similarity criterion.

Answer: (B) SAS

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Q.10

In the figure, BD = 8, BC = 12 and B-D-C. Find:

A(ΔABC) / A(ΔABD) = ?

(A) 2 : 3
(B) 3 : 2
(C) 5 : 3
(D) 3 : 4

Solution:

Triangles ΔABC and ΔABD have the same altitude from A to the line BC.

Therefore, their areas are proportional to their corresponding bases.

A(ΔABC) / A(ΔABD) = BC / BD

Given:

BC = 12,   BD = 8

Substitute the values:

A(ΔABC) / A(ΔABD) = 12 / 8

Dividing numerator and denominator by 4:

= 3 / 2

Therefore,

A(ΔABC) : A(ΔABD) = 3 : 2

Answer: (B) 3 : 2

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Q.1 (B) Solve [1 mark each]

Are the triangles in the figure similar? If yes, then write the test of similarity.

Given:

For ΔABC:

AB = 6,   BC = 8,   AC = 10

For ΔPQR:

PQ = 3,   QR = 4,   PR = 5

Step 1: Compare the corresponding sides.

AB/PQ = 6/3 = 2

BC/QR = 8/4 = 2

AC/PR = 10/5 = 2

Step 2: Compare all three ratios.

AB/PQ = BC/QR = AC/PR

= 2

Thus, all three pairs of corresponding sides are proportional.

Therefore, the triangles are similar by the SSS similarity criterion.

ΔABC ~ ΔPQR

Answer: Yes, the triangles are similar by SSS test.

Q.2

In the figure, line BC ∥ line DE, AB = 2, BD = 3, AC = 4 and CE = x. Find the value of x.

Solution:

Given:

BC ∥ DE

By the Basic Proportionality Theorem (BPT):

AB / BD = AC / CE

Substituting the given values:

2 / 3 = 4 / x

Cross-multiplying:

2x = 3 × 4

2x = 12

Dividing both sides by 2:

x = 6

Answer: x = 6

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Q.3

State whether the following triangles are similar or not. If yes, then write the test of similarity.

Given:

∠P = 30°,   ∠X = 35°

∠Q = 60°,   ∠Y = 60°

Solution:

First, find the third angle of ΔPQR.

∠P + ∠Q + ∠R = 180°

30° + 60° + ∠R = 180°

∠R = 90°

Now, find the third angle of ΔXYZ.

∠X + ∠Y + ∠Z = 180°

35° + 60° + ∠Z = 180°

∠Z = 85°

Therefore:

ΔPQR:   30°, 60°, 90°

ΔXYZ:   35°, 60°, 85°

Only one pair of corresponding angles is equal:

∠Q = ∠Y = 60°

The other corresponding angles are not equal. Therefore, the two triangles are not similar.

Answer: The triangles are not similar.

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Q.4

Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.

Solution:

For two similar triangles:

Area ratio = (Corresponding side ratio)²

Given:

Area ratio = 144 : 49

Therefore, the ratio of corresponding sides is:

√144 : √49

= 12 : 7

Answer: 12 : 7

Chapter 1: Similarity – Activity Solutions

Q.2A). Complete the activity

In the figure, BP ⟂ AC, CQ ⟂ AB, A-P-C, Q lies on AB. Show that ΔAPB and ΔAQC are similar.

Solution:

Since BP ⟂ AC,

∠APB = 90°

Since CQ ⟂ AB,

∠AQC = 90°

Therefore,

∠APB = ∠AQC = 90°    ...(1)

Also, A-P-C and Q lies on AB. Hence, both ∠PAB and ∠QAC represent the same angle at A.

Therefore,

∠PAB = ∠QAC    ...(2)

From (1) and (2),

∠APB = ∠AQC

and

∠PAB = ∠QAC

Thus, two corresponding angles are equal.

Therefore, by the AA similarity test,

ΔAPB ~ ΔAQC

Completed activity:

∠APB = 90°

∠AQC = 90°

∠PAB = ∠QAC [Two corresponding angles are equal]

ΔAPB ~ ΔAQC [AA similarity test]

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Q.2 Observe the figure and complete the activity

Given: ∠B = 75° and ∠D = 75°.

Solution:

We are given:

∠B = 75° and ∠D = 75°

Therefore,

∠B = ∠D

Also, ∠C is common to both triangles.

Therefore,

∠C = ∠C

Thus, we have two pairs of equal angles:

∠B = ∠D

∠C = ∠C

Therefore, the two triangles are similar by the AA similarity test.

ΔABC ~ ΔDEC

Completed activity:

∠B [= ∠D] each of 75°

∠C [= ∠C]

ΔABC ~ ΔDEC

[AA] similarity test

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Q.3 Complete the activity

ΔABC ~ ΔPQR, A(ΔABC) = 80 sq cm and A(ΔPQR) = 125 sq cm.

Solution:

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Therefore,

A(ΔABC) / A(ΔPQR) = (AB / PQ)²

Substituting the given values:

A(ΔABC) / A(ΔPQR) = 80 / 125

Divide both numerator and denominator by 5:

80 / 125 = 16 / 25

Therefore,

A(ΔABC) / A(ΔPQR) = 16 / 25

Hence,

(AB / PQ)² = 16 / 25

Taking square root on both sides:

AB / PQ = √(16 / 25)

AB / PQ = 4 / 5

Therefore,

AB : PQ = 4 : 5

Q.4– 

In the figure, PM = 10 cm, A(ΔPQS) = 100 sq cm and A(ΔQRS) = 110 sq cm. Find NR.

Solution:

Triangles ΔPQS and ΔQRS have the same base QS.

For two triangles having the same base, their areas are proportional to their corresponding heights.

A(ΔPQS) / A(ΔQRS) = PM / NR

Substituting the given values:

100 / 110 = 10 / NR

Simplifying:

10 / 11 = 10 / NR

Cross-multiplying:

10 × NR = 10 × 11

NR = 11 cm

Answer: NR = 11 cm

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Q.2(B) – Q.1

The ratio of the areas of two triangles of equal heights is 2 : 3. The base of the smaller triangle is 6 cm. Find the corresponding base of the larger triangle.

Solution:

We know that:

Area of triangle = ½ × Base × Height

Since the two triangles have equal heights, their areas are directly proportional to their bases.

Therefore,

Area of smaller triangle / Area of larger triangle = Base of smaller triangle / Base of larger triangle

Given:

2 / 3 = 6 / x

Cross-multiplying:

2x = 3 × 6

2x = 18

Dividing both sides by 2:

x = 9

Therefore, the base of the larger triangle is:

Answer: 9 cm

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Q.2(B) – Q.2

ΔABP ~ ΔDEF and A(ΔABP) : A(ΔDEF) = 144 : 81. Find AB : DE.

Solution:

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

A(ΔABP) / A(ΔDEF) = (AB / DE)²

Given:

144 / 81 = (AB / DE)²

Taking square root on both sides:

AB / DE = √(144 / 81)

= 12 / 9

Dividing by 3:

AB / DE = 4 / 3

Therefore:

AB : DE = 4 : 3

Answer: 4 : 3

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Q.2(B) – Q.3

In the figure, segments AC and BD intersect each other at P. If

AP / PC = BP / PD

then prove that ΔABP ~ ΔCPD.

Solution:

Given:

AP / PC = BP / PD

By rearranging the terms:

AP / BP = PC / PD

Also, AC and BD intersect at P.

Therefore, ∠APB and ∠CPD are vertically opposite angles.

Hence,

∠APB = ∠CPD

Now we have:

AP / BP = PC / PD

and

∠APB = ∠CPD

Thus, two corresponding sides are proportional and the included angles are equal.

Therefore, by the SAS similarity criterion:

ΔABP ~ ΔCPD

Hence proved.

Q.4. The base of a triangle is 9 and the height is 5. The base of the second triangle is 10 and height is 6. Find the ratio of the two triangles.

Answer:

Area of first triangle

= ½ × Base × Height

= ½ × 9 × 5

= 45/2

= 22.5 sq. units

Area of second triangle

= ½ × Base × Height

= ½ × 10 × 6

= 30 sq. units

Therefore,

Ratio of the two triangles

= 22.5 : 30

= 22.5*2/30*2

= 45 : 60

= 3 : 4

Hence, the ratio of the two triangles is 3 : 4.

Que. 3 A) Complete the following action.

Q.1 In the figure, show BP ⟂ AC, CQ ⟂ AB, A-P-C, A-B-Q and ΔABP and ΔAQC alike.

Solution:

In ΔABP,

∠APB = 90° (Given BP ⟂ AC)

In ΔAQC,

∠AQC = 90° (Given CQ ⟂ AB)

Therefore,

∠APB = ∠AQC

Also,

∠PAB = ∠QAC (A-P-C and A-B-Q)

Therefore, by AA similarity criterion,

ΔAPB ∼ ΔAQC

Answer:

∠APB = 90°

∠AQC = 90°

∠APB = ∠AQC

∴ ΔAPB ∼ ΔAQC

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Q.2 In the figure, line PQ ∥ DE. Suppose A(ΔPQF) = 20 units and PF = 2DP. Find A(□DPQE).

Solution:

Given,

A(ΔPQF) = 20 square units

PF = 2DP

Let,

DP = x

Therefore,

PF = 2x

DF = DP + PF

DF = x + 2x

DF = 3x

Since PQ ∥ DE,

ΔFDE ∼ ΔFPQ

Therefore,

DF : PF = 3x : 2x

= 3 : 2

For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.

Therefore,

A(ΔFDE) / A(ΔFPQ) = (3/2)²

= 9/4

Given,

A(ΔFPQ) = 20 square units

Therefore,

A(ΔFDE) = 9/4 × 20

= 45 square units

Now,

A(□DPQE) = A(ΔFDE) - A(ΔFPQ)

= 45 - 20

= 25 square units

Answer:

A(□DPQE) = 25 square units.

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Q.3 In the figure X is any point in the interior of triangle. Point X is joined to the vertices of triangle. Seg PQ ∥ DE, Seg QR ∥ EF. Prove that PR ∥ DF.

Solution:

In ΔXDE, PQ ∥ DE.

By the Basic Proportionality Theorem,

XP / PD = XQ / QE ........ (1)

In ΔXEF, QR ∥ EF.

By the Basic Proportionality Theorem,

XQ / QE = XR / RF ........ (2)

From (1) and (2),

XP / PD = XR / RF

Therefore, in ΔXDF, by the Converse of Basic Proportionality Theorem,

PR ∥ DF

Hence proved.

Answer:

XP / PD = XQ / QE ........ (1)

XQ / QE = XR / RF ........ (2)

Therefore,

XP / PD = XR / RF

∴ PR ∥ DF

Q. 4) In the figure XY || AC, if 2AX = 3BX and XY = 9, find the value of AC.

Solution:

Given:

2AX = 3BX

XY = 9

2AX = 3BX

AX/BX = 3/2

Therefore,

(AX + BX)/BX = (3 + 2)/2

              = 5/2

Since AX + BX = AB,

AB/BX = 5/2 ........ (1)

Since XY || AC,

ΔBCA ~ ΔBYX

(AA Test of Similarity)

Therefore,

BA/BX = AC/XY

From (1),

5/2 = AC/9

AC = (5 × 9)/2

AC = 45/2

AC = 22.5

∴ AC = 22.5 units

Answers:

AX/BX = 3/2

(AX + BX)/BX = 5/2

AB/BX = 5/2

Test of Similarity = AA

AC = 22.5 units

(Converse of Basic Proportionality Theorem)

Q.3(B) Solutions

Q.1

In the figure, PQ ⟂ BC and AD ⟂ BC. Find the following ratios:

(a) A(ΔPQB) / A(ΔPBC)

(b) A(ΔABC) / A(ΔADC)

(c) A(ΔPBC) / A(ΔABC)

Solution:

(a) Find A(ΔPQB) / A(ΔPBC)

Triangles ΔPQB and ΔPBC have their bases on the same straight line BC.

Also, both triangles have the same altitude PQ.

We know:

Area of triangle = ½ × Base × Height

Therefore,

A(ΔPQB) / A(ΔPBC) = [½ × QB × PQ] / [½ × BC × PQ]

Cancel ½ and PQ:

A(ΔPQB) / A(ΔPBC) = QB / BC

Answer:

QB : BC

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(b) Find A(ΔABC) / A(ΔADC)

Triangles ΔABC and ΔADC have bases BC and DC, respectively.

Both triangles have the same altitude AD to the line BC.

Therefore, their areas are proportional to their bases.

A(ΔABC) / A(ΔADC) = BC / DC

Answer:

BC : DC

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(c) Find A(ΔPBC) / A(ΔABC)

Triangles ΔPBC and ΔABC have the same base BC.

Their corresponding heights are PQ and AD.

Therefore, when two triangles have the same base, their areas are proportional to their corresponding heights.

A(ΔPBC) / A(ΔABC) = PQ / AD

Answer:

PQ : AD

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Q.2

In the given figure, quadrilateral PQRS has PQ ∥ SR and AR = 5AP. Prove that SR = 5PQ.

9

Solution:

Diagonals PR and QS intersect at A.

Given:

PQ ∥ SR

and

AR = 5AP

Consider triangles ΔPAQ and ΔRAS.

Step 1: Compare the angles.

Since PR and QS intersect at A,

∠PAQ = ∠RAS

These are vertically opposite angles.

Also, PQ ∥ SR and PR is a transversal.

Therefore,

∠APQ = ∠ARS

These are corresponding angles.

Thus, two corresponding angles are equal.

Therefore,

ΔPAQ ~ ΔRAS

by the AA similarity test.

Step 2: Use corresponding sides.

From the similarity of triangles:

AP / AR = PQ / SR

Given:

AR = 5AP

Therefore,

AP / 5AP = PQ / SR

Cancel AP:

1 / 5 = PQ / SR

Cross-multiplying:

SR = 5PQ

Hence proved.

∴ SR = 5PQ

Q.3) In triangle ABC, point D is on side BC such that ∠BAC = ∠ADC. Prove that CA² = CB × CD.

Given:

D lies on BC.

∠BAC = ∠ADC.

To prove:

CA² = CB × CD

Proof:

In ΔABC and ΔADC,

∠BAC = ∠ADC (Given)

∠ACB = ∠ACD (Common angle)

Therefore,

ΔABC ~ ΔADC (AA Test of Similarity)

Hence,

CB/CA = CA/CD

Cross-multiplying,

CB × CD = CA × CA

Therefore,

CA² = CB × CD

Hence proved.

Q.4) In quadrilateral ABCD, side AD || BC. Diagonals AC and BD intersect at P. Prove that PD/AP = PC/BP.

Given:

AD || BC

AC and BD intersect at P.

To prove:

PD/AP = PC/BP

Proof:

In ΔAPD and ΔCPB,

∠APD = ∠CPB (Vertically opposite angles)

∠ADP = ∠CBP (Alternate interior angles)

Therefore,

ΔAPD ~ ΔCPB (AA Test of Similarity)

Hence,

PD/PB = AP/PC

Therefore,

PD/AP = PB/PC

Note: The relation printed in the question appears to have a printing error. The correct result is:

PD/AP = PB/PC.

Q.4 (1) Side of equilateral triangle PQR is 8 cm. Find the area of a triangle whose side is half the side of triangle PQR.

Given:

Side of equilateral triangle PQR = 8 cm

Side of required triangle = 8/2 = 4 cm

Area of equilateral triangle = (√3/4) × side²

Area = (√3/4) × 4²

     = (√3/4) × 16

     = 4√3 cm²

∴ Area of required triangle = 4√3 cm²

Q.4.2) Areas of two similar triangles are equal. Prove that the triangles are congruent.

Given:

ΔABC ~ ΔPQR

Area(ΔABC) = Area(ΔPQR)

To prove:

ΔABC ≅ ΔPQR

Proof:

Since ΔABC ~ ΔPQR,

Area(ΔABC)/Area(ΔPQR)

= (AB/PQ)²

But,

Area(ΔABC) = Area(ΔPQR)

Therefore,

1 = (AB/PQ)²

Hence,

AB/PQ = 1

Therefore,

AB = PQ

Since the triangles are similar,

AB/PQ = BC/QR = CA/RP

Therefore,

BC = QR and CA = RP.

Thus, the corresponding sides of the two triangles are equal.

∴ ΔABC ≅ ΔPQR

by SSS congruence test.

Hence proved.

Q.4 (3) Two triangles are similar. The sides of the smaller triangle are 4 cm, 5 cm and 6 cm. The perimeter of the larger triangle is 90 cm. Find the sides of the larger triangle.

Perimeter of smaller triangle = 4 + 5 + 6 = 15 cm

Perimeter of larger triangle = 90 cm

Scale factor = 90/15 = 6

Therefore,

First side = 4 × 6 = 24 cm

Second side = 5 × 6 = 30 cm

Third side = 6 × 6 = 36 cm

∴ The sides of larger triangle are 24 cm, 30 cm and 36 cm.

Q.5 (1)

In the figure, PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Find the pair of lines and the values of x and y such that ST ∥ QR.

Solution:

Given:

PS = 2,   SQ = 6,   QR = 5

First, find PQ:

PQ = PS + SQ

= 2 + 6

PQ = 8

For ST ∥ QR, by the Basic Proportionality Theorem (BPT):

PS / SQ = PT / TR

Substituting the given values:

2 / 6 = x / y

Simplifying:

1 / 3 = x / y

Therefore:

x : y = 1 : 3

Hence, the required pair of parallel lines is:

ST ∥ QR

Note: With the information given in the question, the individual values of x and y cannot be determined. Only their ratio can be found.

x : y = 1 : 3

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Q.5 (2)

An architect has a model of a building. The length of the actual building is 1 m and the length of its model is 0.75 cm. Find the length and height of the model building whose actual length is 22.5 m and height is 10 m.

Solution:

Step 1: Find the scale

Actual length:

1 m = 100 cm

Model length:

0.75 cm

Therefore:

Model scale = Model length / Actual length

= 0.75 / 100

= 0.0075

Step 2: Find the length of the model

Actual length of the new building:

22.5 m = 22.5 × 100 = 2250 cm

Therefore:

Model length = 2250 × 0.0075

= 16.875 cm

Step 3: Find the height of the model

Actual height:

10 m = 1000 cm

Therefore:

Model height = 1000 × 0.0075

= 7.5 cm

Therefore:

Length of model = 16.875 cm

Height of model = 7.5 cm