RBSE Solutions for Class 10 Maths Chapter 11 Similarity Miscellaneous Exercise is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Miscellaneous Exercise.

## Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Miscellaneous Exercise

Question 1.

In fig. DE || BC, AD = 4 cm, DB = 6 cm (RBSESolutions.com) and AE = 5 cm then EC will be :

(A) 6.5 cm

(B) 7.0 cm

(C) 7.5 cm

(D) 8.0 cm

Solution :

AD = 4 cm

AE = 5 cm

BD = 6 cm

In ∆ABC

BC || DE

∴ \(\frac { AD }{ BD }\) = \(\frac { AE }{ CE }\) (By Basic Prop. Theorem)

⇒ \(\frac { 4 }{ 6 }\) = \(\frac { 5 }{ CE }\)

⇒ CE = \(\frac { 6\times 5 }{ 4 } \)

⇒ CE = 7.5 cm

Hence option (C) is correct.

Question 2.

In fig. AD is (RBSESolutions.com) bisector of ∠A, AB = 6 cm, BD = 8 cm, DC = 6 cm, then AC will be :

(A) 4.0 cm

(B) 4.5 cm

(C) 5 cm

(D) 5.5 cm

Solution :

In ∆ABC

AD is bisector of ∠A

∴ \(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\) (By Basic Prop. Theorem)

⇒ \(\frac { 6 }{ AC }\) = \(\frac { 8 }{ 6 }\)

⇒ 8 AC = 6 × 6

⇒ AC = \(\frac { 6\times 6 }{ 8 } \)

AC = 4.5 cm

Thus, (B) is correct option.

Question 3.

In fig. it DE || BC then x (RBSESolutions.com) will be :

(A) \(\sqrt { 5 }\)

(B) \(\sqrt { 6 }\)

(C) \(\sqrt { 3 }\)

(D) \(\sqrt { 7 }\)

Solution :

In ΔABC

DE || BC

∴ \(\frac { AD }{ BD }\) = \(\frac { AE }{ CE }\)

⇒ \(\frac { x+4 }{ x+3 }\) = \(\frac { 2x-1 }{ x+1 }\) (By cross multiplication method)

⇒ (x + 4)(x + 1) = (x + 3)(2x – 1)

⇒ x^{2} + 4x + x + 4 = 2x^{2} – x + 6x – 3

⇒ x^{2} + 5x + 4 = 2x^{2} + 5x – 3

⇒ x^{2} = 7

⇒ x = ± \(\sqrt { 7 }\)

Negative value of x is not possible.

∴ x = \(\sqrt { 7 }\)

Thus (D) is correct.

Question 4.

In fig. if AB = 34 cm, BD = 4 cm, BC = 10 cm

(A) 5.1 cm

(B) 3.4 cm

(C) 6 cm

(D) 5.3 cm

Solution :

In ΔABC

AD is bisector of (RBSESolutions.com) vertex angle A.

∴ \(\frac { AB }{ AC }\) = \(\frac { BD }{ CD }\)

⇒ \(\frac { AD }{ BD }\) = \(\frac { BD }{ BC-BD }\)

⇒ \(\frac { 3.4 }{ AC }\) = \(\frac { 4 }{ 10-4 }\)

⇒ \(\frac { 3.4 }{ AC }\) = \(\frac { 4 }{ 6 }\)

⇒ AC = \(\frac { 6\times 3.4 }{ 4 } \)

⇒ AC = 5.1 cm

Thus (A) is correct option.

Question 5.

The areas of two similar (RBSESolutions.com) triangle are 25 cm^{2} and 36 cm^{2} respectively. If median of smaller triangle be 10 cm then corresponding median of larger triangle will be :

(A) 12 cm

(B) 15 cm .

(C) 10 cm

(D) 18 cm

Solution :

We know that areas of two similar triangles is equal to the ratio of squares of their corresponding medians :

Thus (A) is correct option.

Question 6.

In a trapezium ABCD, AB || CD and (RBSESolutions.com) the diagonals meet at point O. If AB = 6 cm and DC = 3 cm, then ratio of ar (∆AOB) and ar(∆COD) will be :

(A) 4 : 1

(B) 1 : 2

(C) 2 : 1

(D) 1 : 4

Solution :

∆AOB and ∆COD

∠AOB = ∠COD (Vertically opposite angles)

AB || CD

∠ABO = ∠CDO (Alternate angles)

By A-A similarity

ΔAOB ~ ΔCOD

We know that ratio of two similar (RBSESolutions.com) triangles is equal to ratio- of squares of their corresponding sides.

Thus, \(\frac { ar.\triangle AOB }{ ar.\triangle COD } =\frac { { AB }^{ 2 } }{ { CD }^{ 2 } }\)

⇒ \(\frac { ar.\triangle AOB }{ ar.\triangle COD } =\frac { { 6 }^{ 2 } }{ { 3 }^{ 2 } }\)

⇒ \(\frac { ar.\triangle AOB }{ ar.\triangle COD }\) = \(\frac { 36 }{ 9 }\)

⇒ \(\frac { ar.\triangle AOB }{ ar.\triangle COD }\) = \(\frac { 4 }{ 1 }\)

⇒ \(\frac { ar.\triangle AOB }{ ar.\triangle COD }\) = 4 : 1

Thus, required raito will be 4 : 1.

Thus (A) is correct option.

Question 7.

If in ∆ABC and ∆DEF ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠D = 60°, ∠E = 70°, and ∠F = 50° then which one is correct ?

(A) ∆ABC ~ ∆DEF

(B) ∆ABC ~ ∆EDF

(C) ∆ABC ~ ∆DEF

(D) ∆ABC ~ ∆FED

Solution :

∠A = ∠F = 50°

∠B = ∠E = 70°

∠C = ∠D = 60°

∴ ∆ABC ~ ∆FED

Thus, option (D) is correct.

Question 8.

If ∆ABC ~ ∆DEF and AB = 10 cm (RBSESolutions.com) then ar (∆ABC) will be ar (∆DEF).

(A) 25 : 16

(B) 16 : 25

(C) 4 : 5

(D) 5 : 4

Solution : ∆ABC ~ ∆DEF

\(\frac { ar.\triangle ABC }{ ar.\triangle DEF } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } }\)

= \({ \left( \frac { 10 }{ 8 } \right) }^{ 2 }\)

= \(\frac { 25 }{ 16 }\)

ar. (∆ABC) : ar.(∆DEF) = 25 : 16

Hence (A) is correct.

Question 9.

The point D and E lies on sides AB and AC of ∆ABC (RBSESolutions.com) such that DE || BC and AD = 8 cm, AB = 12 cm and AE = 12 cm, then CE will be :

(A) 6 cm

(B) 18 cm

(C) 9 cm

(D) 15 cm

Solution :

In ∆ABC

DE || BC

∴ \(\frac { AD }{ BD }\) = \(\frac { AE }{ CE }\) (Basic Prop. Theorem)

\(\frac { AD }{ AB-AD }\) = \(\frac { AE }{ CE }\)

⇒ \(\frac { 8 }{ 12-8 }\) = \(\frac { 12 }{ CE }\)

⇒ \(\frac { 8 }{ 4 }\) = \(\frac { 12 }{ CE }\)

⇒ 8 × CE = 12 × 4

⇒ CE = \(\frac { 12\times 4 }{ 8 } \)

⇒ CE = 6 cm

Thus (A) is correct option.

Question 10.

The shadow of a 12 m long vertical rod on (RBSESolutions.com) earth is 8 cm long. At the same time if shadow of tower in 40 m long, then height of pillar will be :

(A) 60 m

(B) 60 cm

(C) 40 cm

(D) 80 cm

Solution :

Let AB is vertical rod and BC is its shadow PQ is tower and CQ is its shadow.

In ∆ABC and ∆PQC

∠ACB = ∠PCQ (Common)

∠ABC = ∠PQC (each 90°)

By AA similarity criterion

∆ABC ~ ∆PQC

∴ \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QC }\) (∵ corresponding sides os (RBSESolutions.com) similar triangle are proportional)

⇒ \(\frac { 12 }{ PQ }\) = \(\frac { 8 }{ 40 }\)

∴ PQ = \(\frac { 40\times 12 }{ 8 } \)

= 5 × 12 = 60 m

Thus, (A) is correct.

Question 11.

If in ΔABCD, is any point on BC such that \(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\) and ∠B = 70°, ∠C = 50°, then find ∠BAD.

Solution :

In ∆ABC

∠B = 70°, ∠C = 50° (given)

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° – (∠B + ∠C)

⇒ ∠A = 180° – (70° + 50°)

⇒ ∠A = 180° – 120°

⇒ ∠A = 60°

Thus, \(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\) (given)

∴ ∆ABD ~ ∆ADC

⇒ ∠BAD and ∠DAC will be equal (RBSESolutions.com) to each other.

2∠BAD = 60°

∠BAD = \(\frac { { 60 }^{ \circ } }{ 2 } \)

∠BAD = 30°

Question 12.

In, ∆ABC DE || BC and AD = 6 cm, DE = 9 cm and AE = 8 cm, then find AC.

Solution :

In ∆ABC

DE || BC

∴ \(\frac { AD }{ DB }\) = \(\frac { AE }{ EC }\) (By Basic Theorem)

⇒ \(\frac { 6 }{ 9 }\) = \(\frac { 8 }{ CE }\)

⇒ 6 × CE = 9 × 8

⇒ CE = \(\frac { 8\times 9 }{ 6 } \)

⇒ CE = 12 cm

∴ AC = AE + CE

= (8 + 12) cm

= 20 cm

Thus, AC = 20 cm

Question 13.

In ∆ABC if AD of is (RBSESolutions.com) bisector of ∠A and AB = 8 cm, BD = 5 cm and DC = 4 cm. then find AC.

Solution :

In ∆ABC

AD, bisect vertex angle A

∴ ∠BAD = ∠CAD

∴ \(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\)

⇒ \(\frac { 8 }{ AC }\) = \(\frac { 5 }{ 4 }\)

⇒ 5 AC = 8 × 4

⇒ AC = \(\frac { 32 }{ 5 }\)

⇒ AC = 6.4 cm

Question 14.

If heights of two triangles are in the ratio 4 : 9, then find the ratio of their areas.

Solution :

We know that ratio of areas of two (RBSESolutions.com) similar triangles is equal to the ratio of square of their corresponding heights.

Let ∆ABC and ∆DEF are similar and AE and DN are their (RBSESolutions.com) corresponding heights.

∴ Ratio of the area of two triangle will be 16 : 81.

We hope the given RBSE Solutions for Class 10 Maths Chapter 11 Similarity Miscellaneous Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.