RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Exercise 11.3.

## Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Ex 11.3

Question 1.

In two triangles ABC and PQR \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\). For similarity of two triangles which angles should be equal, (RBSESolutions.com) name them and also given reason for your answer.

Solution :

In ∆ABC and ∆PQR,

\(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\)

⇒ \(\frac { AB }{ BC }\) = \(\frac { PQ }{ QR }\)

Angle formed by sides AB and BC is ∠B

Angle formed by sides PQ and QR is ∠Q

∴ ∠B = ∠Q is essential for similarity of two triangles.

thus ∠B = ∠Q

\(\frac { AB }{ BC }\) = \(\frac { PQ }{ QR }\)

∴ ∆ABC ~ ∆PQR (SAS similarity theorem)

Question 2.

In ∆ABC and ∆DEF; ∠A = ∠D, ∠B = ∠F. Is ∆ABC ~ ∆DEF? Given (RBSESolutions.com) reason for your answer.

Solution :

No, because when ∠A = ∠D, ∠B = ∠F, then ∆ABC ~ ∆DEF.

Question 3.

If ∆ABC ~ ∆FDE then is it possible?

\(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = \(\frac { CA }{ FD }\) Give reason for your answer.

Solution :

∆ABC ~ ∆FDE (Given)

∴ ∠A = ∠F, ∠B = ∠D and ∠C = ∠E

Side AB ↔ FD, Side BC ↔ DE, side AC ↔ FE

∴ \(\frac { AB }{ FD }\) = \(\frac { BC }{ DE }\) = \(\frac { AC }{ FE }\)

Thus \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = \(\frac { CA }{ FD }\)

is not possible.

Question 4.

If two sides and one angle of a triangle are (RBSESolutions.com) proportional and equal to two sides and one angle of other triangle respectively then two triangle are similar. Is this true ? Give answer with reason.

Solution :

This statement is false because for similarity of triangles two sides and angle included them should be equal.

Question 5.

What do you mean by equiangular triangles ? What relation can be between them?

Solution :

If corresponding angles of two equiangular triangles are same, then both triangles will be similar.

In ∆ABC and ∆DEF

∠A = ∠D

∠B = ∠E

and ∠C = ∠F

∴ ∆ABC ~ ∆DEF

Question 6.

From the following figures of triangles, (RBSESolutions.com) select the similar pair and write in notation form of similarity.

Solution :

(i) In ∆ABC and ∆PQR

∠A = ∠Q = 40°

∠B = ∠P = 60°

∠C = ∠R = 80°

∴ ∆ABC ~ ∆QPR

Thus figure (i) and (viii) will be similar.

(ii) In ∆MNP and ∆XYZ

∠M = ∠Z = 70°

∠P = ∠Y = 30°

∴ ∆MNP ~ ∆XYZ

Thus, figure (ii) and (vii) will be similar.

(iii) In ∆PQR and ∆EFG

∴ ∆PQR and ∆EFG are similar

The figure (iii) and (v) will be similar.

(iv) ∆EFG and ∆LMN

∴ ∆MLN ~ ∆EFG are similar

Hence figure (iv) and (vi) are similar.

Question 7.

In figure, ∆PQR ~ ∆TRS then find in this (RBSESolutions.com) similar pair which angles should be equal ?

Solution :

∆PQR ~ ∆TRS

∴ ∠R = ∠R (common angle)

∠P = ∠RTS (corresponding angle)

∠Q = ∠RST (corresponding angle)

Thus, above angle should be same for similar triangle pair.

Question 8.

You have to select two triangle in figures (RBSESolutions.com) which are similar to each other. If ∠CBE = ∠CAD.

Solution :

In ∆CAD and ∆CBE

∠ACD = ∠BCE (common angle)

∠CAD = ∠CBE (given)

∴ ∆CAD ~ ∆CBE (AA similarity law)

Hence, ∆ADC ~ ∆BEC

Question 9.

In fig. PQ and RS are parallel (RBSESolutions.com) then prove that : ∆POQ ~ ∆SOR.

Solution :

In ∆POQ and ∆SOR

PQ || RS (given)

∠OPQ = ∠OSR (Alternate angle)

∠POQ = ∠ROS (Vertically opposite angles)

and ∠OQP = ∠ORS (Alternate angles)

By AAA similarity law

∴ ∆POQ ~ ∆SOR

Question 10.

A girl with height 90 cm walks (RBSESolutions.com) away from the base of lamp-post at a speed of 1.2 m/sec. lithe lamp is 3.6 m above the ground. Find the length of her shadow after 4 seconds.

Solution :

Let AB = Lamp post

CD = girl

DE = shadow of the girl,

BD = 1.2 × 4 = 4.8 m

Let shadow of girl,

DE = ‘a’ m

In ∆ABE and ∆CDE

∠B = ∆(each 90°)

∠E = ∠E (comman angle)

∆ABE ~ ∆CDE by (AA criterion)

therefore, \(\frac { BE }{ DE }\) = \(\frac { AB }{ CD }\) ⇒ \(\frac { 4.8+a }{ a }\) = \(\frac { 3.6 }{ 0.9 }\)

⇒ 4.8 + a = 4a

⇒ 3 a = 4.8

⇒ a = 1.6 m

Thus after 4 sec. length of girl’s shadow will be 1.6 m

Question 11.

The shadow of a vertical pillar of (RBSESolutions.com) height 12 m is 8 m. At the same time length of shadow 12 m is 8 m. At the same time length of shadow of tower is 56 m then find height of tower.

Solution :

In the figure Length of a vertical pillar AB = 12 m

Length of shadow BC = 8 m

At the same time QC is shadow of other tower PQ.

Here, QC = 56 m

Now, In ∆ABC and ∆PQC

∠B = ∠Q = 90°

∠C = ∠C (Common angle)

∴ ∆ABC ~ ∆PQC

By similarity rule

\(\frac { AB }{ PQ }\) = \(\frac { BC }{ QC }\)

\(\frac { 12 }{ PQ }\) = \(\frac { 8 }{ 56 }\)

PQ = \(\frac { 56\times 12 }{ 8 } \) = 84 m

Thus, height of tower PQ = 84 m

Question 12.

From vertex A of ∆ABC a perpendicular is (RBSESolutions.com) drawn on opposite side BC and get AD^{2} = BD × DC, then show that ABC is a right angled triangle.

Solution :

Given : Perpendicular from vertex A of ∆ABC to side BC.

AD ⊥ BC

AD^{2} = BD × DC

Question 13.

By joining mid-point of three sides of a (RBSESolutions.com) triangle, four triangle so formed are similar to original triangle.

Solution :

Given : D, E, F are mid-point of sides AB, BC and BC of ΔABC.

To prove: ∆ADE ~ ∆DBF ~ ∆EFC ~ ∆DEF ~ ∆ABC

Four triangle formed in figure are similar to original ∆ABC

Proof : Points D and E are mid-point of sides AB and AC.

∴ DE = \(\frac { 1 }{ 2 }\) BC and DE || BC (by mid-point theorem) …(i)

E and F are mid-point of AC and BC sides.

∴ EF = \(\frac { 1 }{ 2 }\) AB and EF || AB …(ii)

Point F and D are mid-point of sector BC and AB.

∴ FD = \(\frac { 1 }{ 2 }\) AC and FD || AC …(iii)

From eq^{n} (i) and (ii)

DE || BC and EF || AB

∴ DEFB is a parallelogram

Diagonal of ||gm DEFB, DE divides into (RBSESolutions.com) two congruent triangles.

∴ ∆DBF = ∆DEF …(iv)

Similarly, diagonal EF, divides the ||gm DECF into two congrwnent triangles.

∴ ∆DEF = ∆DAE …(v)

Simlarly, diagonal EF, divides the ||gm DECF into two congruemt triangles.

∴ ∆DEF = ∆EFC …(iv)

From (iv), (v) and (vi)

∴ ∆DEF = ∆DAE = ∆EFC = ∆DBF

∵ DE || BC

∠C = ∠E

∴ ∠B = ∠D (corresponding angle)

⇒ ∆ABC ~ ∆ADE

Thus, ∆ABC ~ ∆ADE ~ ∆BDF ~ ∆EFC

Question 14.

According to following figure, If AB ⊥ BC, DC ⊥ BC and DE ⊥ AC then (RBSESolutions.com) prove that ∆CED ~ ∆ABC

Solution :

Given

AB ⊥ BC, DC ⊥ BC and DE ⊥ AC

To prove : ∆CED ~ ∆ABC

Proof : AB ⊥ BC and DC ⊥ BC

∴ AB || DC

When AB || DC cuts by transversal AC

∠BAC = ∠ACD

or ∠BAC = ∠ECD

∠B = ∠E = 90° (given) …(ii)

∴ ∠ACB = ∠EDC [due to (i) and (ii)]

⇒ ∆ABC ~ ∆CED.

Question 15.

D is mid-point of side BC of ∆ABC. A line is drawn (RBSESolutions.com) through B and bisects AD at point E and A at C, then prove that \(\frac { EX }{ BE }\) = \(\frac { 1 }{ 3 }\)

Solution :

Given : D is mid-point of side BC of ∆ABC.

BD = DC

and AE = ED

To prove:

\(\frac { EX }{ BE }\) = \(\frac { 1 }{ 3 }\)

Construction : Draw DY || BX meeting AC at Y

Proof : In ∆BCX and ∆DCY

∠BCX = ∠DCY (common)

∠BXC = ∠DYC (corresponding angle)

So, ∆BCX ~ ∆DCY (AA similarity)

We hope the given RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Exercise 11.3, drop a comment below and we will get back to you at the earliest.