# R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles Exercise 7B

R S Aggarwal Solutions for class 10 Maths chapter 7 exercise 7B is available here. These solutions are prepared by subject experts at BYJU’S to help students clear their doubts. Students can freely download R S Aggarwal Class 10 chapter 7 from our website.

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### Access other exercise solutions of Class 10 Maths Chapter 7 Triangles

Exercise 7 A Solutions

Exercise 7 C Solutions

Exercise 7 D Solutions

Exercise 7 E Solutions

Question 1: In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form:

Solution:

Two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.

(i) In âˆ†ABC and âˆ†PQR

âˆ A = âˆ Q = 50Â°

âˆ B = âˆ P = 60Â° and

âˆ C = âˆ R = 70Â°

âˆ†ABC ~ âˆ†QPR (By AAA)

(ii) In âˆ†ABC and âˆ†DEF

In âˆ†ABC and âˆ†DEF

AB = 3 cm, BC = 4.5

DF = 6 cm, DE = 9 cm

âˆ†ABC is not similar to âˆ†DEF

(iii) In âˆ†ABC and âˆ†PQR

In âˆ†ABC and âˆ†PQR

AC = 8 cm BC = 6 cm

Included âˆ C = 80Â°

PQ = 4.5 cm, QR = 6 cm

and included âˆ Q = 80Â°

AC/QR = 8/6 = 4/3

and BC/PQ = 6/4.5 = 4/3

=> AC/QR = BC/PQ

and âˆ C = âˆ Q = 80Â°

âˆ†ABC ~ âˆ†PQR (By SAS)

(iv)In âˆ†DEF and âˆ†PQR

DE = 2.5, DF = 3 and EF = 2

PQ = 4, PR = 6 and QR = 5

DE/QR = 2.5/5 = 1/2

DF/PR = 3/6 = 1/2

and EF/PQ = 2/4 = 1/2

âˆ†DEF ~ âˆ†PQR (By SSS)

(v) In âˆ†ABC and âˆ†MNR

âˆ A = 80Â°, âˆ C = 70Â°

So, âˆ B = 180Â° â€“ (80Â° + 70Â°) = 30Â°

âˆ M = 80Â°, âˆ N = 30Â°, and âˆ R = 180Â° â€“ (80Â° + 30Â°) = 70Â°

Now, in âˆ†ABC

âˆ A = âˆ M â€“ 80Â°, âˆ B = âˆ N = 30Â°

and âˆ C = âˆ R = 70Â°

âˆ†ABC ~ âˆ†MNR (By AAA or AA)

Question 2: In the given figure, âˆ†ODC ~ âˆ†OBA, âˆ BOC = 115Â° and âˆ CDO = 70Â°. Find:

(i) âˆ DOC

(ii) âˆ DCO

(iii) âˆ OAB

(iv) âˆ OBA

Solution:

Here âˆ†ODC ~ âˆ†OBA, so

âˆ D = âˆ B = 70Â°

âˆ C = âˆ A

âˆ COD = âˆ AOB

(i) But âˆ DOC + âˆ BOC = 180Â° (Linear pair)

âˆ DOC + 115Â°= 180Â°

âˆ DOC = 180Â° â€“ 115Â° = 65Â°

(ii) âˆ DOC + âˆ CDO + âˆ DCO = 180Â° (Angles of a triangle)

65Â° + 70Â° + âˆ DCO = 180Â°

135Â° + âˆ DCO = 180Â°

âˆ DCO = 180Â° â€“ 135Â° = 45Â°

(iii) âˆ AOB = âˆ DOC = 65Â° (vertically opposite angles)

âˆ OAB = âˆ DCO = 45Â° (Since âˆ†ODC ~ âˆ†OBA)

(iv) âˆ OBA = âˆ CDO = 70Â° (Since âˆ†ODC ~ âˆ†OBA)

Question 3: In the given figure, âˆ†OAB ~ âˆ†OCD. If AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm, find (i) OA (ii) DO.

Solution:

Since âˆ†OAB ~ âˆ†OCD

AB = 8 cm, BO = 6.4 cm OC = 3.5 cm, CD = 5 cm

Let OD = y and OA = x

Question 4: In the given figure, if âˆ ADE = âˆ B, show that âˆ†ADE ~ âˆ†ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

Solution:

From given figure,

To prove:

find DE

Given: AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm

âˆ ADE = âˆ B (given)

âˆ A = âˆ A (common)

Again,

Question 5: The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm, find AB.

Solution:

Form given statement: âˆ†ABC ~ âˆ†PQR,

PQ = 12 cm

Perimeter of âˆ†ABC = AB + BC + CA = 32 cm

Perimeter of âˆ†PQR = PQ + QR + RP = 24 cm

Now,

Question 6: The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. If the perimeter of âˆ†DEF is 25 cm, find the perimeter of âˆ†ABC.

Solution:

Form given statement: âˆ†ABC ~ âˆ†DEF

BC = 9.1 cm, EF = 6.5 cm and Perimeter of âˆ†DEF = 25 cm

Perimeter of âˆ†ABC is 35 cm

Question 7: In the given figure, âˆ CAB = 90Â° and AD âŠ¥ BC. Show that âˆ†BDA ~ âˆ†BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

Solution:

âˆ CAB = 90Â° and AD âŠ¥ BC

If AC = 75 cm, AB = 1 m or 100 cm, BC = 1.25 m or 125 cm

âˆ BDA = âˆ BAC = 90Â°

âˆ DBA = âˆ CBA [common angles]

By AA

âˆ†BDA ~ âˆ†BAC

And,

Question 8: In the given figure, âˆ ABC = 90Â° and BD âŠ¥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution:

From given:

âˆ ABC = 90Â°, BD âŠ¥ AC.

AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm

In âˆ†ABC and âˆ†BDC,

âˆ ABC = âˆ BDC (each 90Â°)

âˆ BCA = âˆ BCD (common angles)

âˆ†ABC ~ âˆ†BDC (AA axiom)

So, corresponding sides are proportional

AB/BD = BC/CD

=> 5.7/3.8 = BC/5.4

=> BC = 8.1

## R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles Exercise 7B

Class 10 Maths Chapter 7 Triangles Exercise 7B is based on criteria for the similarity of two triangles

• AAA similarity
• AA similarity
• SSS similarity
• SAS similarity