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

R S Aggarwal Solutions for class 10 Maths chapter 7 exercise 7C is available here. This set of important questions helps students to master triangles problems. This exercise is based on the ratio of the areas of two similar triangles. All questions are solved by subject experts at BYJU’S. Students can freely download R S Aggarwal Class 10 chapter 7 and practice more questions.

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Exercise 7 A Solutions

Exercise 7 B Solutions

Exercise 7 D Solutions

Exercise 7 E Solutions

Question 1: âˆ†ABC ~ âˆ†DEF and their areas are respectively 64 cmÂ² and 121 cmÂ². If EF = 15.4 cm, find BC.

Solution:

Area of âˆ†ABC = 64 cmÂ² and

area of âˆ†DEF = 121 cmÂ²

EF = 15.4 cm

Question 2: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Solution:

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16.

BC = 4.5 cm

Question 3: âˆ†ABC ~ âˆ†PQR and ar (âˆ†ABC) = 4ar (âˆ†PQR). If BC = 12 cm, find QR.

Solution:

âˆ†ABC ~ âˆ†PQR

ar (âˆ†ABC) = 4ar (âˆ†PQR)

Question 4: The areas of two similar triangles are 169 cmÂ² and 121 cmÂ² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution:

Areas of two similar triangles are 169 cmÂ² and 121 cmÂ² (given)

Longest side of largest triangle = 26 cm

Let longest side of smallest triangle is x cm

Longest side of smallest triangle is 22 cm

Question 5: âˆ†ABC ~ âˆ†DEF and their areas are respectively 100 cmÂ² and 49 cmÂ². If the altitude of âˆ†ABC is 5 cm, find the corresponding altitude of âˆ†DEF.

Solution:

Area of âˆ†ABC = 100 cmÂ²

area of âˆ†DEF = 49 cmÂ²

Altitude of âˆ†ABC is 5 cm

AL âŠ¥ BC and DM âŠ¥ EF

Let DM = x cm

Or x = 3.5

Altitude of smaller triangle is 3.5 cm

Question 6: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

Solution:

Corresponding altitudes of two similar triangles are 6 cm and 9 cm (given)

We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.

Ratio in the areas of two similar triangles = (6)Â² : (9)Â² = 36 : 81 = 4 : 9

Question 7: The areas of two similar triangles are 81 cmÂ² and 49 cmÂ² respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other.

Solution:

Areas of two similar triangles are 81 cmÂ² and 49 cmÂ²

Altitude of the first triangle = 6.3 cm

Let altitude of second triangle = x cm

Area of âˆ†ABC = 81 cmÂ² and area of âˆ†DEF = 49cmÂ²

Altitude AL = 6 â€“ 3 cm

Let altitude DM = x cm

Altitude of second triangle is 4.9 cm

Question 8: The areas of two similar triangles are 100 cmÂ² and 64 cmÂ² respectively. If a median of the smaller triangle is 5.6 cm, find the corresponding median of the other.

Solution:

Areas of two similar triangles are 100 cmÂ² and 64 cmÂ²

Median DM of âˆ†DEF = 5.6 cm

Let median AL of âˆ†ABC = x

Corresponding median of the other triangle is 7 cm.

Question 9: In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of âˆ†APQ is 1/16 of the area of âˆ†ABC.

Solution:

In âˆ†ABC, PQ is a line which meets AB in P and AC in Q.

Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm QC = 4.5 cm

Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3

=> AP/PB = AQ/QC

From figure: AB = AP + PB = 1+3 = 4 cm

AC = AQ + QC = 1.5 + 4.5 = 6 cm

In âˆ†APQ and âˆ†ABC,

AP/AB = AQ/AC

angle A (common)

âˆ†APQ and âˆ†ABC are similar triangles.

Now,

Which implies,

area of âˆ†APQ = 1/16 of the area of âˆ†ABC

Hence Proved.

Question 10: In the given figure, DE || BC. If DE = 3 cm, BC = 6 cm and ar (âˆ†ADE) = 15 cmÂ², find the area of âˆ†ABC.

Solution:

DE || BC

DE = 3 cm, BC = 6 cm