## RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

**Exercise 4C**

**Question 1.**

**Solution:**

Given : Area of ∆ABC = 64 cm²

and area of ∆DEF =121 cm²

EF = 15.4 cm

**Question 2.**

**Solution:**

**Question 3.**

**Solution:**

∆ABC ~ ∆PQR

ar (∆ABC) = 4ar (∆PQR),

**Question 4.**

**Solution:**

Areas of two similar triangles are 169 cm² and 121 cm²

Longest side of largest triangle = 26 cm

Let longest side of smallest triangle = x

∆s are similar

**Question 5.**

**Solution:**

Area of ∆ABC = 100 cm²

and area of ∆DEF = 49 cm²

**Question 6.**

**Solution:**

Given : Corresponding altitudes of two similar triangles are 6 cm and 9 cm

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)² : (9)² = 36 : 81 = 4 : 9 (Dividing by 9)

**Question 7.**

**Solution:**

The areas of two similar triangles are 81 cm² and 49 cm²

Altitude of the first triangle = 6.3 cm

Let altitude of second triangle = x cm

The areas of two similar triangles are in the ratio of the squares on their corresponding altitude,

Let area of ∆ABC = 81 cm²

and area of ∆DEF = 49cm²

Altitude AL = 6 – 3 cm

Let altitude DM = x cm

**Question 8.**

**Solution:**

Areas of two similar triangles are 100 cm² and 64 cm²

Let area of ∆ABC = 100 cm²

and area of ∆DEF = 64 cm²

Median DM of ∆DEF = 5.6 cm

Let median AL of ∆ABC = x

The areas of two similar triangles is proportional to the squares of their corresponding median.

**Question 9.**

**Solution:**

Given : In ∆ABC, PQ is a line which meets AB in P and AC in Q.

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

**Question 10.**

**Solution:**

In ∆ABC,

DE || BC

DE = 3 cm, BC = 6 cm

area (∆ADE) = 15 cm²

**Question 11.**

**Solution:**

Given : In right ∆ABC, ∠A = 90°

AD ⊥ BC

BC = 13 cm, AC = 5 cm

To find : Ratio in area of ∆ABC and ∆ADC

In ∆ABC and ∆ADC

∠C = ∠C (common)

∠BAC = ∠ADC (each 90°)

∆ABC ~ ∆ADC (AA axiom)

**Question 12.**

**Solution:**

In the given figure ∆ABC,

DE || BC and DE : BC = 3 : 5.

In ∆ABC and ∆ADE,

DE || BC

∆ABC ~ ∆ADE

**Question 13.**

**Solution:**

In ∆ABC, D and E are the midpoints of sides AB and AC respectively.

DE || BC and DE = \(\frac { 1 }{ 2 }\) BC

∆ADE ~ ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

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