## RS Aggarwal Class 6 Solutions Chapter 16 Triangles Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16A
- RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B

**Objective questions**

**Mark against the correct answer in each of following.**

**Question 1.**

**Solution:**

(c) ∵ It has three sides and three angles i.e. six.

**Question 2.**

**Solution:**

(b) ∵ Sum of three angles of a triangle is 180°.

**Question 3.**

**Solution:**

(b) ∵ Largest angle

\(=\frac { { 180 }^{ O }\times 4 }{ 2+3+4 } =\frac { { 180 }^{ O }\times 4 }{ 9 } \)

= 80°

**Question 4.**

**Solution:**

(d) ∵ A triangle has 180° and if two angles are complementary i.e. sum of two angles is 90°, then third angle will be 180° – 90° = 90°.

**Question 5.**

**Solution:**

(c) ∵ Sum of three angles is 180° and sum of two equal angles = 70° + 70° = 140°, then third angle will be 180°- 140° = 40°.

**Question 6.**

**Solution:**

(c) ∵ A scalene triangle has different sides.

**Question 7.**

**Solution:**

In an isosceles ∆ABC, ∠B = ∠C and bisector of ∠B and ∠C meet at O and ∠A = 40°

**Question 8.**

**Solution:**

Side of a triangle are in the ratio 3:2:5 and perimeter = 30 m

Length of longest side = \(\frac { 30\times 5 }{ 3+2+5 } \)

= \(\frac { 30\times 5 }{ 10 } \) cm

= 15 cm (b)

**Question 9.**

**Solution:**

Two angles of a triangle are 30° and 25° But sum of three angles of a triangle – 180°

Third angle = 180° – (30 + 25°)

= 180° – 55° = 125° (d)

**Question 10.**

**Solution:**

Each angles of an equilateral triangle = 60°

as each angle of an equilateral triangle are equal

Each angle = \(\\ \frac { 180 }{ 3 } \) = 60° (c)

**Question 11.**

**Solution:**

In the figure, P lies on AB

Its lies on the ∆ABC (c)

Hope given RS Aggarwal Solutions Class 6 Chapter 16 Triangles Ex 16B are helpful to complete your math homework.

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