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## Download PDF of RS Aggarwal Solutions for Class 6 Chapter 16 Triangles.

### Access answers to chapter 16 – Triangles

**Exercise 16A PAGE NO: 196**

**1. Take three noncollinear points A, B, and C on a page of your notebook. Join AB, BC and CA. What figure do you get?**

**Name: (i) the side opposite to âˆ C**

** (ii) the angle opposite to the side BC**

** (iii) the vertex opposite to the side CA**

** (iv) the side opposite to the vertex B**

**Solution**

By joining three noncollinear points we get a triangle A, B, C

(i) The side opposite to âˆ C is AB

(ii) The angle opposite to the side BC is âˆ A

(iii) The vertex opposite to the side CA is B

(iv) The side opposite to the vertex B is AC

**2. The measures of two angles of a triangle are 72 ^{0} and 58^{0}. Find the measure of the third angle.**

**Solution**

Given the measure of two angles of a triangle are 72^{0} and 58^{0}

Let the third angle be x

Sum of the measures of all angles of a triangle = 180^{0}

âˆ´ x + 72^{0} + 58^{0} = 180^{0}

x + 130^{0} = 180^{0}

x = 180^{0} – 130^{0}

x = 50^{0}

Hence, the measure of third angle in a triangle is 50^{0}

**3. The angles of a triangle are in the ratio 1 : 3 : 5. Find the measure of each of the angles.**

**Solution**

Given the angle of a triangle are in the ratio 1: 3: 5

Let the measures of the angles of a triangle be 1x, 2x and 3x

We know that sum of the measures of all angles of a triangle is 180^{0}

^{ }âˆ´ 1x + 3x + 5x = 180^{0}

9x = 180^{0}

x = 180^{0} / 9

x = 20^{0}

1x = 1 Ã— 20^{0} = 20^{0}

3x = 3 Ã— 20^{0} = 60^{0}

5x = 5 Ã— 20^{0} = 100^{0}

âˆ´ The measures of the angles are 20^{0}, 60^{0} and 100

**4. One of the acute angles of a right triangle is 50 ^{0}. Find the other acute angle.**

**Solution**

We know that a triangle whose one angle measures 90^{0} is called a right angled triangle.

Given that one of the acute angle of a right triangle is 50^{0}

Let the third angle be x

We know that sum of the measures of all angles of a triangle is 180^{0}

âˆ´ 90^{0} + 50^{0} + x = 180^{0}

140^{0} + x = 180^{0}

x = 180^{0} – 140^{0}

x = 40^{0}

**5. One of the angles of a triangle is 110 ^{0} and the other two angles are equal. What is the measure of each of these equal angles?**

**Solution**

Given one of the angle of a triangle is 110^{0} and the other two angles are equal

We know that sum of the measures of all angles of a triangle is 180^{0}

âˆ´ âˆ A + âˆ B + âˆ C = 180^{0}

110^{0} + âˆ B + âˆ C = 180^{0}

110^{0} + âˆ B + âˆ C = 180^{0}

110^{0} + âˆ B + âˆ B = 180^{0}

110^{0} + 2âˆ B = 180^{0}

2âˆ B = 180^{0} – 110^{0}

2âˆ B = 70^{0}

âˆ B = 70^{0} / 2

âˆ B = 35^{0}

Hence, âˆ C = 35^{0}

âˆ´ The measure of each angles are

âˆ A = 110^{0} âˆ B = 35^{0} âˆ C = 35^{0}

**6. If one angle of a triangle is equal to the sum of other two, show that the triangle is a right triangle.**

**Solution**

Given âˆ A = âˆ B + âˆ C

We know that sum of the measures of all angles of a triangle is 180^{0}

** **âˆ´ âˆ A + âˆ B + âˆ C = 180^{0}

âˆ B + âˆ C + âˆ B + âˆ C = 180^{0}

2âˆ B + 2âˆ C = 180^{0}

2(âˆ B + âˆ C) = 180^{0}

âˆ B + âˆ C = 180^{0} / 2

âˆ B + âˆ C = 90^{0}

âˆ A = 90^{0}

âˆ´ This shows that the triangle is a right triangle.

**Exercise 16B PAGE NO: 197**

**OBJECTIVE QUESTIONS**

**Mark () against the correct answer in each of the following:**

**1. How many parts does a triangle have?**

**(a) 2**

**(b) 3**

**(c) 6**

**(d) 9**

**Solution**

A triangle has 6 parts, three sides and three angles

Option (c) is the correct answer.

**2. With the angles given below, in which case the construction of triangle is possible?**

**(a) 30 ^{0}, 60^{0}, 70^{0}**

**(b) 50 ^{0}, 70^{0}, 60^{0}**

**(c) 40 ^{0}, 80^{0}, 65^{0}**

**(d) 72 ^{0}, 28^{0}, 90^{0}**

**Solution**

We know that sum of the measures of all angles of a triangle is 180^{0}

(a) 30^{0} + 60^{0} + 70^{0} = 160^{0} (which is not equal to sum of angles of a triangle)

(b) 50^{0} + 70^{0} + 60^{0} = 180^{0} (which is equal to sum of angles of a triangle)

(c) 40^{0} + 80^{0} + 65^{0} = 185^{0} (which is not equal to sum of angles of a triangle)

(d) 72^{0} + 28^{0} + 90^{0} = 190^{0} (which is not equal to sum of angles of a triangle)

Option (b) is the correct answer

**3. The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is **

**(a) 60 ^{0}**

**(b) 80 ^{0}**

**(c) 76 ^{0}**

**(d) 84 ^{0}**

**Solution**

Given the angles of a triangle are in the ratio 2: 3: 4

Let the measure of the given angles be 2x, 3x and 4x

Sum of the measures of all angles of triangle is 180^{0}

âˆ´ 2x + 3x + 4x = 180^{0}

9x = 180^{0}

x = 180^{0} / 9

x = 20^{0}

2x = 2 Ã— 20^{0} = 40^{0}\

3x = 3 Ã— 20^{0} = 60^{0}

4x = 4 Ã— 20^{0} = 80^{0}

Hence, the largest angle is 80^{0}

Option (b) is the correct answer.

**4. The two angles of a triangle are complementary. The third angle is**

**(a) 60 ^{0}**

**(b) 45 ^{0}**

**(c) 36 ^{0}**

**(d) 90 ^{0=}**

**Solution**

Given two angles of a triangle are complementary if their sum is 90^{0}

Let the two angles be x and y such that (x + y) = 90^{0}

Let the third angle be z

Sum of the measures of all angles of triangle is 180^{0}

x + y + z = 180^{0}

90^{0} + z = 180^{0}

z = 180^{0} – 90^{0}

z = 90^{0}

Option (d) is the correct answer.

**5. One of the base angles of an isosceles triangle is 70 ^{0}. The vertical angle is**

**(a) 60 ^{0}**

**(b) 80 ^{0}**

**(c) 40 ^{0}**

**(d) 35 ^{0}**

**Solution**

Given the base angle of an isosceles triangle is 70^{0}

Let âˆ A = 70^{0}

Since the triangle is an isosceles triangle, we know that the angles opposite to the equal sides of an isosceles triangle are equal

âˆ´ âˆ B = 70^{0}

Let the third angle be C

Sum of the measures of all angles of triangle is 180^{0}

âˆ A + âˆ B + âˆ C = 180^{0}

70^{0} + 70^{0} + âˆ C = 180^{0}

140^{0} + âˆ C = 180^{0}

âˆ C = 180^{0} – 140^{0}

âˆ C = 40^{0}

Option (c) is the correct answer.

**6. A triangle having sides of different lengths is called**

**(a) an isosceles triangle**

**(b) an equilateral triangle**

**(c) a scalene triangle**

**(d) a right triangle**

**Solution**

A triangle having sides of different lengths is called a scalene triangle

Option (c) is the correct answer.

### RS Aggarwal Solutions for Class 6 Chapter 16 Triangles

Chapter 16 – Triangles consists of 2 exercises. Each and every question of RS Aggarwal textbook has been solved thoroughly. Letâ€™s have a glance at the topics included in Triangles.

- Congruent triangles
- Interior and exterior of a triangle
- Various types of triangles
- Naming triangles by considering their angles
- Angle sum property of a triangle

Also, access RS Aggarwal Solutions for Class 6 Chapter 16 Exercises

### Chapter Brief of RS Aggarwal Solutions for Class 6 Maths Chapter 16 – Triangles

A polygon with three line segments and three vertices is known as a Triangle. Types of triangles are equilateral, isosceles, scalene, acute and obtuse triangles. Traffic signs, roof, staircase and ladder, buildings, monuments, towers etc. are examples of triangles used in daily life.