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The Triangles and its Properties

    Two Special Triangles : Equilateral and Isosceles

    Exercise 6.3

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7th class > The Triangles and its Properties > Exercise 6.3

Exercise 6.3

1. Find the value of the unknown x in the following diagrams:

a

(i)

Solution:

The sum of the internal angles of a triangle is °.

In △ABC, ∠BAC + ∠ACB + ∠ABC = 180° ⇒ x + ° + ° = 180°

⇒ x + ° = °

Subtract 110° from both sides and simplify.

⇒ x = ° − ° = °

The value of x is 70°.

b

(ii)

Solution:

In △PQR, ∠QRP + ∠RPQ + ∠PQR = 180°

⇒ x + ° + ° = °

⇒ x + ° = 180°

Subtract 120° from both sides and simplify.

⇒ x = ° − ° = °

The value of x is 60°.

c

(iii)

Solution:

In △XYZ, ∠ZXY + ∠XYZ + ∠YZX = 180°

⇒ ° + ° + = °

⇒ x + ° = 180°

Subtract 140° from both sides and simplify.

⇒ x = 180° − 140° = °

The value of x is 40°.

d

(iv)

Solution:

In the given triangle, ° + x + x = °.

50° + = 180°

Subtract 50° from both sides and simplify.

⇒ 2x = 180° − 50° = °

Divide both sides by 2 and simplify.

⇒ x = 130°2 = °

The value of x is 65°.

e

(v)

Solution:

In the given triangle, x + x + x = °.

= 180°

Divide both sides by 3 and simplify.

⇒ x = 180°3 ⇒ x = °

The value of x is 60°.

f

(vi)

Solution:

In the given triangle, °+ x + = 180°.

⇒ 90° + = 180°

Subtract 90° from both sides and simplify.

⇒ 3x = 180° − 90° = °

Divide both sides by 3 and simplify.

⇒ x = 90°3 = °

The value of x is 30°.

2. Find the values of the unknowns x and y in the following diagrams:

a

(i)

Solution:

The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

x + ° = °

Subtract 50° from both sides of the equation.

⇒ x = 120° − 50° = °

The value of x is 70°.

The sum of the internal angles of a triangle is °.

In the given triangle, 50° + ° + = 180°.

⇒ ° + y = °

Subtract 120° from both sides and simplify.

⇒ y = 180° − 120° = °

The value of y is 60°.

b

(ii)

Solution:

Since the vertical opposite angles are equal, y = °.

The value of y is 80°.

In the given triangle, 50° + ° + = 180°.

⇒ ° + x = 180°

Subtract 130° from both sides and simplify.

⇒ x = 180° − 130° = °

The value of y is 50°.

c

(iii)

Solution:

The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

50° + ° =

Hence, x = °

The value of x is 110°.

In the given triangle, ° + 60° + = 180°.

Hence, ° + y = 180°

Subtract 110° from both sides and simplify.

y = 180° − 110° = °

The value of y is 70°.

d

(iv)

Solution:

Since the vertical opposite angles are equal, x = °

The value of x is 60°.

In the given triangle, ° + 30° + = 180°

⇒ ° + y = 180°.

Subtract 90 °from both sides and simplify.

⇒ y = 180° − 90° = °

The value of y is 90°.

e

(v)

Solution:

Since the vertical opposite angles are equal, y = °.

The value of y is 90°.

In the given triangle, x + + ° = 180°.

Hence, 90° + = 180°

Subtract 90° from both sides and simplify.

⇒ 2x = ° − °

⇒ 2x = 90°

Divide both sides by 2 and simplify.

⇒ x = 90°2 = °

The value of x is 45°.

f

(vi)

Solution:

Since the vertical opposite angles are equal, y = .

In the given triangle, x + + = 180°.

Hence, x = 180°

Divide both sides by 3 and simplify.

⇒ x = 180°3 = °

The value of x is 60° and the value of y is 60°.