# Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre

**Solution:**

Given, two segments of a circle are formed by a chord of length 5 cm subtending an angle of 90° at the centre.

We have to find the difference between the areas of two segments of the circle.

Let r be the radius of the circle.

AB is the chord which is equal to 5 cm.

The chord AB subtends an angle of 90° at the centre of the circle.

Considering triangle OAB,

AB² = OA² + OB²

(5)² = r² + r²

25 = 2r²

r² = 25/2

Taking square root,

r = 5/√2 cm

Area of the circle = πr²

= (22/7)(5/√2)²

= (22/7)(25/2)

= 39.285 cm²

Area of minor segment = area of sector - area of triangle

Area of triangle = (1/2) × base × height

Area of triangle OAB = (1/2) × OB × OA

= (1/2) × (5/√2) × (5/√2)

= 25/4

= 6.25 cm²

Area of sector = πr²θ/360°

= (22/7)(25/2)(90°/360°)

= (22/7)(25/2)(1/4)

= 9.82 cm²

Area of minor segment = 9.82 - 6.25

= 3.57 cm²

Area of major segment = area of circle - area of minor segment

= 39.285 - 3.57

= 35.71 cm²

Difference between two segments = area of major segment - area of minor segment

= 35.71 - 3.57

= 32.14 cm²

Therefore, the difference between two segments is 32.14 cm²

**✦ Try This: **Find the difference of the areas of two segments of a circle formed by a chord of length 6 cm subtending an angle of 60° at the centre.

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 12

**NCERT Exemplar Class 10 Maths Exercise 11.4 Problem 19**

## Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre

**Summary:**

The difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre is 32.14 cm²

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