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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