## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 18 of Formative Assessment (Unit Test)

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18
##### A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 22/7 and √3 = 1.732.]

ACB = 60°

Chord AB subtends an angle of 60° at the centre

In triangle ABC, AC = BC

So, CAB = CBA

ACB + CAB + CBA = 180°

60° + 2CAB = 180°

2CAB = 180° - 60° = 120°

CAB = 120°/2 = 60°

CAB = CBA = 60°

ΔABC is a equilateral triangle

Length of side of an equilateral triangle = radius of circle = 30 cm

Area of equilateral triangle = √3/4 × side2 = 1.732/4 × 30 × 30 cm2

= 389.7 cm2

Area of sector ACB = = 3.14 × 30 × 30 × 60°/360° = 471.45 cm2

Area of minor Segment = Area of sector ACB – Area of ΔABC

= 471.45 cm2 – 389.7 cm2 = 81.75 cm2

Area of circle = R2 = 3.14 × 30 × 30 cm2 = 2828.57 cm2

Area of major segment = Area of circle – Area of minor segment

= 2826 cm2 – 81.75 cm2

= 2744.25 cm2

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