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A chord of a circle of radius $30 c m$ makes an angle of $600$ at the center of the circle.

The area of the major segment is ____ $c m 2$ . $π = 3.14 and 3 = 1.732$

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Detailed Solution

A chord of a circle of radius

30 c m

makes an angle of

600

at the center of the circle.
The area of the major segment is 2744.7

c m 2

.
Given:

r = 30 c m

and

θ = 600

Consider the following figure-
Question Image

Area of triangle with angle

θ = 12 (side) 2 s i n θ

where,

θ

is the sector angle.
Here,

∠ A O B = 600

, area of sector

O A C B O

is-

= 60360 × π r 2

= 16 × 3.14 × 30 × 30

= 471 c m 2

Area of

Δ A O B

is-

= 12 r 2 s i n θ

= 12 × 30 × 30 × s i n 60 °

= 225 × 1.732

= 389.7 c m 2

Area of sector of a circle having an angle

θ

= θ 360 π r 2

where,

r

is the radius of the circle and

θ

is the sector angle.
Area of the minor segment is-

= A r e a o f t h e s e c t o r O A C B O − (A r e a o f Δ A O B)

= 471 − 389.7

= 81.3 c m 2

Now, area of the major segment is-

= A r e a o f t h e c i r c l e − A r e a o f t h e m i n o r s e g m e n t

= 3.14 × 30 × 30 − 81.3

= 2744.7 c m 2

Hence,

2744.7 c m 2

is the required area.

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