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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.

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32.14 square cm

38.16 square cm

30.18 square cm

None of these

answer is A.

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Detailed Solution

Given that, circle formed by a chord of length 5 cm subtending an angle 90° at the centre.
We know that, the area of circle is

π r 2

A B 2 = O A 2 + O B 2

\Rightarrow

52 = r 2 + r 2

\Rightarrow

2 r 2 = 25

\Rightarrow

r = 52 cm

Calculating the area of

Δ O A B

A r Δ O A B = 12 × B × H

⇒ 12 × 52 × 52

⇒ 254 c m 2

Calculating the area of circle,

π r 2 = 227 × 52 × 52

⇒ 39.28 c m 2

Calculating the area of minor segment by subtracting the area of ΔOAB from the area of sector,

90 ° 360 ° × π r 2 − 254 ⇒ 14 × 39.28 − 254

⇒ 3.57 c m 2

To find the major segment, subtract the Area of minor segment from the Area of circle.

39.28 − 3.57 = 35.71

Instruction To find the required difference, subtract the Area of minor segment from the Area of major segment.

35.71 − 3.57 = 32.14

Therefore, difference of the area of two segments of circle is 32.14

{cm}^{2}

.
Hence the correct option is 1.

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