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In figure, a chord AB of a circle, with centre $O$ and radius $10 cm$ , that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. $(U s e π = 3.14$ )

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28.57 c m^{2}

27.88 c m^{2}

26.55 c m^{2}

None of these

answer is A.

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Detailed Solution

Given that, AB is a chord of the circle with center O and radius 10 cm which makes 90° angle at the center of the circle.
Question Image

We know that, area of circle of radius r units is equal to

π r 2

.
Area of sector which makes

θ

angle at the center of the circle of radius r units is equal to

π r 2 θ 360 °

.
Area of the triangle made by the chord of the circle which makes an angle

θ

at the center of the circle is equal to

12 r 2 sin θ

.
Substitute 10 for r and 90° for

θ

into

π r 2 θ 360 °

to find the area of the sector,

Area A O B P − π 102 × 90 ° 360 °

⇒ 100 π 4

⇒ 25 π

c m^{2}

Substitute 10 for r and 90° for

θ

into

12 r 2 sin θ

to find the area of triangle AOB.

A r e a A O B = 12102 sin 90 °

⇒ 12 × 100 × 1

⇒

c m^{2}

Subtract area of triangle AOB from area of sector AOBP to determine the area of minor segment AQBP .
Area

A Q B P = A r e a A O B P − A r e a A O B

⇒ 25 π − 50

⇒ 25 × 227 − 50

⇒ 5507 − 50

⇒

A r e a A Q B P = 78.57 − 50

⇒

28.57

c m^{2}

Therefore, the area of the minor segment AQBP of the circle is

28.57 c m 2

.
Hence the correct option is 1.

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