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ABCP is a quadrant of a circle of radius $20 cm$ . With AC as diameter, a semicircle is drawn. Find the area of the shaded portion.

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510 c m^{2}

500 c m^{2}

514.29 c m^{2}

505 c m^{2}

answer is A.

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Detailed Solution

Given that, ABCP is a quadrant of a circle of radius 20 cm. with AC as diameter, a semicircle is drawn.
We know that, the area of sector is

θ 360 × π r 2

It is given that ABCP is a quadrant, so it makes 90°, to find the diameter of the semi-circle, use Pythagoras theorem in

Δ A B C

.
Calculating the side AC,

A C 2 = A B 2 − B C 2

\Rightarrow

A C 2

2 A B 2

\Rightarrow

A C 2 = 2 × 202

\Rightarrow

A C = 2 × 20

So the radius is

102

cm

.
The area shaded region is

A r A B C Q A − A r A B C P A

.
To find the area of ABCQA, calculating the area of

Δ A B C

and the area of semicircle ACQA and add them, we get,

A r (A B C Q A) = A r (Δ A B C) + A r (A C Q A)

⇒ 12 × b × h + π r 2

⇒ 12 × 20 × 20 + 227 × 102 × 102

\Rightarrow

200 + 628.57

\Rightarrow

828.57

c m^{2}

Calculating the area of (ABCPA),

A r A B C P A = 90 ° 360 ° × π r 2

⇒ 14 × 227 × 20 × 20

⇒ 314.28 c m 2

Calculating the area of shaded region by subtracting the area of region ABCPA from the area of the region ABCQA,

A r A B C Q A − A r A B C P A = 828.57 − 314.28

\Rightarrow

514.29

c m^{2}

Therefore, the area of the shaded region is

514.29 c m 2

. Hence the correct option is 1.

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