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Q.

In figure, ABCD is a square of side $14 cm$ . Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.

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a

112 c m^{2}

b

98 c m^{2}

c

114 c m^{2}

d

118 c m^{2}

answer is A.

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Detailed Solution

Given that, side of the square is of length 14 cm and four semicircles are drawn at each side of the square.
In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. - Sarthaks eConnect | Largest Online Education Community

In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. - Sarthaks eConnect | Largest Online Education Community

We know that area of square of side X units is equal to X

2

and area of semicircle of radius r units is equal to

π r 2 2

.
Substitute 14 for X into X

2

and solve for the area of square.

A s = 142

⇒

196

c m^{2}

Divide 14 by 2 to determine the radius of each semicircle.

r = 142

⇒

7

c m^{2}

Area of all the semicircles is same. Substitute 7 for r into

π r 2 2

to determine the area of semicircle.

A s c = π 72 2

⇒ 49 π 2

c m^{2}

Let I be the upper un-shaded region, II be the lower un-shaded region, III be the left un-shaded region and IV be the right un-shaded region.
Subtract the area of left and right semicircles from the area of square to determine the sum of regions I and II .

A I + I I = A 5 − 2 × A S C

⇒ 196 − 2 × 49 π 2

⇒ 196 − 49 π

⇒ 49 4 − π

......(1)
Subtract the area of left and right semicircles from the area of square to determine the sum of regions I and IV .

A I I I + I V = A S − 2 × A S C

⇒ 196 − 2 × 49 π 2

⇒ 196 − 49 π

⇒ 49 4 − π

.............(2)
Let

A'

be the area of the unshaded region. Add equation (1) and (2) to determine the area of the unshaded region.

A' = A I + I I + A I I I + I V

⇒

49 4 − π + 49 4 − π

⇒

98 4 − π

Subtract the area of un-shaded region from the area of square to determine the area of the shaded region.

A = A s − A'

⇒ 196 − 98 4 − π

⇒ 98 2 − 4 + π

⇒ 98 π − 2

Substitute

227

for

π

in the obtained equation and simplify.

A = 98 π − 2

⇒ 98 227 − 2

⇒ 98 × 87

⇒

112

c m^{2}

Therefore, the area of the shaded region is 112

c m^{2}

.
Hence the correct option is 1.

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