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In the given figure, $ABC$ is a quadrant of a circle of radius $14 cm$ and a semicircle is drawn with $BC$ as diameter. What is the area of the shaded region?

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{98 cm}^{2}

{38 cm}^{2}

{48 cm}^{2}

{58 cm}^{2}

answer is A.

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Detailed Solution

Given,

∵

BC is hypotenuse of right angle

△ ABC

here

AB = AC = 14

.
Pythagoras theorem states that the square of the hypotenuse side in a right-angled triangle equals the sum of the squares of the other two sides.
According to Pythagoras' theorem

BC = \sqrt{1 4^{2} + 1 4^{2}}

\Rightarrow BC = 14 \sqrt{2} = 2 \times radius

\Rightarrow radius = 7 \sqrt{2}

So, the area of semicircle of diameter BC

= \frac{π r^{2}}{2}

\Rightarrow

The area of semicircle of diameter BC

= \frac{1}{2} \times \frac{22}{7} \times (7 \sqrt{2})^{2}

\Rightarrow

The area of semicircle of diameter BC

= {154 cm}^{2}

Area of quadrant of radius

\frac{AB}{AC}

= \frac{π r^{2}}{4}

\Rightarrow

Area of quadrant

= \frac{1}{4} \times \frac{22}{7} \times 14 \times 14

\Rightarrow

Area of quadrant

= {154 cm}^{2}

Area of

△ ABC = \frac{1}{2} \times h \times b

\Rightarrow

Area of

△ ABC

= \frac{1}{2} \times 14 \times 14 = {98 cm}^{2}

Area of the shaded region = area of semicircle of diameter BC - {Area of the quadrant - Area of

△ ABC

\Rightarrow

Area of shaded region

= 154 - {154 - 98}

\Rightarrow

Area of shaded region

= {98 cm}^{2}

So, option 1 is correct.

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