Figure below shows a kite in which $ B C D $ is the shape of a quadrant of a circle of radius $ 42 \mathrm{~cm} . A B C D $ is a square and $ \Delta C E F $ is an isosceles right angled triangle whose equal sides are $ 6 \mathrm{~cm} $ long. Find the area of the shaded region."

Given:

\( B C D \) is the shape of a quadrant of a circle of radius \( 42 \mathrm{~cm} \).

\( A B C D \) is a square and \( \Delta C E F \) is an isosceles right angled triangle whose equal sides are \( 6 \mathrm{~cm} \) long. 

To do: 

We have to find the area of the shaded region.

Solution:

Length of the side of the square $ABCD = 42\ cm$

$BCD$ is a quadrant in which $\angle BCD = 90^o$

Radius $= 42\ cm$

$\triangle CEF$ is an isosceles right-angled triangle in which $CE = CF = 6\ cm$

Therefore,

Area of the shaded region $=$ Area of the quadrant $\mathrm{BCD}+$ Area of $\triangle \mathrm{CEF}$

$=\frac{1}{4} \pi r^{2}+\frac{1}{2} \mathrm{CF} \times \mathrm{CE}$

$=\frac{1}{4} \times \frac{22}{7} \times (42)^2+\frac{1}{2} \times 6^2$

$=1386+18$

$=1404 \mathrm{~cm}^{2}$

The area of the shaded region is $1404\ cm^2$.

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Updated on: 10-Oct-2022

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