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In the figure below, $ A B C D $ is a trapezium with $ A B \| D C, A B=18 \mathrm{~cm}, D C=32 \mathrm{~cm} $ and the distance between $ A B $ and $ D C $ is $ 14 \mathrm{~cm} $. Circles of equal radii $ 7 \mathrm{~cm} $ with centres $ A, B, C $ and $ D $ have been drawn. Then, find the area of the shaded region of the figure. (Use $ \pi=22 / 7) $.
"
Given:
\( A B C D \) is a trapezium with \( A B \| D C, A B=18 \mathrm{~cm}, D C=32 \mathrm{~cm} \) and the distance between \( A B \) and \( D C \) is \( 14 \mathrm{~cm} \).
Circles of equal radii \( 7 \mathrm{~cm} \) with centres \( A, B, C \) and \( D \) have been drawn.
To do:
We have to find the area of the shaded region of the figure.
Solution:
Radius of each sector at the corner of the trapezium $= 7\ cm$
Sum of the angles of a quadrilateral $=360^{\circ}$
This implies,
The four sectors complete a circle.
Therefore,
Area of the circle $=\pi(7)^{2}$
$=\frac{22}{7} \times 49$
$=154 \mathrm{~cm}^{2}$
Area of the trapezium $=\frac{1}{2}(\mathrm{AB}+\mathrm{DC}) \times h$
$=\frac{1}{2}(18+32) \times 14$
$=\frac{1}{2} \times 50 \times 14$
$=350 \mathrm{~cm}^{2}$
Therefore,
Area of the shaded region $=$ Area of the trapezium $-$ Sum of the areas of the four sectors
$=350-154$
$=196 \mathrm{~cm}^{2}$
The area of the shaded region is $196\ cm^2$.
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