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 $196\ cm^2$.