Question
In the following figure, from a rectangular region $ABCD$ with $AB = 20\ cm$, a right triangle $AED$ with $AE = 9\ cm$ and $DE = 12\ cm$, is cut off. On the other end, taking $BC$ as diameter, a semicircle is added on outside' the region. Find the area of the shaded region.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

In right triangle $AED$
$ A D^2=A E^2+D E^2 $
$ =(9)^2+(12)^2 $
$= 81 + 144$
$= 225$
$\therefore\text{AD}^2=225$
$\Rightarrow\text{AD}=15\text{cm}$
We know that the opposite sides of a rectangle are equal
$AD = BC\  15\ cm$
Area of the shaded region = Area of rectangle − Area of triangle $AED$ + Area of semicircle
$=\text{AB}\times\text{BC}-\frac{1}{2}\times\text{AE}\times\text{DE}+\frac{1}{2}\pi\Big(\frac{\text{BC}}{2}\Big)^2$
$=20\times15-\frac{1}{2}\times9\times12+\frac{1}{2}\times\frac{22}{7}\times\big(\frac{15}{2}\big)^2$
$=300-54+88.3982$
$=334.3928\text{cm}^2$
Hence, the area of shaded region is $334.39 \text{cm}^2$

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