Question
In the following figure, from a rectangular region $A B C D$ with $A B=20 cm$, a right triangle $A E D$ with $A E=9 cm$ and $D E$ $=12 cm$, 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. [Use $\pi=\frac{22}{7}$ ]

Answer

In right triangle $A E D$
$A D^2=A E^2+D E^2$
$=(9)^2+(12)^2$
$=81+144$
$=225$
$\therefore AD^2=225$
$\Rightarrow AD=15 cm$
We know that the opposite sides of a rectangle are equal
$A D=B C 15 cm$
Area of the shaded region $=$ Area of rectangle - Area of triangle $A E D+$ 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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