In the following figure, from a rectangular region ABCD with AB =… - Vidyadip

Question

In the following figure, from a rectangular region ABCD with AB = 20cm, a right triangle AED with AE = 9cm and DE = 12cm, 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.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$

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Answer

In right triangle $AED$
$AD^2 = AE^2 + DE^2$
$= (9)^2 + (12)^2$
$= 81 + 144$
$= 225$
$\therefore\text{AD}^2=225$
$\Rightarrow\text{AD}=15\text{cm}$
We know that the opposite sides of a rectangle are equal
$AD = BC 15cm$
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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