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. [Use π =$\frac{22}{7}$] [CBSE 2014]

Answer 1

In right triangle AED
AD² = AE² + DE²
= (9)² + (12)²
= 81 + 144
= 225
∴ AD² = 225
⇒ AD = 15 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
$$\begin{gathered} =\mathrm{AB}\times \mathrm{BC}-\frac{1}{2}\times \mathrm{AE}\times \mathrm{DE}+\frac{1}{2}\mathrm{\pi }{\left(\frac{\mathrm{BC}}{2}\right)}^2 \\ =20\times 15-\frac{1}{2}\times 9\times 12+\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{15}{2}\right)}^2 \\ =300-54+88.3928 \\ =334.3928{\mathrm{cm}}^2 \end{gathered}$$
=AB×BC−12×AE×DE+12π(BC2)2=20×15−12×9×12+12×3.14(152)2=300−54+88.31=334.31 cm2
Hence, the area of shaded region is 334.39 cm²