Question

In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region. [CBSE 2014]

Answer 1

We know that the opposite sides of a rectangle are equal
AD = BC = 70 cm
In right triangle AED
AE² = AD² − DE²
= (70)² − (42)²
= 4900 − 1764
= 3136
∴ AE² = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle AED + Area of semicircle)
$$\begin{gathered} =\mathrm{AB}\times \mathrm{BC}-\left[\frac{1}{2}\times \mathrm{AE}\times \mathrm{DE}+\frac{1}{2}\mathrm{\pi }{\left(\frac{\mathrm{BC}}{2}\right)}^2\right] \\ =80\times 70-\left[\frac{1}{2}\times 56\times 42+\frac{1}{2}\times \frac{22}{7}{\left(\frac{70}{2}\right)}^2\right] \\ =5600-3101 \\ =2499{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 2499 cm²