Question

In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, $△AED={90}^{\circ }$ and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.

Answer 1

ABCD is a rectangle with AB = 80 cm and BC = 70 cm.

Area of the rectangle = l $\times$ b = 80 $\times$ 70 = 5600 $c{m}^2$.

In rectangle AB = CD = 80 cm and BC = AD = 70 cm

Given AED is a right angled triangle
and
DE = 42 cm

then
$A{D}^2=A{E}^2+E{D}^2$
${70}^2=A{E}^2+{42}^2$.
$A{E}^2=4900-1764=3136$
AE = 56 cm.
Area of a triangle AED $=\left(\frac{1}{2}\right)\times AE\times DE=\left(\frac{1}{2}\right)\times 56\times 42=1176c{m}^2$

Given BC is a diameter of the semi circle. BC = 70 cm radius of the semi circle = 35 cm.

Area of the semi circle $=\left(\frac{1}{2}\right)\pi r^2=\left(\frac{1}{2}\right)\times \frac{22}{7}\times 35\times 35=1925c{m}^2$.

Area of the shaded region = Area of the rectangle - (Area of a triangle AED + Area of the semi-circle)

Area of the shaded region $=5600c{m}^2-(1176c{m}^2+1925c{m}^2)=2499c{m}^2$.