Question

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 sshaded region. (Use $\pi =\frac{22}{7}$)

Answer 1

AED is an right angle triangle

$A{D}^2=A{E}^2+E{D}^2$

⇒ $A{D}^2=(92+122)$

= (81 + 144)

= 225 $c{m}^2$

⇒ AD = 15 cm

Area of the rectangular region ABCD

= AB $\times$ AD

= (20 $\times$ 15)

= 300 $c{m}^2$

Area of ∆AED

= 12$\times$AE $\times$ DE = (12$\times$ 9 $\times$12) cm2 = 54 $c{m}^2$

In a rectangle

AD = BC = 15 cm

Since, BC is the diameter of the circle, radius of the circle = 15 cm

Area of the semi-circle = $\frac{\pi r^2}{2}$

= ($\frac{1}{2}$ × 3.14 $\times$ 15$\times$ 15) = 88.3125$c{m}^2$

Area of the shaded region = Area of the rectangle + Area of the semi-circle − Area of the triangle

= (300 + 88.3125 − 54)$c{m}^2$

= 334.3125 $c{m}^2$