Question

In the given 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 $\pi$ = 3.14]

Answer 1

AED is a right angle triangle

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

⇒ $A{D}^2=(9^2+{12}^2)$

= $(81+144)$

= $225c{m}^2$

⇒ AD = 15 cm

Area of the rectangular region ABCD

= $AB\times \times AD=(20\times 15)=300c{m}^2$

Area of ∆AED

= $\frac{1}{2}\times AE\times DE=(\frac{1}{2}\times 9\times 12)c{m}^2=54c{m}^2$

In a rectangle

AD = BC = 15 cm

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

Area of the semi-circle $=\frac{1}{2}\times \pi \times r^2$

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

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

$=(300+88.3125-54)=334.3125c{m}^2$ .