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 π = 3.14]

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

${\displaystyle =\text{AB}\times \text{BC}-\frac{1}{2}\times \text{AE}\times \text{DE}+\frac{1}{2}\pi {\left(\frac{\text{BC}}{2}\right)}^2}$

${\displaystyle =20\times 15-\frac{1}{2}\times 9\times 12+\frac{1}{2}\times 3.14\left(\frac{15}{2}\right)}$^2

=  300 -  54   + 88.31

= 334. 31 cm²

Hence, the area of shaded region is 334.31 cm²