Question

Find the area of the shaded region in figure.

Answer 1

Join GH and FE such that EFGH is the rectangle.

Here, breadth of the rectangle ABCD = BC = 12 m

∴ Breadth of the inner rectangle EFGH = EF

= [12 – (4 + 4)] m

= 4 m

which is equal to the diameter of the semi-circle EJF = 4 m

 ∴ Radius of semi-circle EJF, (r) = 2 m

Length of inner rectangle EFGH = EH

= [26 – (5 + 5)] m

= 16 m

∴ Area of two semi-circles EJF and HIG

= `2((π"r"^2)/2)`

= `2 xx (π(2)^2)/2 "m"^2`

= 4π m²

Now, area of inner rectangle EFGH

= EH × EF

= (16 × 4) m²

= 64 m²

And area of outer rectangle ABCD

= (26 × 12) m²

= 312 m²

∴ Area of shaded region = Area of outer rectangle – (Area of two semi-circles + Area of inner rectangle)

= [312 – (4π + 64)] m²

= (248 – 4π) m²