Find the area of the shaded region in figure.

Explanation

Join GH and FE
Here, breadth of the rectangle BC = 12m
`:.` Breadth of the inner rectangle EFGH = `12 - (4 + 4 ) = 4cm`
which is equal to the diameter of the semi-circle EJF, d = 4m
`:.` Radius of semi-circle EJF , r= 2m
`:.` Length of inner rectangle EFGH = `26 - (5 + 5) = 16m`
`:.` Area of two semi-circles EJF and HIG = `2((pir^(2))/(2))= 2xxpi (2)^(2)/(2)= 4pim`
Now, area of inner rectangle EFGH = `EHxxFG= 16 xx 4= 64m^(2)`
and area of outer rectangle ABCD = `26 xx 12 = 312m^(2)`
`:.` Area of shaded region = Area of outer rectangle - (Area of two semi - circles + Area of inner rectangle)
= `312 - (64+4pi) = (248 = 4pi)m^(2)`