Find the area of the shaded region in figure.

#### Solution

Label the figure as above.

Area of the shaded region = Area of the rectangle ABCD – (Area of the rect. EFGH + (Area of the semi-circle EFJ + Area of the semi-circle GHI)

Length and breadth of outer rect. ABCD are 26 m and 12 m respectively.

∴ Area of the rect. ABCD = Length × Breadth

= AB × BC

= 26 × 12

= 312 m^{2}

From the figure, length and breadth of inner rect. EFGH are (26 - 5 - 5) m and (12 - 4 - 4) m, i.e. 16 m and 4 m respectively.

∴ Area of the rect. EFGH = Length × Breadth

= EF × FG

= 16 × 4

= 64 m^{2}

Breadth of the inner rectangle = Diameter of the semi-circle EJF = d = 4m

∴ Radius of semi-circle EJF = r = 2 m

Area of the semi-circle EFJ = Area of the semi-circle GHI

= `(pir^2)/2`

= `(4π)/2` = 2π m^{2}

∴ Area of shaded region = 312 – (64 + 2π + 2π)m^{2}

= ( 248 – 4π) m^{2}