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 m2
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 m2
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 m2
∴ Area of shaded region = 312 – (64 + 2π + 2π)m2
= ( 248 – 4π) m2