Solution:- In the given figure
Side of square= 10 cm
area of square = side x side m= 10 x 10 = 100 sq cm
Now , diameter of semi circle AOD = 2r = 10cm
⇒ radius of semi circle AOD = 5 cm
∴ area of AOD = [tex]\frac{\pi\ r^2}{2}=\frac{\pi(5)^2}{2} =\pi \frac{25}{2}=12.5\pi cm^2[/tex]
similarly Area of semicircle BOC = [tex]12.5\pi cm^2[/tex]
Area of white portion in figure 1 = area of square - [ area of semicircle BOD + area of semicircle AOD]
=[tex]=100-2(12.5\pi) =100-25\pi[/tex]
Total area of white portion in the given figure=[tex]100-25\pi[/tex]
∴Area of Shaded portion = area of square - area of total white portion
= [tex]100-(200-50\pi )=100-200+50\pi =50\pi -100\ cm^2[/tex]
So the area of shaded portion is [tex]50\pi -100\ cm^2[/tex]
Now perimeter of AD(which is a semicircle) = [tex]\pi\ r=3.14\cdot\ 5=15.7cm[/tex](as perimeter of semicircle=πr)
So the perimeter of shaded region = 4×perimeter of AD=4×15.7=62.8cm
