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 = ![\frac{\pi\ r^2}{2}=\frac{\pi(5)^2}{2} =\pi \frac{25}{2}=12.5\pi cm^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%5C%20r%5E2%7D%7B2%7D%3D%5Cfrac%7B%5Cpi%285%29%5E2%7D%7B2%7D%20%3D%5Cpi%20%5Cfrac%7B25%7D%7B2%7D%3D12.5%5Cpi%20cm%5E2)
similarly Area of semicircle BOC =
Area of white portion in figure 1 = area of square - [ area of semicircle BOD + area of semicircle AOD]
=![=100-2(12.5\pi) =100-25\pi](https://tex.z-dn.net/?f=%3D100-2%2812.5%5Cpi%29%20%3D100-25%5Cpi)
Total area of white portion in the given figure=![100-25\pi](https://tex.z-dn.net/?f=100-25%5Cpi)
∴Area of Shaded portion = area of square - area of total white portion
= ![100-(200-50\pi )=100-200+50\pi =50\pi -100\ cm^2](https://tex.z-dn.net/?f=100-%28200-50%5Cpi%20%29%3D100-200%2B50%5Cpi%20%3D50%5Cpi%20-100%5C%20cm%5E2)
So the area of shaded portion is ![50\pi -100\ cm^2](https://tex.z-dn.net/?f=50%5Cpi%20-100%5C%20cm%5E2)
Now perimeter of AD(which is a semicircle) =
(as perimeter of semicircle=πr)
So the perimeter of shaded region = 4×perimeter of AD=4×15.7=62.8cm