Problem: To find area of shaded portion (AOB) in the below figure.
Description of Figure:
- ABCD is a square of side 14 cm
- AOCD and BODC are quadrants
Problem: To find area of shaded portion (AOB) in the below figure.
Description of Figure:
Let's observe picture below . The area of the shaded region is :
$A=14^2-\left(2\cdot \frac {14^2 \cdot \pi \cdot 30°}{360°}+\frac{14^2 \cdot \sqrt {3}}{4}\right)$
I would start by ignoring the circular arcs $CO$ and $DO$ and replacing them with straight lines from $C$ to $O$ and from $D$ to $O$. Then I'd calculate:
The area of the shaded region is then the first minus the second and the third.
You could ease computation even further by cutting the square in half vertically down the middle and calculating the area of the resulting half-shaded-area $-$ then your triangle is right-angled, and all you have to do at the end is multiply by $2$.