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

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Hint: forget the half area at the right, consider then the area of the half rectangle minus the area of the truncated circle. –  Raymond Manzoni Jan 15 '12 at 14:57

Let's observe picture below . The area of the shaded region is :

$A=14^2-\left(2\cdot \frac {14^2 \cdot \pi \cdot 30°}{360°}+\frac{14^2 \cdot \sqrt {3}}{4}\right)$

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thank u so much..... its very easy and well explained as well.. –  Javed Akram Jan 17 '12 at 3:30

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 square
• The area of the circular segments $ADO$ and $BCO$
• The area of the triangle $DOC$

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 $2$.

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