A rectangle is overlapping the circle. The centre of the rectangle and circle are the same point. The two sides of the rectangle is tangent to the circle. The dimension of the rectangle are 6 by 12. The sides 6 are both tangent to the circle.
How do I find the area of the intersected part comprise of the circle and rectangle.
Try: The diagonal of the rectangle we can use Pythagoras theorem
Let d be the diagonal of the rectangle,
$d=\sqrt{6^2+12^2}=6\sqrt{5}$
So the radius, r, of the circle is $r=3\sqrt{5}$
I can only find parts of the areas. The rest it is quite difficult to tackle.
The radius of the circle is $6$.
Let $ABCD$ be the rectangle with $AD=BC=6$ and $AB=CD=12$. Let $O$ be their common centre.
Suppose that the circle cuts $AB$ at $E$ and $F$ such that $E$ is closer to $A$ than $B$. Also, suppose that the circle cuts $CD$ at $G$ and $H$ such that $H$ is closer to $D$ than $C$.
We have $OE=OH=EH=6$. So $\angle EOH=60^\circ$. The area of sector $OEH$ is $\pi (6^2)\div 6=6\pi$. The area of sector $OFG$ is the same.
The area of $\triangle OEF$ is $\frac{1}{2}(6)^2\sin 120^\circ=9\sqrt{3}$. So is the area of $OGH$.
The area of the required region is $2(6\pi+9\sqrt{3})$.