A square with sides of length $6\,\text{cm}$ is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown. The area of the shaded region is \(\underline{\hspace{1cm}}\) $\text{cm}^2$.
Show Hint
When two equal circles overlap, remember the lens formula
$A_{\cap}=2r^2\cos^{-1}\!\left(\tfrac{d}{2r}\right)-\tfrac{d}{2}\sqrt{4r^2-d^2}$.
For "two semi-circles on perpendicular sides", the centers are $d=r\sqrt{2}$ apart and the lens simplifies to $\frac{9\pi}{2}-9$ for $r=3$.
Setup.
Let the square be $[0,6]\times[0,6]$. The two semi-circles have radius $r=3$ with centers at $(0,3)$ (left side) and $(3,0)$ (bottom side). Their full circles are
\[
C_1:\ x^2+(y-3)^2=9,
C_2:\ (x-3)^2+y^2=9.
\]
The shaded region consists of the parts of these semi-circles excluding their common lens (the lens is white in the figure).
Step 1: Area of the two semi-circles.
Each semi-circle area $=\tfrac12\pi r^2=\tfrac12\pi(3^2)=\tfrac{9\pi}{2}$.
Sum of two semi-circles:
\[
A_{\text{semi-sum}}=\frac{9\pi}{2}+\frac{9\pi}{2}=9\pi.
\]
Step 2: Area of their overlap (circular lens).
Distance between centers:
\[
d=\sqrt{(3-0)^2+(0-3)^2}=3\sqrt{2}.
\]
For two equal circles of radius $r$ and separation $d$, the overlap area is
\[
A_{\cap}=2r^2\cos^{-1}\!\left(\frac{d}{2r}\right)-\frac{d}{2}\sqrt{4r^2-d^2}.
\]
Here $r=3,\ d=3\sqrt{2}\Rightarrow \frac{d}{2r}=\frac{\sqrt{2}}{2}$, so $\cos^{-1}(\sqrt{2}/2)=\frac{\pi}{4}$. Thus
\[
A_{\cap}=2(3^2)\left(\frac{\pi}{4}\right)-\frac{3\sqrt{2}}{2}\sqrt{36-18}
= \frac{18\pi}{4}-\frac{3\sqrt{2}}{2}\cdot 3\sqrt{2}
= \frac{9\pi}{2}-9.
\]
Step 3: Shaded area (union minus the lens twice).
The shaded part is the two semi-circles with the overlap removed from both, i.e.
\[
A_{\text{shaded}} = A_{\text{semi-sum}} - 2A_{\cap}
= 9\pi - 2\!\left(\frac{9\pi}{2}-9\right)
= 9\pi - 9\pi + 18
= \boxed{18\ \text{cm}^2}.
\]
