Rhombus cake with the inscribed circle removed

Text

A cake has the shape of a rhombus, with interior angles of $60^\circ$ and $120^\circ$. A waiter draws the circle inscribed in the rhombus, of radius $3$, and removes it from the cake. Compute the area of the remaining part of the cake.

Solution

The radius of the circle inscribed in a rhombus is half the height: $r=\dfrac{h}{2}$, so $h=2r=6$. With side $\ell$, the height is $h=\ell\sin60^\circ$: $\ell\cdot\frac{\sqrt3}{2}=6\ \Longrightarrow\ \ell=4\sqrt3.$ Area of the rhombus (base times height): $\mathcal A_\text{rhombus}=\ell\cdot h=4\sqrt3\cdot6=24\sqrt3\approx41{,}57.$ Area of the circle: $9\pi\approx28{,}27$. The remaining part is $\mathcal A=24\sqrt3-9\pi\approx13{,}30.$ $\boxed{\mathcal A=24\sqrt3-9\pi\approx13{,}30.}$