Annulus between the incircle and circumcircle of a square

Text

A square has side $10$ cm. Compute the area of the annulus between the inscribed and circumscribed circles of the square.

Solution

For a square of side $\ell$: inradius $r=\dfrac{\ell}{2}=5$, circumradius $R=\dfrac{\ell\sqrt2}{2}=5\sqrt2$. $\mathcal A=\pi(R^2-r^2)=\pi(50-25)=25\pi\approx78.54\ \text{cm}^2.$ $\boxed{\mathcal A=25\pi\approx78.54\ \text{cm}^2}$