Annulus between the incircle and circumcircle of an equilateral triangle

Text

An equilateral triangle has side $12$ cm. Compute the area of the annulus between the inscribed and circumscribed circles of the triangle.

Solution

For an equilateral triangle of side $\ell$: circumradius $R=\dfrac{\ell}{\sqrt3}$, inradius $r=\dfrac{\ell}{2\sqrt3}=\dfrac R2$. With $\ell=12$: $R^2=\dfrac{144}{3}=48$, $\ r^2=\dfrac{48}{4}=12$. $\mathcal A=\pi(R^2-r^2)=\pi(48-12)=36\pi\approx113.10\ \text{cm}^2.$ $\boxed{\mathcal A=36\pi\approx113.10\ \text{cm}^2}$