Area of a circular segment from an equilateral triangle

Text

In a circle of radius $r=6$ cm consider the chord equal to the side of the inscribed equilateral triangle. Compute the area of the circular segment between that chord and its (minor) arc.

Solution

The side of the inscribed equilateral triangle subtends a central angle of $120^\circ$. Area of the $120^\circ$ sector: $\ \mathcal A_\text{sec}=\dfrac{120}{360}\pi r^2=\dfrac{\pi r^2}{3}=12\pi$. Area of the isosceles triangle with sides $r,r$ and angle $120^\circ$: $\ \dfrac12 r^2\sin120^\circ=\dfrac12\cdot36\cdot\dfrac{\sqrt3}{2}=9\sqrt3$. Segment $=\mathcal A_\text{sec}-\text{triangle}=12\pi-9\sqrt3\approx37.70-15.59=22.11\ \text{cm}^2.$ $\boxed{\mathcal A=12\pi-9\sqrt3\approx22.11\ \text{cm}^2}$