[ Area of a circle, sector, and segment $]$ [

The height of an equilateral triangle with side length $b$ is used as the diameter to construct a circle. Find the area of the part of the triangle that lies inside the circle.

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Let a circle with center $O$ and radius $R$, constructed on the height $A M$ of an equilateral triangle $A B C$, intersect the sides $A B$ and $A C$ at points $D$ and $E$ respectively. Then

$$
R=\frac{1}{2} A M=\frac{b \sqrt{3}}{4}, O E=O M=O D=R, \angle E O D=2 \angle D A E=120^{\circ} .
$$

Let $S 1$ be the area of sector $D O E$, $S 2$ be the area of the circle with radius $R$, and $S$ be the desired area of the part of the triangle enclosed within the circle. Then

$$
S 2=\pi R 2, S 1=\frac{1}{3} S 2=\frac{1}{3} \pi R 2 .
$$

Therefore,

$$
\begin{aligned}
& S=S 1+2 S_{\triangle A O D}=\frac{1}{3} \pi R 2+2 \cdot \frac{1}{2} R 2 \sin \angle A O D= \\
& =\frac{1}{3} \pi R 2+R 2 \sin 60^{\circ}=\frac{1}{3} \pi R 2+\frac{R^{2} \sqrt{3}}{2}=\frac{1}{6} R 2(2 \pi+3 \sqrt{3})= \\
& =\frac{1}{6} \cdot\left(\frac{b \sqrt{3}}{4}\right) 2 \cdot(2 \pi+3 \sqrt{3})=\frac{b^{2}(2 \pi+3 \sqrt{3})}{32} .
\end{aligned}
$$