The diameter of a circle with radius $r$ is the base of an equilateral triangle. Find the part of the area of the triangle that lies outside the circle.

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Let the circle with center $O$, constructed on the side $B C$ of an equilateral triangle $A B C$ as its diameter, intersect its sides $A B$ and $A C$ at points $M$ and $N$ respectively. Then $C M$ is the height of triangle $A B C$. Therefore, $M$ is the midpoint of $A B$ and triangle $M O B$ is similar to triangle $A C B$ with a similarity ratio of $\frac{1}{2}$. Consequently,

$$
S_{\triangle M O B}=\frac{1}{4} S_{\triangle A B C}=\frac{1}{4} \cdot \frac{(2 r)^{2} \sqrt{3}}{4}=\frac{r^{2} \sqrt{3}}{4}
$$

Similarly,

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-37.jpg?height=126&width=468&top_left_y=2687&top_left_x=795)

Since $\angle M O N=60^{\circ}$, the area of sector $M O N$ is one-sixth of the area of the circle with radius $r$, i.e., $S_{\text {sect. }}=\frac{\pi r^{2}}{6}$. Therefore, the desired area is

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-38.jpg?height=122&width=840&top_left_y=59&top_left_x=609)

## Answer

$\frac{r^{2}(3 \sqrt{3}-\pi)}{6}$