[ Areas of Curvilinear Figures Area of a Circle, Sector, and Segment [The area of a figure is equal to the sum of the areas of the figures into which it is divided]

In triangle $ABC$, the lateral sides $AB$ and $BC$ are equal to $a$, and the angle $ABC$ is $120^{\circ}$. A circle is inscribed in triangle $ABC$, touching side $AB$ at point $D$. A second circle has its center at point $B$ and passes through point $D$. Find the area of that part of the inscribed circle that lies inside the second circle.
The area of a sector with an angle of $120^{\circ}$ is one-third of the area of the circle; with an angle of $60^{\circ}$ — one-sixth.

## Solution

Notice that

$$
A C=a \sqrt{3}, B D=\frac{A B+B C+A C}{2}-A C=a\left(1-\frac{\sqrt{3}}{2}\right)
$$

Let $M$ be the point of tangency of the inscribed circle with side $B C$. By subtracting the area of triangle $D B M$ from the area of sector $D B M$, we get that

$$
\frac{1}{3} \pi B D^{2}-\frac{1}{2} B D^{2} \sin 120^{\circ}=\frac{a^{2}(7-4 \sqrt{3})\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)}{4}
$$

Let $O$ be the center of the inscribed circle. Then

$$
O M=r=B M \tan 60^{\circ}=\frac{a \sqrt{3}(2-\sqrt{3})}{2}
$$

By subtracting the area of triangle $D O M$ from the area of sector $D O M$ ( $\angle D O M=60^{\circ}$ ), we get that

$$
\frac{1}{6} \pi r^{2}-\frac{1}{2} r^{2} \sin 60^{\circ}=\frac{a^{2}(7-4 \sqrt{3})\left(\frac{\pi}{2}-\frac{3 \sqrt{3}}{4}\right)}{4}
$$

The desired area is the sum of the obtained differences, i.e.,

$$
\frac{a^{2}(7-4 \sqrt{3})\left(\frac{5 \pi}{6}-\sqrt{3}\right)}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-28.jpg?height=315&width=1018&top_left_y=2235&top_left_x=520)

## Answer

$\frac{a^{2}(7-4 \sqrt{3})\left(\frac{5 \pi}{6}-\sqrt{3}\right)}{4}$.