Given a triangle $K L M$ with base $K M$, equal to $\frac{\sqrt{3}}{2}$, and side $K L$, equal to 1. A circle is drawn through points $K$ and $L$, with its center lying on the height $L F$, dropped to the base $K M$. It is known that $F M=\frac{\sqrt{3}}{6}$, and point $F$ lies on $K M$. Find the area of the circle bounded by this circle.
Drop a perpendicular $OA$ from the center $O$ of the given circle to the chord $KL$ and consider the right triangles $FKL$ and $AOL$.

## Solution

Since point $F$ lies on the segment $KM$, then

$$
KF = KM - FM = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}
$$

Drop a perpendicular $OA$ from the center $O$ of the given circle with radius $R$ to the chord $KL$. Then $A$ is the midpoint of $KL$. Let $\angle FLK = \alpha$. From the right triangle $FKL$, we find that $\sin \alpha = \frac{FK}{KL} = \frac{\sqrt{3}}{3}$. Then $\cos \alpha =$

From the right triangle $OAL$, we find that

$$
R = OL = \frac{AL}{\cos \alpha} = \frac{\frac{1}{2}}{\sqrt{\frac{2}{3}}} = \frac{\sqrt{3}}{2 \sqrt{2}}
$$

Therefore, the area of the circle is $\pi R^2 = \frac{3}{8} \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-31.jpg?height=509&width=506&top_left_y=520&top_left_x=776)

## Answer

$\frac{3}{8} \pi$