Given a rhombus with an acute angle $\alpha$. What fraction of the area of the rhombus is the area of the circle inscribed in it?

#
The height of the rhombus is equal to the diameter of the circle inscribed in the rhombus.

## Solution

Let the side of the rhombus $ABCD$ be $a$ and $\angle BAC = \alpha$. Then the area of the rhombus is $a^2 \sin \alpha$.

Dropping a perpendicular $BM$ from vertex $B$ of the obtuse angle to side $AD$, we have

$$
BM = AB \sin \alpha = a \sin \alpha
$$

If $r$ is the radius of the circle inscribed in the rhombus, then

$$
2r = BM = a \sin \alpha
$$

Therefore,

$$
r = \frac{1}{2} a \sin \alpha
$$

and the area of the circle inscribed in the rhombus is

$$
\pi r^2 = \frac{\pi}{4} \cdot a^2 \sin^2 \alpha
$$

Thus, the desired ratio is

$$
\frac{\frac{\pi}{4} a^2 \sin^2 \alpha}{a^2 \sin \alpha} = \frac{1}{4} \pi \sin \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-40.jpg?height=435&width=897&top_left_y=1229&top_left_x=589)

Answer

$\frac{1}{4} \pi \sin \alpha$