10.289. A circle is inscribed in a rhombus with side $a$ and an acute angle of $60^{\circ}$. Determine the area of the quadrilateral whose vertices are the points of tangency of the circle with the sides of the rhombus.
Solution.

The radius of the circle inscribed in the rhombus $ABCD$ (Fig. 10.82) is $R=\frac{a \sqrt{3}}{4}$, since $\angle A=60^{\circ}$. The quadrilateral $KLMN$ is a rectangle, as its angles subtend the diameter of the circle.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0771.jpg?height=442&width=1169&top_left_y=66&top_left_x=94)

Fig. 10.82

Fig. 10.83

Its area $S=MN \cdot LM$, where $MN=R$ (the leg opposite the $30^{\circ}$ angle), and $LM=R \sqrt{3}$. Thus, we have $S=R^{2} \sqrt{3}=\frac{3 a^{2} \sqrt{3}}{16}$.

Answer: $\frac{3 a^{2} \sqrt{3}}{16}$.