10.139. Semicircles are described on the sides of a rhombus, turned inward. Determine the area of the resulting rosette if the diagonals of the rhombus are equal to $a$ and $b$.
## Solution.

The area of the obtained rosette will be equal to the sum of the areas of four shaded (Fig. 10.125) parts. The area of each shaded part is equal to the difference between the area of a semicircle and the area of a right triangle with sides $\frac{a}{2}$ and $\frac{b}{2}$. The side of the rhombus can be found as the hypotenuse: $c=\frac{1}{2} \sqrt{a^{2}+b^{2}}$. Then the radius of the semicircle $R=\frac{1}{2} c=\frac{1}{4} \sqrt{a^{2}+b^{2}}$. The area of the semicircle $S_{1}=\frac{1}{2} \pi R^{2}=\frac{\pi}{32}\left(a^{2}+b^{2}\right)$. The area of the right triangle $S_{2}=\frac{1}{2} \cdot \frac{a}{2} \cdot \frac{b}{2}=\frac{a b}{8}$. The area of the rosette $S=4\left(S_{1}-S_{2}\right)=$ $=4\left(\frac{\pi}{32}\left(a^{2}+b^{2}\right)-\frac{a b}{8}\right)=\frac{\pi\left(a^{2}+b^{2}\right)-4 a b}{8}$.

Answer: $\frac{\pi\left(a^{2}+b^{2}\right)-4 a b}{8}$.