10.344. Calculate the area of the common part of two rhombuses, the lengths of the diagonals of the first of which are 4 and $6 \mathrm{~cm}$, and the second is obtained by rotating the first by $90^{\circ}$ around its center.
Solution.

The required area $S$ is equal to $4\left(S_{\triangle A O B}-S_{\triangle A F K}\right)$ (Fig. 10.127). We find $S_{\triangle A O B}=0.5 \cdot 3 \cdot 2=3 \, \text{cm}^{2}$. The side of the rhombus is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$ (cm). In $\triangle A O B$, the segment $O K$ is the angle bisector; then, using the formula

$$
l_{c}=\frac{\sqrt{a b(a+b+c)(a+b-c)}}{a+b}, \text{ we have } O K=\frac{\sqrt{6(5+\sqrt{13})(5-\sqrt{13})}}{5}=\frac{6 \sqrt{2}}{5} \, \text{cm.}
$$

Next, $S_{\triangle A F K}=\frac{1}{2} A F \cdot K H$, where $K H=\frac{O K}{\sqrt{2}}=\frac{6}{5}$ cm, $A F=3-2=1$ cm; therefore, $S_{\triangle A F K}=0.6 \, \text{cm}^{2}$. Finally, $S=4(3-0.6)=9.6 \, \text{cm}^{2}$.

Answer: $9.6 \, \text{cm}^{2}$.