1.61. Calculate the area of the common part of two rhombi, 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.
1.61. The desired area $S$ is equal to $4\left(S_{\triangle A O B}-S_{\triangle A F K}\right)$ (Fig. 1.48). We find $S_{\triangle A O B}=0.5 \cdot 3 \cdot 2=3$ ( $\left.\mathrm{cm}^{2}\right)$. 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

$$
l_{c}=\frac{\sqrt{a b(a+b+c)(a+b-c)}}{a+b}
$$

(see relation (1.38)), we get

$$
O K=\frac{\sqrt{6(5+\sqrt{13})(5-\sqrt{13})}}{5}=\frac{6 \sqrt{2}}{5}(\mathrm{cm})
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-441.jpg?height=658&width=683&top_left_y=1088&top_left_x=243)

Fig. 1.48

Now we find $S_{\triangle A F K}=\frac{1}{2} A F \cdot K H$, where $K H=\frac{O K}{\sqrt{2}}=\frac{6}{5}$ (cm), $A F=3-2=1$ (cm); thus, $S_{\triangle A F K}=0.6 \quad \mathrm{~cm}^{2}$. Therefore, $S=4(3-0.6)=9.6 \mathrm{~cm}^{2}$.

Answer: $9.6 \mathrm{~cm}^{2}$.