1.60. Given a rhombus $A B C D$, the diagonals of which are 3 and 4 cm. From the vertex of the obtuse angle $B$, heights $B E$ and $B F$ are drawn. Calculate the area of the quadrilateral $B F D E$.
1.60. The area of the rhombus $S=0.5 \times 3 \cdot 4=6=AD \cdot BE$ (Fig. 1.47). Next, from $\triangle AOD$ we find $AD=\sqrt{2^{2}+1.5^{2}}=2.5$ (cm) and, therefore, $BE=6: 2.5=2.4$ (cm).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-440.jpg?height=386&width=513&top_left_y=1474&top_left_x=628)

Fig. 1.47

Then from $\triangle BDE$ we get $DE=\sqrt{BD^{2}-BE^{2}}=\sqrt{3^{2}-2.4^{2}}=1.8$ (cm). Therefore,

$$
S_{BEDF}=2 S_{\triangle BED}=1.8 \cdot 2.4=4.32\left(\mathrm{cm}^{2}\right)
$$

Answer: $4.32 \mathrm{~cm}^{2}$.