10.314. Given a rhombus $A B C D$, the diagonals of which are 3 and 4 cm. From the vertex of the obtuse angle $B$, two heights $B E$ and $B F$ are drawn. Calculate the area of the quadrilateral $B F D E$.
## Solution.

The area of the rhombus $S=0.5 \cdot 3 \cdot 4=6=A D \cdot B E$ (Fig. 10.103). From $\triangle A O D$, we find $A D=\sqrt{2^{2}+1.5^{2}}=2.5$ (cm) $\Rightarrow B E=6: 2.5=2.4$ (cm). From $\triangle B D E$, we have $D E=\sqrt{B D^{2}-B E^{2}}=\sqrt{3^{2}-2.4^{2}}=1.8(\mathrm{~cm})$.

We obtain $S_{B E D F}=2 S_{\triangle B E D}=1.8 \cdot 2.4=4.32\left(\mathrm{~cm}^{2}\right)$.

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

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0788.jpg?height=463&width=584&top_left_y=160&top_left_x=160)

Fig. 10.103

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0788.jpg?height=569&width=456&top_left_y=63&top_left_x=755)

Fig. 10.104