10.119. The perimeter of a rhombus is $2 p$; the lengths of the diagonals are in the ratio $m: n$. Calculate the area of the rhombus.
Solution.

Using Fig. 10.111, we have $B D=m x$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-612.jpg?height=254&width=468&top_left_y=97&top_left_x=778)

Fig. 10.112 $A C=n x . B C^{2}=\left(\frac{B D}{2}\right)^{2}+\left(\frac{A C}{2}\right)^{2}=\left(\frac{m x}{2}\right)^{2}+$ $+\left(\frac{n x}{2}\right)^{2}=\frac{x^{2}\left(m^{2}+n^{2}\right)}{4}$

Perimeter

$$
2 p=4 B C=4 \sqrt{\frac{x^{2}\left(m^{2}+n^{2}\right)}{4}}=2 x \sqrt{m^{2}+n^{2}}
$$

From this, $x=\frac{p}{\sqrt{m^{2}+n^{2}}}$.

Area of the rhombus $S=\frac{1}{2} B D \cdot A C=\frac{1}{2} m n x^{2}=\frac{1}{2} \frac{m n p^{2}}{\left(m^{2}+n^{2}\right)}$.

Answer: $\frac{m n p^{2}}{2\left(m^{2}+n^{2}\right)}$.