10.083. The sum of the lengths of the diagonals of a rhombus is $m$, and its area is $S$. Find the side of the rhombus.
Solution.

$A C+B D=m ; A C=x, B D=y . S=\frac{1}{2} d_{1} d_{2}$ (Fig. 10.081). Solving the system $\left\{\begin{array}{l}x+y=m, \\ 2 S=x y,\end{array}\right.$ we get $x_{1}=\frac{m+\sqrt{m^{2}-8 S}}{2}, y_{1}=\frac{m-\sqrt{m^{2}-8 S}}{2}$ or $x_{2}=\frac{m-\sqrt{m^{2}-8 S}}{2}, y_{2}=\frac{m+\sqrt{m^{2}-8 S}}{2}$.

From this:

$B C^{2}=\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{2}\right)^{2}=\left(\frac{m+\sqrt{m^{2}-8 S}}{2 \cdot 2}\right)^{2}+\left(\frac{m-\sqrt{m^{2}-8 S}}{2 \cdot 2}\right)^{2}=\frac{m^{2}-4 S}{4}$.

$B C=\frac{\sqrt{m^{2}-4 S}}{2}$.

Answer: $\quad \frac{\sqrt{m^{2}-4 S}}{2}$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-596.jpg?height=402&width=500&top_left_y=91&top_left_x=110)

Fig. 10.81

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-596.jpg?height=402&width=564&top_left_y=91&top_left_x=688)

Fig. 10.82