10.332. Determine the sides of a right-angled triangle, given that its perimeter is $2 p$, and the area is $m^{2}$.
## Solution.

Let $a$ and $b$ be the legs, and $c$ be the hypotenuse of the given triangle. Using the properties of a right triangle, we can set up the following system:

$$
\left\{\begin{array}{l}
a+b+c=2 p \\
\frac{a b}{2}=m^{2}, \\
a^{2}+b^{2}=c^{2}
\end{array} \quad \text { We have: } a+b=2 p-c, a^{2}+2 a b+b^{2}=(2 p-c)^{2} .\right. \text { Therefore, }
$$

$c^{2}+4 m^{2}=4 p^{2}-4 p c+c^{2} ; c=\frac{p^{2}-m^{2}}{p}$. Then $a+b=$ $=2 p-c=\frac{p^{2}+m^{2}}{p}, a b=2 m^{2}$. Thus, $a$ and $b$ are the roots of the equation $x^{2}-\frac{p^{2}+m^{2}}{p} x+2 m^{2}=0$. Therefore, $p x^{2}-\left(p^{2}+m^{2}\right) x+2 m^{2} p=0 ;$ $x=\frac{p^{2}+m^{2} \pm \sqrt{\left(p^{2}+m^{2}\right)^{2}-8 p^{2} m^{2}}}{2 p} \cdot$ The legs of the given triangle are $\frac{p^{2}+m^{2}+\sqrt{\left(p^{2}+m^{2}\right)^{2}-8 p^{2} m^{2}}}{2 p}$ and $\frac{p^{2}+m^{2}-\sqrt{\left(p^{2}+m^{2}\right)^{2}-8 p^{2} m^{2}}}{2 p}$. Answer: $\frac{p^{2}-m^{2}}{p} ; \frac{p^{2}+m^{2} \pm \sqrt{\left(p^{2}+m^{2}\right)^{2}-8 p^{2} m^{2}}}{2 p}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0799.jpg?height=470&width=672&top_left_y=82&top_left_x=90)

Fig. 10.119

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0799.jpg?height=477&width=524&top_left_y=70&top_left_x=727)

Fig. 10.120