A geometry problem by Interliser 727

With legs $a,b$ and hypotenuse $c = 20$ we are given that $a + b + c = 46 \Longrightarrow a + b = 26$ . Since $a^{2} + b^{2} = c^{2} = 20^{2} = 400$ we see that

$(a + b)^{2} = a^{2} + b^{2} + 2ab \Longrightarrow 26^{2} = 400 + 2ab \Longrightarrow 676 - 400 = 2ab \Longrightarrow ab = 138$ .

Finally, the desired area is $\dfrac{1}{2}ab = \boxed{69}$ .

Note: With $b = \dfrac{138}{a}$ we have that $a + b = 26 \Longrightarrow a + \dfrac{138}{a} = 26 \Longrightarrow a^{2} - 26a + 138 = 0 \Longrightarrow a = \dfrac{26 \pm \sqrt{26^{2} - 4 \times 138}}{2} = 13 \pm \sqrt{31}$ .

So the side lengths are $13 + \sqrt{31}$ and $13 - \sqrt{31}$ .