2008 AMC 10A Problem 18

Let the legs be $y, z$ and the hypotenuse $x.$ Then $y^2 + z^2 = x^2,$ $y + z = 32 - x,$ and $yz = 40.$ Squaring the second equation, $$\begin{aligned} &(32 - x)^2 \\ &\quad = y^2 + z^2 + 2yz = x^2 + 80. \end{aligned}$$ This gives $1024 - 64x = 80,$ so $x = \dfrac{59}{4}.$ Thus, the correct answer is B.