Question

一个直角三角形周长为 $32$ ，面积为 $20$ 。它的斜边长是多少？

A right triangle has perimeter $32$ and area $20.$ What is the length of its hypotenuse?

$\dfrac{57}{4}$

$\dfrac{59}{4}$

$\dfrac{61}{4}$

$\dfrac{63}{4}$

$\dfrac{65}{4}$

设两条直角边为 $y, z$ ，斜边为 $x$ 。则 $y^2 + z^2 = x^2$ ， $y + z = 32 - x$ ，且 $yz = 40$ 。

将第二个等式平方，得到 $\begin{aligned} &(32 - x)^2 \\ &\quad = y^2 + z^2 + 2yz = x^2 + 80. \end{aligned}$

这给出 $1024 - 64x = 80$ ，所以 $x = \dfrac{59}{4}$ 。

所以正确答案是 B。

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.