AMC10 2008 A

AMC10 2008 A · Q18

AMC10 2008 A · Q18. It mainly tests Systems of equations, Pythagorean theorem.

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

一个直角三角形的周长为 32，面积为 20。求其斜边长度。

(A) $\frac{57}{4}$ $\frac{57}{4}$

(B) $\frac{59}{4}$ $\frac{59}{4}$

(D) $\frac{63}{4}$ $\frac{63}{4}$

(E) $\frac{65}{4}$ $\frac{65}{4}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $x$ be the length of the hypotenuse, and let $y$ and $z$ be the lengths of the legs. The given conditions imply that $$ y^2+z^2=x^2,\quad y+z=32-x,\quad \text{and}\quad yz=40. $$ Thus $$ (32-x)^2=(y+z)^2=y^2+z^2+2yz=x^2+80, $$ from which $1024-64x=80$, and $x=\frac{59}{4}$. Note: Solving the system of equations yields leg lengths of $$ \frac{1}{8}\left(69+\sqrt{2201}\right)\ \text{and}\ \frac{1}{8}\left(69-\sqrt{2201}\right), $$ so a triangle satisfying the given conditions does in fact exist.

答案（B）：设 $x$ 为斜边长度，$y$ 和 $z$ 为两条直角边的长度。已知条件推出 $$ y^2+z^2=x^2,\quad y+z=32-x,\quad \text{且}\quad yz=40。 $$ 因此 $$ (32-x)^2=(y+z)^2=y^2+z^2+2yz=x^2+80， $$ 由此得到 $1024-64x=80$，并且 $x=\frac{59}{4}$。注：解该方程组可得两条直角边的长度为 $$ \frac{1}{8}\left(69+\sqrt{2201}\right)\ \text{和}\ \frac{1}{8}\left(69-\sqrt{2201}\right)， $$ 因此确实存在满足所给条件的三角形。

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