AMC10 2005 A

AMC10 2005 A · Q4

AMC10 2005 A · Q4. It mainly tests Pythagorean theorem, Area & perimeter.

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

一个对角线长度为 $x$ 的矩形，长是宽的两倍。求该矩形的面积。

(A) $\frac{1}{4}x^2$ $\frac{1}{4}x^2$

(B) $\frac{2}{5}x^2$ $\frac{2}{5}x^2$

(D) $x^2$ $x^2$

(E) $\frac{3}{2}x^2$ $\frac{3}{2}x^2$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $w$ be the width of the rectangle. Then the length is $2w$, and $x^2 = w^2 + (2w)^2 = 5w^2$. The area is consequently $w(2w) = 2w^2 = \frac{2}{5}x^2$.

设矩形的宽为 $w$。则长为 $2w$，且 $x^2 = w^2 + (2w)^2 = 5w^2$。因此面积为 $w(2w) = 2w^2 = \frac{2}{5}x^2$。

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