AMC12 2005 A
AMC12 2005 A · Q3
AMC12 2005 A · Q3. 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，其长是宽的两倍。该矩形的面积是多少？
(A)
\frac{1}{4}x^2
\frac{1}{4}x^2
(B)
\frac{2}{5}x^2
\frac{2}{5}x^2
(C)
\frac{1}{2}x^2
\frac{1}{2}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, so the length is $2w$. By the Pythagorean Theorem, $w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w$. The area of the rectangle is $(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}$.
Let $w$ be the width, so the length is $2w$. By the Pythagorean Theorem, $w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w$. The area of the rectangle is $(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}$.
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