2005 AMC 10A Problems/Problem 4

Problem

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

$\textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2$

Solution 1

Let's set the length to $2$ and the width to $1$ , so the rectangle has area $2 \cdot 1 = 2$ and diagonal $x = \sqrt{1^2+2^2} = \sqrt{5}$ (by Pythagoras' theorem).

Now we can plug this value into the answer choices and test which one will give our desired area of $2$ . (Note that all of the answer choices contain $x^2$ , so keep in mind that $\left(\sqrt{5}\right)^2 = 5$ .)

We eventually find that $\frac{2}{5} \cdot \left(\sqrt{5}\right)^2 = 2$ , so the correct answer is $\boxed{\textbf{(B) } \frac{2}{5}x^2}$ .

-JinhoK

Solution 2

Call the length $2l$ and the width $l$ , so the area of the rectangle is $2l \cdot l = 2l^2$ . As in Solution 1, $x$ is the hypotenuse of the right triangle with legs $2l$ and $l$ , so by Pythagoras' theorem, $x = \sqrt{\left(2l\right)^2+l^2} = \sqrt{4l^2+l^2} = \sqrt{5l^2}$ . This becomes $l^2 = \frac{x^2}{5}$ , and therefore the area is $2l^2 = \boxed{\textbf{(B) } \frac{2}{5}x^2}$ .

-mobius247

Video Solution 1

https://youtu.be/X8QyT5RR-_M

Video Solution 2

https://youtu.be/5Bz7PC-tgyU

~Charles3829

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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