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?

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

Video Solution

CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M

Video Solution 2

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

~Charles3829

Solution 1

Let's set our length to $2$ and our width to $1$ .

We have our area as $2*1 = 2$ and our diagonal: $x$ as $\sqrt{1^2+2^2} = \sqrt{5}$ (Pythagoras Theorem)

Now we can plug this value into the answer choices and test which one will give our desired area of $2$ .

All of the answer choices have our $x$ value squared, so keep in mind that $\sqrt{5}^2 = 5$

Through testing, we see that ${2/5}*\sqrt{5}^2 = 2$

So our correct answer choice is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$

-JinhoK

Solution 2

Call the length $2l$ and the width $l$ .

The area of the rectangle is $2l*l = 2l^2$

$x$ is the hypotenuse of the right triangle with $2l$ and $l$ as legs. By the Pythagorean theorem, $(2l)^2+l^2 = x^2$

$4l^2 + l^2 = x^2$ $,$ $5l^2 = x^2$ $,$ and $l^2 = \frac{x^2}{5}$ .

Therefore, the area is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$

-mobius247

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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