2005 AMC 10A Problems/Problem 4: Difference between revisions

Revision as of 09:25, 12 August 2018

Problem

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

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

Solution

Let the width of the rectangle be $w$ . Then the length is $2w$ .

Using the Pythagorean Theorem:

$x^{2}=w^{2}+(2w)^{2}$

$x^{2}=5w^{2}$

The area of the rectangle is $2w^2=\frac{2}{5}x^2$

$(B)$

@@ Line 5: / Line 5: @@
 ==Solution==
-Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>
+Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>.
 Using the [[Pythagorean Theorem]]:

Page

Toolbox

Search

2005 AMC 10A Problems/Problem 4: Difference between revisions

Revision as of 09:25, 12 August 2018

Problem

Solution

See Also