Difference between revisions of "2005 AMC 10A Problems/Problem 4"

Revision as of 13:03, 7 December 2020

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

This problem does not make sense.

Video Solution

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

2005 AMC 10A (Problems • Answer Key • Resources)
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@@ Line 5: / Line 5: @@
 ==Solution==
-Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>.
+This problem does not make sense.
-Using the [[Pythagorean Theorem]]:
-<math>x^{2}=w^{2}+(2w)^{2}</math>
-<math>x^{2}=5w^{2}</math>
-The [[area]] of the [[rectangle]] is <math>2w^2=\frac{2}{5}x^2</math>
-<math>(B)</math>
 ==Video Solution==

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Difference between revisions of "2005 AMC 10A Problems/Problem 4"

Revision as of 13:03, 7 December 2020

Contents

Problem

Solution

Video Solution

See also