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

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==Solution==
==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==
==Video Solution==

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

See also

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

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