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

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==Problem==
 
==Problem==
A rectangle with a [[diagonal]] of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle?  
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A rectangle with a [[diagonal]] of length <math>x</math> and the length is twice as long as the width. What is the area of the rectangle?  
  
 
<math> \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 </math>
 
<math> \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 </math>

Revision as of 14:04, 7 December 2020

Problem

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

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

Video Solution

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

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AMC 10 Problems and Solutions

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