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

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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?  
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?  


<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>


==Solution==
==Solution==
Let the width of the rectangle be <math>w</math>.
Let the width of the rectangle be <math>w</math>. Then the length is <math>2w</math>
 
Then the length is <math>2w</math>


Using the [[Pythagorean Theorem]]:
Using the [[Pythagorean Theorem]]:


<math>(x^{2})=(w^{2})+(2w)^{2}</math>
<math>x^{2}=w^{2}+(2w)^{2}</math>


<math>x^{2}=5w^{2}</math>
<math>x^{2}=5w^{2}</math>
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<math>2w=\frac{2x}{\sqrt{5}}</math>
<math>2w=\frac{2x}{\sqrt{5}}</math>


So the area of the rectangle is <math> w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \Rightarrow B</math>
So the [[area]] of the [[rectangle]] is <math> w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \Longrightarrow \mathrm{(B)}</math>
==See Also==
==See Also==
*[[2005 AMC 10A Problems]]
*[[2005 AMC 10A Problems]]
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*[[2005 AMC 10A Problems/Problem 5|Next Problem]]
*[[2005 AMC 10A Problems/Problem 5|Next Problem]]
[[Category:Introductory Geometry Problems]]

Revision as of 09:37, 2 August 2006

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:

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

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

$w=\frac{x}{\sqrt{5}}$

$2w=\frac{2x}{\sqrt{5}}$

So the area of the rectangle is $w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \Longrightarrow \mathrm{(B)}$

See Also

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