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>
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==Video Solution==
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CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M
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==Solution==
 
==Solution==
Let the width of the rectangle be <math>w</math>.
 
  
Then the length is <math>2w</math>
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Let's set our length to <math>2</math> and our width to <math>1</math>.
  
Using the [[Pythagorean Theorem]]:
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We have our area as <math>2*1 = 2</math> and our diagonal: <math>x</math> as <math>\sqrt{1^2+2^2} = \sqrt{5}</math> (Pythagoras Theorem)
  
<math>(x^{2})=(w^{2})+(2w)^{2}</math>
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Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>.
  
<math>x^{2}=5w^{2}</math>
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* All of the answer choices have our <math>x</math> value squared, so keep in mind that <math>\sqrt{5}^2 = 5</math>
  
<math>w=\frac{x}{\sqrt{5}}</math>
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Through testing, we see that <math>{2/5}*\sqrt{5}^2 = 2</math>
  
<math>2w=\frac{2x}{\sqrt{5}}</math>
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So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</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>
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-JinhoK
==See Also==
 
*[[2005 AMC 10A Problems]]
 
  
*[[2005 AMC 10A Problems/Problem 3|Previous Problem]]
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}}
  
*[[2005 AMC 10A Problems/Problem 5|Next Problem]]
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{{MAA Notice}}

Latest revision as of 20:24, 22 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


Solution

Let's set our length to $2$ and our width to $1$.

We have our area as $2*1 = 2$ and our diagonal: $x$ as $\sqrt{1^2+2^2} = \sqrt{5}$ (Pythagoras Theorem)

Now we can plug this value into the answer choices and test which one will give our desired area of $2$.

  • All of the answer choices have our $x$ value squared, so keep in mind that $\sqrt{5}^2 = 5$

Through testing, we see that ${2/5}*\sqrt{5}^2 = 2$

So our correct answer choice is $\mathrm{(B) \ } \frac{2}{5}x^2\qquad$

-JinhoK

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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