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

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<math>x</math> is the hypotenuse of the right triangle with <math>2l</math> and <math>l</math> as legs. By the Pythagorean theorem, <math>(2l)^2+l^2 = x^2</math>
 
<math>x</math> is the hypotenuse of the right triangle with <math>2l</math> and <math>l</math> as legs. By the Pythagorean theorem, <math>(2l)^2+l^2 = x^2</math>
  
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<math>4l^2 + l^2 = x^2</math><math>,</math> <math>5l^2 = x^2</math><math>,</math>and <math>l^2 = \frac{x^2}{5}</math>.
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<math>4l^2 + l^2 = x^2</math><math>,</math> <math>5l^2 = x^2</math><math>,</math> and <math>l^2 = \frac{x^2}{5}</math>.
  
 
Therefore, the area is <math>\frac{2}{5}x^2\ \Longrightarrow \mathrm{(B) \ } </math>
 
Therefore, the area is <math>\frac{2}{5}x^2\ \Longrightarrow \mathrm{(B) \ } </math>

Revision as of 13:27, 31 May 2021

Problem

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

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

Video Solution

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


Solution 1

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

Solution 2

Call the length $2l$ and the width $l$.

The area of the rectangle is $2l*l = 2l^2$

$x$ is the hypotenuse of the right triangle with $2l$ and $l$ as legs. By the Pythagorean theorem, $(2l)^2+l^2 = x^2$

$4l^2 + l^2 = x^2$$,$ $5l^2 = x^2$$,$ and $l^2 = \frac{x^2}{5}$.

Therefore, the area is $\frac{2}{5}x^2\ \Longrightarrow \mathrm{(B) \ }$

-mobius247

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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