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

Revision as of 13:11, 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, (2 = x^2 $We have our area as$ 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$ (Error compiling LaTeX. Unknown error_msg)2$.

All of the answer choices have our$ (Error compiling LaTeX. Unknown error_msg)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$

-mobius247

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 30: / Line 30: @@
 The area of the rectangle is <math>2l*l = 2l^2</math>
-<math>x</math> is the hypotenuse of the right triangle with 2l and l as legs. By the Pythagorean theorem, <math>x</math> <math>\sqrt{1^2+2^2} = \sqrt{5}</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, (2 = x^2<math>
-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)
+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)
-Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>.
+Now we can plug this value into the answer choices and test which one will give our desired area of </math>2<math>.
-* All of the answer choices have our <math>x</math> value squared, so keep in mind that <math>\sqrt{5}^2 = 5</math>
+* All of the answer choices have our </math>x<math> value squared, so keep in mind that </math>\sqrt{5}^2 = 5<math>
-Through testing, we see that <math>{2/5}*\sqrt{5}^2 = 2</math>
+Through testing, we see that </math>{2/5}*\sqrt{5}^2 = 2<math>
-So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</math>
+So our correct answer choice is </math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad$
 -mobius247

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