Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
m (Improved formatting of answer choices)
 
(39 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==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>
 +
\textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2
 +
</math>
  
−
==Solution==
+
==Solution 1==
−
Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>.
+
Let's set the length to <math>2</math> and the width to <math>1</math>, so the rectangle has area <math>2 \cdot 1 = 2</math> and diagonal <math>x = \sqrt{1^2+2^2} = \sqrt{5}</math> (by Pythagoras' theorem).
  
−
Using the [[Pythagorean Theorem]]:
+
Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>. (Note that all of the answer choices contain <math>x^2</math>, so keep in mind that <math>\left(\sqrt{5}\right)^2 = 5</math>.)
  
−
<math>x^{2}=w^{2}+(2w)^{2}</math>
+
We eventually find that <math>\frac{2}{5} \cdot \left(\sqrt{5}\right)^2 = 2</math>, so the correct answer is <math>\boxed{\textbf{(B) } \frac{2}{5}x^2}</math>.
  
−
<math>x^{2}=5w^{2}</math>
+
-JinhoK
  
−
The [[area]] of the [[rectangle]] is <math>2w^2=\frac{2}{5}x^2</math>
+
==Solution 2==
 +
Call the length <math>2l</math> and the width <math>l</math>, so the area of the rectangle is <math>2l \cdot l = 2l^2</math>. As in Solution 1, <math>x</math> is the hypotenuse of the right triangle with legs <math>2l</math> and <math>l</math>, so by Pythagoras' theorem, <math>x = \sqrt{\left(2l\right)^2+l^2} = \sqrt{4l^2+l^2} = \sqrt{5l^2}</math>. This becomes <math>l^2 = \frac{x^2}{5}</math>, and therefore the area is <math>2l^2 = \boxed{\textbf{(B) } \frac{2}{5}x^2}</math>.
  
−
<math>(B)</math>
+
-mobius247
  
−
==See Also==
+
==Video Solution 1==
 +
https://youtu.be/X8QyT5RR-_M
  
−
{{AMC10 box|year=2005|ab=A|before=Problem 3|num-a=5}}
+
==Video Solution 2==
 +
https://youtu.be/5Bz7PC-tgyU
 +
 
 +
~Charles3829
 +
 
 +
==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}}
  
−
[[Category:Introductory Algebra Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:02, 1 July 2025

Problem

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

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

Solution 1

Let's set the length to $2$ and the width to $1$, so the rectangle has area $2 \cdot 1 = 2$ and diagonal $x = \sqrt{1^2+2^2} = \sqrt{5}$ (by Pythagoras' theorem).

Now we can plug this value into the answer choices and test which one will give our desired area of $2$. (Note that all of the answer choices contain $x^2$, so keep in mind that $\left(\sqrt{5}\right)^2 = 5$.)

We eventually find that $\frac{2}{5} \cdot \left(\sqrt{5}\right)^2 = 2$, so the correct answer is $\boxed{\textbf{(B) } \frac{2}{5}x^2}$.

-JinhoK

Solution 2

Call the length $2l$ and the width $l$, so the area of the rectangle is $2l \cdot l = 2l^2$. As in Solution 1, $x$ is the hypotenuse of the right triangle with legs $2l$ and $l$, so by Pythagoras' theorem, $x = \sqrt{\left(2l\right)^2+l^2} = \sqrt{4l^2+l^2} = \sqrt{5l^2}$. This becomes $l^2 = \frac{x^2}{5}$, and therefore the area is $2l^2 = \boxed{\textbf{(B) } \frac{2}{5}x^2}$.

-mobius247

Video Solution 1

https://youtu.be/X8QyT5RR-_M

Video Solution 2

https://youtu.be/5Bz7PC-tgyU

~Charles3829

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_4&oldid=253053"