Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | A rectangle with a | + | 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> \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> | + | <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> | |
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==Solution 1== | ==Solution 1== | ||
| + | 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). | ||
| − | + | 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>.) | |
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| − | + | 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>. | |
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-JinhoK | -JinhoK | ||
==Solution 2== | ==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>. | ||
| − | + | -mobius247 | |
| − | + | ==Video Solution 1== | |
| + | https://youtu.be/X8QyT5RR-_M | ||
| − | + | ==Video Solution 2== | |
| + | https://youtu.be/5Bz7PC-tgyU | ||
| − | + | ~Charles3829 | |
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==See also== | ==See also== | ||
Latest revision as of 17:02, 1 July 2025
Problem
A rectangle with a diagonal of length 
is twice as long as it is wide. What is the area of the rectangle?
Solution 1
Let's set the length to 
and the width to 
, so the rectangle has area 
and diagonal 
(by Pythagoras' theorem).
Now we can plug this value into the answer choices and test which one will give our desired area of . (Note that all of the answer choices contain 
, so keep in mind that 
.)
We eventually find that 
, so the correct answer is 
.
-JinhoK
Solution 2
Call the length 
and the width 
, so the area of the rectangle is 
. As in Solution 1, is the hypotenuse of the right triangle with legs and , so by Pythagoras' theorem, 
. This becomes 
, and therefore the area is 
.
-mobius247
Video Solution 1
Video Solution 2
~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. 
