Difference between revisions of "2005 AMC 12A Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | Let <math>w</math> be the width, so the length is <math>2w</math>. By the [[Pythagorean Theorem]], <math>w^2 + | + | Let <math>w</math> be the width, so the length is <math>2w</math>. By the [[Pythagorean Theorem]], <math>w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w</math>. The area of the rectangle is <math>(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}</math>. |
== See also == | == See also == | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:15, 3 July 2013
Problem
A rectangle with diagonal length 
is twice as long as it is wide. What is the area of the rectangle?
Solution
Let 
be the width, so the length is 
. By the Pythagorean Theorem, 
. The area of the rectangle is 
.
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
