Difference between revisions of "2005 AMC 12A Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | A [[rectangle]] with diagonal 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> | <math> | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:16, 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
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
