2005 AMC 10A Problems/Problem 4
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 |
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