Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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==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> \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> |
Revision as of 15:12, 1 November 2024
Problem
.A rectangle with a diagonal of length is twice as long as it is wide. What is the area of the rectangle?
Video Solution
CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M
Video Solution 2
~Charles3829
Solution 1
Let's set our length to and our width to .
We have our area as and our diagonal: as (Pythagoras Theorem)
Now we can plug this value into the answer choices and test which one will give our desired area of .
- All of the answer choices have our value squared, so keep in mind that
Through testing, we see that
So our correct answer choice is
-JinhoK
Solution 2
Call the length and the width .
The area of the rectangle is
is the hypotenuse of the right triangle with and as legs. By the Pythagorean theorem,
and .
Therefore, the area is
-mobius247
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.