Difference between revisions of "2005 AMC 10A Problems/Problem 4"
(→Solution 2) |
(→Solution 2) |
||
| Line 34: | Line 34: | ||
<math>4l^2 + l^2 = x^2</math>, <math>5l^2 = x^2</math>, <math>l^2 = \frac{x^2}{5}</math>. | <math>4l^2 + l^2 = x^2</math>, <math>5l^2 = x^2</math>, <math>l^2 = \frac{x^2}{5}</math>. | ||
| − | Therefore, the area is <math>\frac{2}{5}x^2\\Longrightarrow \mathrm{(B) \ } </math> | + | Therefore, the area is <math>\frac{2}{5}x^2\ \Longrightarrow \mathrm{(B) \ } </math> |
-mobius247 | -mobius247 | ||
Revision as of 13:25, 31 May 2021
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
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, 
, 
, 
.
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. 
