Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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==Video Solution== | ==Video Solution== | ||
CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M | CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M | ||
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| + | ==Solution== | ||
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| + | Let's set our length to <math>2</math> and our width to <math>1</math>. | ||
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| + | We have our area as <math>2*1 = 2</math> and our diagonal: <math>x</math> as <math>\sqrt{1^2+2^2} = \sqrt{5}</math> | ||
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| + | Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>. | ||
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| + | *All of the answer choices have our <math>x</math> value squared, so keep in mind that <math>\sqrt{5}^2 = 5</math>* | ||
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| + | Through testing, we see that <math>{2/5}*\sqrt{5}^2 = 2</math> | ||
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| + | So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</math> | ||
==See also== | ==See also== | ||
Revision as of 19:23, 22 December 2020
Contents
Problem
A rectangle with a diagonal of length 
and the length is twice as long as the width. What is the area of the rectangle?
Video Solution
CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M
Solution
Let's set our length to 
and our width to 
.
We have our area as 
and our diagonal: as 
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 
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. 
