Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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Call the length <math>2l</math> and the width <math>l</math>. | Call the length <math>2l</math> and the width <math>l</math>. | ||
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| + | The area of the rectangle is <math>2l</math> * <math>l</math> = <math>2L^2</math> | ||
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> (Pythagoras Theorem) | 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> (Pythagoras Theorem) | ||
Revision as of 13:07, 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 * = 
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
-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 | ||
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