Difference between revisions of "2005 AMC 10A Problems/Problem 4"
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Let's set our length to <math>2</math> and our width to <math>1</math>. | Let's set our length to <math>2</math> and our width to <math>1</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> | + | 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) |
Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>. | Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>. | ||
− | *All of the answer choices have our <math>x</math> value squared, so keep in mind that <math>\sqrt{5}^2 = 5</math> | + | * All of the answer choices have our <math>x</math> value squared, so keep in mind that <math>\sqrt{5}^2 = 5</math> |
Through testing, we see that <math>{2/5}*\sqrt{5}^2 = 2</math> | Through testing, we see that <math>{2/5}*\sqrt{5}^2 = 2</math> | ||
So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</math> | So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</math> | ||
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+ | -JinhoK | ||
==See also== | ==See also== |
Revision as of 20:24, 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 (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
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.