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 | + | A rectangle with a [[diagonal]] of length <math>x</math> and the length is twice as long as the width. What is the area of the rectangle? |
<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math> | <math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M | ||
+ | |||
==Solution== | ==Solution== | ||
− | |||
− | + | 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> (Pythagoras Theorem) | ||
− | <math> | + | Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>. |
− | <math>x | + | * 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> | |
− | + | So our correct answer choice is <math>\mathrm{(B) \ } \frac{2}{5}x^2\qquad</math> | |
− | |||
− | + | -JinhoK | |
− | + | ==See also== | |
+ | {{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest 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 |
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