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> | + | A rectangle with a [[diagonal]] of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? |
| − | <math> \ | + | <math> \textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2 </math> |
==Video Solution== | ==Video Solution== | ||
CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M | CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M | ||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/5Bz7PC-tgyU | ||
| − | ==Solution== | + | ~Charles3829 |
| + | |||
| + | ==Solution 1== | ||
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>. | ||
| Line 20: | Line 24: | ||
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>\ | + | So our correct answer choice is <math>\boxed{\textbf{(B) }\frac{2}{5}x^2}</math> |
-JinhoK | -JinhoK | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Call the length <math>2l</math> and the width <math>l</math>. | ||
| + | |||
| + | The area of the rectangle is <math>2l*l = 2l^2</math> | ||
| + | |||
| + | <math>x</math> is the hypotenuse of the right triangle with <math>2l</math> and <math>l</math> as legs. By the Pythagorean theorem, <math>(2l)^2+l^2 = x^2</math> | ||
| + | |||
| + | <math>4l^2 + l^2 = x^2</math><math>,</math> <math>5l^2 = x^2</math><math>,</math> and <math>l^2 = \frac{x^2}{5}</math>. | ||
| + | |||
| + | Therefore, the area is <math>\boxed{\textbf{(B) }\frac{2}{5}x^2}</math> | ||
| + | |||
| + | -mobius247 | ||
==See also== | ==See also== | ||
Latest revision as of 14:12, 1 November 2024
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
Video Solution 2
~Charles3829
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, 
and 
.
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. 
