Difference between revisions of "2005 AMC 10A Problems/Problem 4"

Latest revision as of 14:12, 1 November 2024

Problem

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$\textbf{(A) } \frac{1}{4}x^2\qquad \textbf{(B) } \frac{2}{5}x^2\qquad \textbf{(C) } \frac{1}{2}x^2\qquad \textbf{(D) } x^2\qquad \textbf{(E) } \frac{3}{2}x^2$

Video Solution

CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M

Video Solution 2

https://youtu.be/5Bz7PC-tgyU

~Charles3829

Solution 1

Let's set our length to $2$ and our width to $1$ .

We have our area as $2*1 = 2$ and our diagonal: $x$ as $\sqrt{1^2+2^2} = \sqrt{5}$ (Pythagoras Theorem)

Now we can plug this value into the answer choices and test which one will give our desired area of $2$ .

All of the answer choices have our $x$ value squared, so keep in mind that $\sqrt{5}^2 = 5$

Through testing, we see that ${2/5}*\sqrt{5}^2 = 2$

So our correct answer choice is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$

-JinhoK

Solution 2

Call the length $2l$ and the width $l$ .

The area of the rectangle is $2l*l = 2l^2$

$x$ is the hypotenuse of the right triangle with $2l$ and $l$ as legs. By the Pythagorean theorem, $(2l)^2+l^2 = x^2$

$4l^2 + l^2 = x^2$ $,$ $5l^2 = x^2$ $,$ and $l^2 = \frac{x^2}{5}$ .

Therefore, the area is $\boxed{\textbf{(B) }\frac{2}{5}x^2}$

-mobius247

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.

@@ Line 2: / Line 2: @@
 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> \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> \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>
-==Solution==
+==Video Solution==
-Let the width of the rectangle be <math>w</math>.  Then the length is <math>2w</math>
+CHECK OUT Video Solution: https://youtu.be/X8QyT5RR-_M
-Using the [[Pythagorean Theorem]]:
+==Video Solution 2==
+https://youtu.be/5Bz7PC-tgyU
-<math>x^{2}=w^{2}+(2w)^{2}</math>
+~Charles3829
-<math>x^{2}=5w^{2}</math>
+==Solution 1==
-The [[area]] of the [[rectangle]] is <math>2w^2=\frac{5}{2}x^2</math>
+Let's set our length to <math>2</math> and our width to <math>1</math>.
-==See Also==
+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)
-*[[2005 AMC 10A Problems]]
-*[[2005 AMC 10A Problems/Problem 3|Previous Problem]]
+Now we can plug this value into the answer choices and test which one will give our desired area of <math>2</math>.
-*[[2005 AMC 10A Problems/Problem 5|Next Problem]]
+* 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>\boxed{\textbf{(B) }\frac{2}{5}x^2}</math>
+-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==
+{{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}}
- [[Category:Introductory Geometry Problems]]
 {{MAA Notice}}

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