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

Revision as of 13:23, 13 August 2019

Problem

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

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

Solution

Let the width of the rectangle be $w$ . Then the length is $2w$ .

Using the Pythagorean Theorem:

$x^{2}=w^{2}+(2w)^{2}$

$x^{2}=5w^{2}$

The area of the rectangle is $2w^2=\frac{2}{5}x^2$

$(B)$

2005 AMC 10A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 <math>(B)</math>
-==See Also==
+==See also==
+{{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}}
-{{AMC10 box|year=2005|ab=A|before=Problem 3|num-a=5}}
+[[Category:Introductory Number Theory Problems]]
-[[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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

Revision as of 13:23, 13 August 2019

Problem

Solution

See also