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2005 AMC 12B Problems/Problem 3: Difference between revisions

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== Solution ==
== Solution ==
<asy>
size(5cm);
pen f = fontsize(10);
pair A = (0,0);
pair B = (0,1);
pair C = (2,1);
pair DD = (2,0);
D(A--B--C--DD--cycle);
D(A--C);
MP("2w",(A+DD)/2,plain.N,f);
MP("w",(C+DD)/2,plain.E,f);
MP("x",(A+C)/2,plain.NW,f);
</asy>
Using the Pythagorean theorem, we have
<math>w^2+(2w)^2=x^2 \Rightarrow 5w^2 = x^2 \Rightarrow w^2 = x^2/5.</math>
<math>\therefore \mbox{Area} = w \cdot 2w = 2w^2 = \boxed{\frac25x^2}</math>.


== See also ==
== See also ==
* [[2005 AMC 12B Problems]]
* [[2005 AMC 12B Problems]]

Revision as of 20:47, 17 April 2009

Problem

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

$\mathrm{(A)}\ \frac14x^2 \qquad \mathrm{(B)}\ \frac25x^2 \qquad \mathrm{(C)}\ \frac12x^2 \qquad \mathrm{(D)}\ x^2 \qquad \mathrm{(E)}\ \frac32x^2$

Solution

[asy] size(5cm); pen f = fontsize(10); pair A = (0,0); pair B = (0,1); pair C = (2,1); pair DD = (2,0); D(A--B--C--DD--cycle); D(A--C); MP("2w",(A+DD)/2,plain.N,f); MP("w",(C+DD)/2,plain.E,f); MP("x",(A+C)/2,plain.NW,f); [/asy] Using the Pythagorean theorem, we have

$w^2+(2w)^2=x^2 \Rightarrow 5w^2 = x^2 \Rightarrow w^2 = x^2/5.$

$\therefore \mbox{Area} = w \cdot 2w = 2w^2 = \boxed{\frac25x^2}$.

See also

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