Difference between revisions of "2005 AMC 12B Problems/Problem 3"

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

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

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

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Difference between revisions of "2005 AMC 12B Problems/Problem 3"

Revision as of 20:47, 17 April 2009

Problem

Solution

See also

@@ Line 11: / Line 11: @@
 == 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 ==
 * [[2005 AMC 12B Problems]]