Difference between revisions of "1962 AHSME Problems/Problem 16"

Revision as of 10:09, 17 April 2014

Problem

Given rectangle $R_1$ with one side $2$ inches and area $12$ square inches. Rectangle $R_2$ with diagonal $15$ inches is similar to $R_1$ . Expressed in square inches the area of $R_2$ is:

$\textbf{(A)}\ \frac{9}2\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 9\sqrt{10}\qquad\textbf{(E)}\ \frac{27\sqrt{10}}{4}$

Solution

Clearly, the other side of $R_1$ has length $\frac{12}2=6$ . Now call the sides of $R_2$ a and b, with $b>a$ . We know that $\frac{b}a=\frac31=3$ , because the two rectangles are similar. We also know that $a^2+b^2=15^2=225$ . But $b=3a$ , so substituting gives $a^2+(3a)^2=225$ $10a^2=225$ $a^2=\frac{45}2$ $a=\frac{3\sqrt{10}}2$ $b=\frac{9\sqrt{10}}2$ $[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}$

@@ Line 5: / Line 5: @@
 ==Solution==
-{{solution}}
+Clearly, the other side of <math>R_1</math> has length <math>\frac{12}2=6</math>.
+Now call the sides of <math>R_2</math> a and b, with <math>b>a</math>. We know that <math>\frac{b}a=\frac31=3</math>, because the two rectangles are similar. We also know that <math>a^2+b^2=15^2=225</math>. But <math>b=3a</math>, so substituting gives
+<cmath>a^2+(3a)^2=225</cmath>
+<cmath>10a^2=225</cmath>
+<cmath>a^2=\frac{45}2</cmath>
+<cmath>a=\frac{3\sqrt{10}}2</cmath>
+<cmath>b=\frac{9\sqrt{10}}2</cmath>
+<cmath>[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}</cmath>

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Difference between revisions of "1962 AHSME Problems/Problem 16"

Revision as of 10:09, 17 April 2014

Problem

Solution