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

Latest revision as of 21:16, 3 October 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 $a^2+(3a)^2=225$ $10a^2=225$ $a^2=\frac{45}2$ $a=\frac{3\sqrt{10}}2$ $b=\frac{9\sqrt{10}}2$ $[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}$

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 <cmath>b=\frac{9\sqrt{10}}2</cmath>
 <cmath>[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}</cmath>
+==See Also==
+{{AHSME 40p box|year=1962|before=Problem 15|num-a=17}}
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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

Latest revision as of 21:16, 3 October 2014

Problem

Solution

See Also