1962 AHSME Problems/Problem 16

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 \[a^2+(3a)^2=225\] \[10a^2=225\] \[a^2=\frac{45}2\] \[a=\frac{3\sqrt{10}}2\] \[b=\frac{9\sqrt{10}}2\] \[[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}\]

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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