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

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==Solution==
 
==Solution==
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{{solution}}
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Clearly, the other side of <math>R_1</math> has length <math>\frac{12}2=6</math>.
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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
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<cmath>a^2+(3a)^2=225</cmath>
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<cmath>10a^2=225</cmath>
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<cmath>a^2=\frac{45}2</cmath>
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<cmath>a=\frac{3\sqrt{10}}2</cmath>
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<cmath>b=\frac{9\sqrt{10}}2</cmath>
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<cmath>[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}</cmath>
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 15|num-a=17}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 22:16, 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 \[a^2+(3a)^2=225\] \[10a^2=225\] \[a^2=\frac{45}2\] \[a=\frac{3\sqrt{10}}2\] \[b=\frac{9\sqrt{10}}2\] \[[R_2]=ab=\boxed{\frac{135}2\textbf{ (C)}}\]

See Also

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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