1953 AHSME Problems/Problem 46

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Problem 46

Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution

Let $x<y$ be the sides of the rectangle. The length of the diagonal is $\sqrt{x^2+y^2}$, and the length of the two adjacent sides is $x+y$. Then the distance the boy saves is $x+y-\sqrt{x^2+y^2}$. Setting this equal to $\frac12y$, we have \[x+y-\sqrt{x^2+y^2}=\frac12y\] \[x+\frac12y=\sqrt{x^2+y^2}\] \[x^2+xy+\frac14y^2=x^2+y^2\] \[xy=\frac34y^2\] \[\frac xy=\frac34,\] so the answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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