1961 AHSME Problems/Problem 14

Problem

A rhombus is given with one diagonal twice the length of the other diagonal. Express the side of the rhombus is terms of $K$ , where $K$ is the area of the rhombus in square inches.

$\textbf{(A)}\ \sqrt{K}\qquad \textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad \textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad \textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad \textbf{(E)}\ \text{None of these are correct}$

Solution

$[asy] draw((40,0)--(0,20)--(-40,0)--(0,-20)--(40,0)); draw((-40,0)--(40,0)); draw((0,3)--(3,3)--(3,0)); draw((0,-20)--(0,20)); label("$\frac{a}{2}$",(0,10),W); label("$a$",(-15,0),N); [/asy]$

Let one of the diagonals of the rhombus be $a$ units long, so the other diagonal is $2a$ units long. That means the area of the rhombus is $\frac{1}{2} \cdot 2a \cdot a = a^2$ , so $K = a^2$ .

From the Pythagorean Theorem, one of the side lengths of the rhombus is $\sqrt{(\frac{a}{2})^2 + a^2} = \sqrt{\frac{5a^2}{4}} = \frac{\sqrt{5K}}{2}$ , so the answer is $\boxed{\textbf{(E)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1961 AHSME Problems/Problem 14

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