1967 AHSME Problems/Problem 11

Problem

If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$

Solution

The least possible value of the diagonal is when the side lengths are as close as possible. In this case, the perimeter is divisible by $4$ , so we can make a square. Since the perimeter is $20$ , $20$ divided by $4$ is $5$ . So when $5$ is the side length of the square, the diagonal is $5\sqrt2$ or $\sqrt50$ . $\fbox{B}$

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1967 AHSME Problems/Problem 11

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