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

For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by: $AC = \sqrt{l^2 + w^2}$ To minimize $AC$ , $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$ . So, the minimum $AC$ is: $AC = \sqrt{5^2 + 5^2} = \boxed{\textbf{(B) } \sqrt{50}}$ ~ proloto

1967 AHSC (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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