Difference between revisions of "1967 AHSME Problems/Problem 11"

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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
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Revision as of 01:39, 16 August 2023

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 $\sqrt50$. $\fbox{B}$

See also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
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