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1962 AHSME Problems/Problem 6

Revision as of 12:23, 12 February 2018 by Hi3142 (talk | contribs) (Solution)

Problem

A square and an equilateral triangle have equal perimeters. The area of the triangle is $9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is:

$\textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these}$

Solution

To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is (x^2*$\sqrt{3}$)/4. This has to be equal to $9 \sqrt{3}$ so solving for x brings the answer to a side length to 6. Knowing this gives us a perimeter of 18 so each side length of the square is 4.5. Now we solve for the diagonal getting the answer of $\boxed{(D) \ 9\sqrt{2}/{2}}$

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AHSME Problems and Solutions

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