1962 AHSME Problems/Problem 6

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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 D, $\\frac{9\sqrt{2}}\{2}$ (Error compiling LaTeX. Unknown error_msg).

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1962 AHSME Problems/Problem 6

Problem

Solution