Difference between revisions of "1962 AHSME Problems/Problem 6"
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| − | 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 <math>9 \sqrt{3}</math> 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}. | + | 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 <math>9 \sqrt{3}</math> 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, <math>\\frac{9\sqrt{2}}{2}</math>. |
Revision as of 15:27, 17 March 2014
Problem
A square and an equilateral triangle have equal perimeters. The area of the triangle is 
square inches. Expressed in inches the diagonal of the square is:
Solution
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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 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, 
.
