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 | + | To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is (x^2<math>\sqrt{3}</math>)/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 15:28, 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
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
)/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, 
.
