Difference between revisions of "1962 AHSME Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is | + | To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is <math>\dfrac{x^2\sqrt{3}}{4}</math>. This has to be equal to <math>9 \sqrt{3}</math>, which means that <math>x=6</math>, or the side length of the triangle is <math>6</math>. Thus, the triangle (and the square) have a perimeter of <math>18</math>. It follows that each side of the square is <math>\dfrac{18}{4}=4.5</math>. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is <math>\boxed{\text{(D)} \dfrac{9\sqrt{2}}{2}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 14:55, 18 April 2018
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
To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is . This has to be equal to , which means that , or the side length of the triangle is . Thus, the triangle (and the square) have a perimeter of . It follows that each side of the square is . If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is .
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AHSME Problems and Solutions |
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