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Difference between revisions of "1962 AHSME Problems/Problem 6"

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==Solution==
 
==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*<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>.
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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 <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>.
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 5|num-a=7}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

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 $9 \sqrt{3}$ 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 $\dfrac{x^2\sqrt{3}}{4}$. This has to be equal to $9 \sqrt{3}$, which means that $x=6$, or the side length of the triangle is $6$. Thus, the triangle (and the square) have a perimeter of $18$. It follows that each side of the square is $\dfrac{18}{4}=4.5$. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is $\boxed{\text{(D)} \dfrac{9\sqrt{2}}{2}}$.

See Also

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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