Difference between revisions of "1978 AHSME Problems/Problem 7"

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Draw a perpendicular through the midpoint of the line of length <math>12</math> such that it passes through a vertex. We now have created <math>2</math> <math>30-60-90</math> triangles. Using the ratios, we get that the hypotenuse is <math>6 \times \frac {2}{\sqrt{3}}</math> <math>= \frac {12}{\sqrt {3}}</math> <math>= 4\sqrt{3}</math>                   
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== Problem 7==
<math>\boxed {E}</math>
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Opposite sides of a regular hexagon are <math>12</math> inches apart. The length of each side, in inches, is
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<math>\textbf{(A) }7.5\qquad
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\textbf{(B) }6\sqrt{2}\qquad
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\textbf{(C) }5\sqrt{2}\qquad
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\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad
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\textbf{(E) }4\sqrt{3}    </math>
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== Solution ==
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Draw a perpendicular through the midpoint of the line of length <math>12</math> such that it passes through a vertex. We now have created <math>2</math> <math>30-60-90</math> triangles. Using the ratios, we get that the hypotenuse is <math>6 \times \frac {2}{\sqrt{3}}</math> <math>= \frac {12}{\sqrt {3}}</math> <math>= 4\sqrt{3} \rightarrow \boxed {\textbf{(E)}}</math>
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==See Also==
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{{AHSME box|year=1978|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 00:06, 22 February 2024

Problem 7

Opposite sides of a regular hexagon are $12$ inches apart. The length of each side, in inches, is

$\textbf{(A) }7.5\qquad \textbf{(B) }6\sqrt{2}\qquad \textbf{(C) }5\sqrt{2}\qquad  \textbf{(D) }\frac{9}{2}\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$


Solution

Draw a perpendicular through the midpoint of the line of length $12$ such that it passes through a vertex. We now have created $2$ $30-60-90$ triangles. Using the ratios, we get that the hypotenuse is $6 \times \frac {2}{\sqrt{3}}$ $= \frac {12}{\sqrt {3}}$ $= 4\sqrt{3} \rightarrow \boxed {\textbf{(E)}}$


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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
All AHSME Problems and Solutions

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