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1978 AHSME Problems/Problem 7: Difference between revisions

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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>                   
== Problem 7==
<math>\boxed {E}</math>
Opposite sides of a regular hexagon are <math>12</math> inches apart. The length of each side, in inches, is
 
<math>\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}    </math>
 
 
== Solution ==
 
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>
 
 
==See Also==
{{AHSME box|year=1978|num-b=6|num-a=8}}
{{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 (Problems • Answer Key • Resources)
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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