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} | + | == Problem 7== |
− | + | 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 inches apart. The length of each side, in inches, is
Solution
Draw a perpendicular through the midpoint of the line of length such that it passes through a vertex. We now have created triangles. Using the ratios, we get that the hypotenuse is
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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