Difference between revisions of "1978 AHSME Problems/Problem 7"
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<math>\textbf{(A) }7.5\qquad | <math>\textbf{(A) }7.5\qquad | ||
\textbf{(B) }6\sqrt{2}\qquad | \textbf{(B) }6\sqrt{2}\qquad | ||
| − | \textbf{(C) }5\sqrt{2 | + | \textbf{(C) }5\sqrt{2}\qquad |
| − | |||
\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad | \textbf{(D) }\frac{9}{2}\sqrt{3}\qquad | ||
\textbf{(E) }4\sqrt{3} </math> | \textbf{(E) }4\sqrt{3} </math> | ||
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
