Difference between revisions of "1998 AHSME Problems/Problem 15"
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| − | == Problem | + | == Problem== |
A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon? | A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon? | ||
<math> \mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6 </math> | <math> \mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6 </math> | ||
| − | + | ==Solution== | |
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| + | <math>A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}</math> | ||
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| + | <math>A_{hex} = \frac{6s_h^2\sqrt{3}}{4}</math> since a regular hexagon is just six equilateral triangles. | ||
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| + | Setting the areas equal, we get: | ||
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| + | <math>s_t^2 = 6s_h^2</math> | ||
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| + | <math>\left(\frac{s_t}{s_h}\right)^2 = 6</math> | ||
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| + | <math>\frac{s_t}{s_h} = \sqrt{6}</math>, and the answer is <math>\boxed{C}</math>. | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1998|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:29, 5 July 2013
Problem
A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
Solution
since a regular hexagon is just six equilateral triangles.
Setting the areas equal, we get:
, and the answer is 
.
See also
| 1998 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
