Difference between revisions of "1962 AHSME Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | {{ | + | The formula for the area of a regular dodecagon is <math>3r^2</math>. The answer is <math>\boxed{\textbf{(A)}}</math>. |
| + | (If you don't know this formula, it's pretty easy to figure out that the area of a square inscribed in a circle is <math>2r^2</math>, and all the choices except <math>3r^2</math> are less than <math>2r^2</math>. Remember, the more sides a regular polygon has, the closer its area gets to <math>\pi r^2</math>.) | ||
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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 17|num-a=19}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:17, 3 October 2014
Problem
A regular dodecagon (
sides) is inscribed in a circle with radius 
inches. The area of the dodecagon, in square inches, is:
Solution
The formula for the area of a regular dodecagon is 
. The answer is 
.
(If you don't know this formula, it's pretty easy to figure out that the area of a square inscribed in a circle is 
, and all the choices except are less than . Remember, the more sides a regular polygon has, the closer its area gets to 
.)
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
