Difference between revisions of "1962 AHSME Problems/Problem 18"

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The formula for the area of a regular dodecagon is <math>3r^2</math>. The answer is <math>\boxed{\textbf{(A)}}</math>.
 
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>.)
 
(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==
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{{AHSME 40p box|year=1962|before=Problem 17|num-a=19}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 21:17, 3 October 2014

Problem

A regular dodecagon ($12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:

$\textbf{(A)}\ 3r^2\qquad\textbf{(B)}\ 2r^2\qquad\textbf{(C)}\ \frac{3r^2\sqrt{3}}{4}\qquad\textbf{(D)}\ r^2\sqrt{3}\qquad\textbf{(E)}\ 3r^2\sqrt{3}$

Solution

The formula for the area of a regular dodecagon is $3r^2$. The answer is $\boxed{\textbf{(A)}}$. (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 $2r^2$, and all the choices except $3r^2$ are less than $2r^2$. Remember, the more sides a regular polygon has, the closer its area gets to $\pi r^2$.)

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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