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

Revision as of 10:25, 17 April 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 $\pi r^2$ .)

@@ Line 5: / Line 5: @@
 ==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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Difference between revisions of "1962 AHSME Problems/Problem 18"

Revision as of 10:25, 17 April 2014

Problem

Solution