1962 AMC 12 Problem 18

Attempt Problem 18 of the 1962 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1962 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

A regular dodecagon (

12

sides) is inscribed in a circle with radius

r

inches. The area of the dodecagon, in square inches, is:

3r^2

2r^2

\dfrac{3r^2\sqrt3}{4}

r^2\sqrt3

3r^2\sqrt3

Divide the dodecagon into

12

triangles with vertex at the center

Each central angle is

30^\circ

, so use

\frac12r^2\sin30^\circ

Each of the

12

central triangles has area

\frac12r^2\sin30^\circ=\frac{r^2}{4}.

Their total area is

12(\frac{r^2}{4})=3r^2.

Thus, the correct answer is A.

Problem 18 in other years