1958 AMC 12 Problem 16

Attempt Problem 16 of the 1958 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 1958 AMC 12 solutions, or check the answer key.

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

16.

The area of a circle inscribed in a regular hexagon is 100π.100\pi. The area of the hexagon is:
600600
300300
2002200\sqrt2
2003200\sqrt3
1205120\sqrt5
Answer: D
Concepts:incircle, incenter, and inradiusregular polygonarea
Small hint:
The circle’s radius is the apothem of the regular hexagon
Big hint:
For apothem r,r, the hexagon side is 2rtan⁡30∘2r\tan30^\circ
Solution:
The circle has radius r=10,r=10, which is the hexagon’s apothem. Thus the side length is 2rtan⁡30∘=203. 2r\tan30^\circ=\frac{20}{\sqrt3}. Using one-half the product of perimeter and apothem, the hexagon’s area is 12(6⋅203)(10)=2003. \frac12\left(6\cdot\frac{20}{\sqrt3}\right)(10) =200\sqrt3.
Therefore, the correct answer is D.
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