2006 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2006 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

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Concepts:regular polygon distance formula area

Difficulty rating: 1740

16.

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0, 0)$ and $(7, 1),$ respectively. What is its area?

$20\sqrt{3}$

$22\sqrt{3}$

$25\sqrt{3}$

$27\sqrt{3}$

$50$

Solution:

The distance is $AC = \sqrt{7^2 + 1^2} = \sqrt{50}.$ In a regular hexagon with side $s,$ the distance between vertices two apart is $s\sqrt3,$ so $s^2 \cdot 3 = 50,$ giving $s^2 = \dfrac{50}{3}.$

The hexagon's area is $\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3.$

Thus, the correct answer is C.

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