2014 AMC 10B Problem 13

Attempt Problem 13 of the 2014 AMC 10B 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 2014 AMC 10B solutions, or check the answer key.

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13.

Six regular hexagons surround a regular hexagon of side length 11 as shown. What is the area of ABC?\triangle{ABC}?

23 2\sqrt{3}

33 3\sqrt{3}

1+32 1+3\sqrt{2}

2+23 2+2\sqrt{3}

3+23 3+2\sqrt{3}

Answer: B
Concepts:regular polygonequilateral triangletriangle area
Difficulty rating: 1480
Solution:

Follow the 6060^\circ grid formed by the unit hexagons and place A=(0,0)A=(0,0). The marked vertices may then be written as B=(3,3)B=(3,\sqrt3) and C=(3,3)C=(3,-\sqrt3).

Thus BC=23BC=2\sqrt3, and AB=AC=32+(3)2=23. \begin{aligned} AB=AC&=\sqrt{3^2+(\sqrt3)^2}\\ &=2\sqrt3. \end{aligned} Hence ABC\triangle ABC is equilateral, with area (23)234=33. \frac{(2\sqrt3)^2\sqrt3}{4}=3\sqrt3.

Thus, the correct answer is B .

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