Difference between revisions of "2006 AMC 12B Problems/Problem 16"

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== Solution ==
 
== Solution ==
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To find the area of the regular hexagon, we only need to calculate the side length.
 
To find the area of the regular hexagon, we only need to calculate the side length.
  

Revision as of 19:47, 16 February 2013

Problem

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

$\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$

Solution

To find the area of the regular hexagon, we only need to calculate the side length.

Drawing in points $A$, $B$, and $C$, and connecting $A$ and $C$ with an auxiliary line, we see two 30-60-90 triangles are formed.

Points $A$ and $C$ are a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.

The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}$.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions
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