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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.
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a distance of <math>\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}</math> apart.  Half of this distance is the length of the longer leg of the right triangles.  Therefore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
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Drawing in points <math>A</math>, <math>B</math>, and <math>C</math>, and connecting <math>A</math> and <math>C</math> with an auxiliary line, we see two 30-60-90 triangles are formed.
 
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Points <math>A</math> and <math>C</math> are a distance of <math>\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}</math> apart.  Half of this distance is the length of the longer leg of the right triangles.  Therefore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
 
  
 
The apothem is thus <math>\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}</math>.
 
The apothem is thus <math>\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}</math>.
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 16:41, 16 June 2018

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. 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 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AMC 12 Problems and Solutions

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