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

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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>.
  
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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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} \implies \mathrm{(C)}</math>.
  
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>.
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== Solution 2 ==
  
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>.
+
Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this very problem, <math>\frac{5\sqrt{6}}{3}</math> into the hexagon area formula, <math>\frac{3  (5 \sqrt{2} )^2 \sqrt{3} }{2}=25\sqrt{3}</math>
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:05, 16 August 2021

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 $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}$.

Solution 2

Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this very problem, $\frac{5\sqrt{6}}{3}$ into the hexagon area formula, $\frac{3  (5 \sqrt{2} )^2 \sqrt{3} }{2}=25\sqrt{3}$

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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