2006 AMC 12B Problems/Problem 16

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 problem, $\frac{5\sqrt{6}}{3}$ into the hexagon area formula, $\frac{3 (5 \sqrt{2} )^2 \sqrt{3} }{2}=25\sqrt{3} \implies \mathrm{(C)}$ <-- 5 root 6 over 3 instead of 5 root 2 btw

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2006 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 2

See also