2014 AMC 10B Problems/Problem 13

Problem

Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle{ABC}$ ?

$[asy] draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0)); draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0)); draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0)); draw((15,8.66025404)--(10, 17.3205081)--(15, 25.9807621)--(25, 25.9807621)--(30, 17.3205081)--(25, 8.66025404)--(15, 8.66025404)); draw((15,-8.66025404)--(10, -17.3205081)--(15, -25.9807621)--(25, -25.9807621)--(30, -17.3205081)--(25, -8.66025404)--(15, -8.66025404)); label("A", (0,0), W); label("B", (30, 17.3205081), NE); label("C", (30, -17.3205081), SE); draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); //(Diagram Creds-DivideBy0) [/asy]$

$\textbf {(A) } 2\sqrt{3} \qquad \textbf {(B) } 3\sqrt{3} \qquad \textbf {(C) } 1+3\sqrt{2} \qquad \textbf {(D) } 2+2\sqrt{3} \qquad \textbf {(E) } 3+2\sqrt{3}$

Solution

We note that the $6$ triangular sections in $\triangle{ABC}$ can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as $\dfrac{6\sqrt{3}}{4}$ . The area of $\triangle{ABC}$ , which is equivalent to two of these hexagons together, is $\boxed{\textbf{(B)} 3\sqrt{3}}$ .

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2014 AMC 10B Problems/Problem 13

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