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Difference between revisions of "2014 AMC 10B Problems/Problem 13"

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We note that the <math>6</math> triangular sections in <math>\triangle{ABC}</math> 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 <math>\dfrac{6\sqrt{3}}{4}</math>. The area of <math>\triangle{ABC}</math>, which is equivalent to two of these hexagons together, is <math>\boxed{\textbf{(B)}  3\sqrt{3}}</math>.
 
We note that the <math>6</math> triangular sections in <math>\triangle{ABC}</math> 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 <math>\dfrac{6\sqrt{3}}{4}</math>. The area of <math>\triangle{ABC}</math>, which is equivalent to two of these hexagons together, is <math>\boxed{\textbf{(B)}  3\sqrt{3}}</math>.
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==Solution 2==
 
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The area of triangle ABC consists of a regular hexagon and 6 triangles. Each one of these triangles can be split into two 30-60-90 triangles. Since the side length of the regular hexagon is 1, and the side lengths of a 30-60-90 triangle are in a ratio of 1:\sqrt{3}:2, the side length of triangle ABC is 4(\sqrt{3}/2)=2\sqrt{3}. Using the formula for area of an equilateral triangle, the area of triangle ABC is (((2\sqrt{3})^2)\sqrt{3})/2, which equals <math>\boxed{\textbf{(B)}  3\sqrt{3}}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:41, 22 February 2015

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 $\boxed{\textbf{(B)} 3\sqrt{3}}$.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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