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
Contents
Problem
Six regular hexagons surround a regular hexagon of side length as shown. What is the area of ?
Solution
We note that the triangular sections in 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 . The area of , which is equivalent to two of these hexagons together, is .
Solution 2
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 .
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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