Difference between revisions of "2014 AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
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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>. | ||
==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 18:29, 20 February 2014
Problem
Six regular hexagons surround a regular hexagon of side length 
as shown. What is the area of 
?
(diagram needed)
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
