Difference between revisions of "2006 AMC 12B Problems/Problem 16"
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== Solution == | == Solution == | ||
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To find the area of the regular hexagon, we only need to calculate the side length. | To find the area of the regular hexagon, we only need to calculate the side length. | ||
Revision as of 18:47, 16 February 2013
Problem
Regular hexagon has vertices and at and , respectively. What is its area?
Solution
This solution needs a diagram To find the area of the regular hexagon, we only need to calculate the side length.
Drawing in points , , and , and connecting and with an auxiliary line, we see two 30-60-90 triangles are formed.
Points and are a distance of apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is .
The apothem is thus , yielding an area of .
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |