Difference between revisions of "2006 AMC 12B Problems/Problem 16"
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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. | ||
− | + | a distance of <math>\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}</math> apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>. | |
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The apothem is thus <math>\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}</math>. | The apothem is thus <math>\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}</math>. |
Revision as of 16:41, 16 June 2018
Problem
Regular hexagon has vertices and at and , respectively. What is its area?
Solution
To find the area of the regular hexagon, we only need to calculate the side length. 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 |
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