Difference between revisions of "2002 AMC 10B Problems/Problem 17"
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The area of the triangle <math>ADG</math> can be computed as <math>\frac{DG \cdot AP}2</math>. We will now find <math>DG</math> and <math>AP</math>. | The area of the triangle <math>ADG</math> can be computed as <math>\frac{DG \cdot AP}2</math>. We will now find <math>DG</math> and <math>AP</math>. | ||
− | Clearly, <math>PFG</math> is a right isosceles triangle with hypotenuse of | + | Clearly, <math>PFG</math> is a right isosceles triangle with hypotenuse of length <math>2</math>, hence <math>PG=\sqrt 2</math>. |
The same holds for triangle <math>QED</math> and its leg <math>QD</math>. The length of <math>PQ</math> is equal to <math>FE=2</math>. | The same holds for triangle <math>QED</math> and its leg <math>QD</math>. The length of <math>PQ</math> is equal to <math>FE=2</math>. | ||
Hence <math>GD = 2 + 2\sqrt 2</math>, and <math>AP = PD = 2 + \sqrt 2</math>. | Hence <math>GD = 2 + 2\sqrt 2</math>, and <math>AP = PD = 2 + \sqrt 2</math>. |
Revision as of 13:43, 11 July 2012
Problem
A regular octagon has sides of length two. Find the area of .
Solution
The area of the triangle can be computed as . We will now find and .
Clearly, is a right isosceles triangle with hypotenuse of length , hence . The same holds for triangle and its leg . The length of is equal to . Hence , and .
Then the area of equals .
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |