Difference between revisions of "2002 AMC 10B Problems/Problem 17"
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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>. | ||
− | Then the area of <math>ADG</math> equals <math>\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = 4+3\sqrt 2 | + | Then the area of <math>ADG</math> equals <math>\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}</math>. |
== See Also == | == See Also == |
Latest revision as of 13:27, 21 May 2021
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 | |
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All AMC 10 Problems and Solutions |
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