Difference between revisions of "2002 AMC 10B Problems/Problem 17"
Mathcool2009 (talk | contribs) m |
(→Solution) |
||
Line 38: | Line 38: | ||
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 = \boxed{\textbf{(C)}4+3\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 = \boxed{\textbf{(C)}=4+3\sqrt 2}</math>. |
== See Also == | == See Also == |
Latest revision as of 22:42, 23 November 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.