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 == | ||
{{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
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 | |
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 |
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