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2002 AMC 10B Problems/Problem 17

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Problem

A regular octagon $ABCDEFGH$ has sides of length two. Find the area of $\triangle ADG$ .

$\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \q...$

Solution

The area of the triangle $ADG$ can be computed as $\frac{DG \cdot AP}2$ . We will now find $DG$ and $AP$ .

Clearly, $PFG$ is a right isosceles triangle with hypotenuse of lenght $2$ , hence $PG=\sqrt 2$ . The same holds for triangle $QED$ and its leg $QD$ . The length of $PQ$ is equal to $FE=2$ . Hence $GD = 2 + 2\sqrt 2$ , and $AP = PD = 2 + \sqrt 2$ .

Then the area of $ADG$ equals $\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{4+3\sqrt 2}$ .

See Also

2002 AMC 10B (Problems • 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

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