2002 AMC 10B Problems/Problem 17

Revision as of 14:43, 11 July 2012 by Swimmerstar (talk | contribs) (Solution)

Problem

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

$\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2$

Solution

[asy] unitsize(1cm); defaultpen(0.8); pair[] A = new pair[8]; A[0]=(0,0); for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1)); draw( A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle ); label("$A$",A[0],SW); label("$B$",A[1],SE); label("$C$",A[2],SE); label("$D$",A[3],NE); label("$E$",A[4],NE); label("$F$",A[5],NW); label("$G$",A[6],NW); label("$H$",A[7],SW); filldraw( A[0]--A[3]--A[6]--cycle, lightgray, black ); pair P = intersectionpoint( A[3]--A[6], A[0]--A[5] ); draw( A[0]--P ); draw( P -- A[5], dashed ); label("$P$",P,NE); draw( A[1]--A[4], dashed ); pair Q = intersectionpoint( A[3]--A[6], A[1]--A[4] ); label("$Q$",Q,NW); [/asy]

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 length $2$, 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 $\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = 4+3\sqrt 2 \Rightarrow$ $\boxed{(C)}$.

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions
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