# Difference between revisions of "2002 AMC 10B Problems/Problem 17"

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

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