2002 AMC 10B Problem 17

Draw the octagon with side $AB$ flat along the bottom. Going around, $CD$ is the right vertical side, $EF$ the top, $GH$ the left vertical side, and $BC$, $DE$, $FG$, $HA$ are the four slanted sides.

1. See the octagon inside a square. Extend the four flat sides (bottom, right, top, left). They form a square, and the octagon is that square with four small corner triangles cut off. Each slanted side of the octagon is the long side of one corner triangle.
2. Size of a corner triangle. Each corner triangle has a right angle and two equal legs (a $45$-$45$-$90$ triangle). Its long side is $2$, so each leg is $\frac{2}{\sqrt2} = \sqrt2$ (check: $\sqrt2^2 + \sqrt2^2 = 4 = 2^2$).
3. Side of the big square. Along the bottom edge: a leg, then side $AB$, then another leg:
$$\sqrt2 + 2 + \sqrt2 = 2 + 2\sqrt2.$$
4. The base $DG$. $D$ is the top end of the right vertical side and $G$ is the top end of the left vertical side. They are at the same height, so $DG$ is flat and stretches across the whole square: $DG = 2 + 2\sqrt2$.
5. The height from $A$. $A$ lies on the bottom edge of the square. $D$ is above the bottom edge by one leg plus the vertical side $CD$: $\sqrt2 + 2$. So the height of the triangle, from $A$ straight up to line $DG$, is $2 + \sqrt2$.
6. Area.
$$\frac12 \cdot (2 + 2\sqrt2)(2 + \sqrt2) = \frac12\,(4 + 2\sqrt2 + 4\sqrt2 + 4) = \frac12\,(8 + 6\sqrt2) = 4 + 3\sqrt2.$$

The answer is $\boxed{\textbf{(C)}\ 4 + 3\sqrt2}$.

"A regular octagon is a square with its corners cut off" is the one picture that gives every length in an octagon. Choosing $DG$ as the base was the smart move: it is parallel to a side of the square, so the height is just a sum of lengths you already know, with no need for angles or trigonometry.

Put the octagon on a grid with the big square's bottom-left corner at $(0, 0)$. Then $A = (\sqrt2, 0)$, $G = (0,\ 2 + \sqrt2)$ and $D = (2 + 2\sqrt2,\ 2 + \sqrt2)$. The base $DG$ is flat with length $2 + 2\sqrt2$, and the height is the difference in heights, $2 + \sqrt2$. Half their product is again $4 + 3\sqrt2$.