2002 AMC 10B Problem 17

Draw the octagon with $AB$ horizontal at the bottom; then $CD$ is vertical on the right, $EF$ horizontal on top, and $GH$ vertical on the left, while $BC, DE, FG, HA$ are the four slanted sides. Extending the four axis-parallel sides produces a square, and each slanted side is the hypotenuse of a cut-off isosceles right triangle. Hypotenuse $2$ means legs $\frac{2}{\sqrt2} = \sqrt2$, so the square has side
$$\sqrt2 + 2 + \sqrt2 = 2 + 2\sqrt2.$$

Now locate the triangle's vertices. $D$ is the top of the right vertical side and $G$ is the top of the left vertical side, so $DG$ is horizontal and spans the full width of the square: $DG = 2 + 2\sqrt2$. Both $D$ and $G$ sit at height $\sqrt2 + 2$ above the square's bottom edge (a corner leg plus a vertical side), and $A$ lies on that bottom edge. So the altitude from $A$ to $DG$ is $2 + \sqrt2$.

$$[ADG] = \tfrac12 (2 + 2\sqrt2)(2 + \sqrt2) = \tfrac12\left(4 + 2\sqrt2 + 4\sqrt2 + 4\right) = \tfrac12(8 + 6\sqrt2) = 4 + 3\sqrt2.$$

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

A regular octagon is a square with four congruent isosceles right triangles trimmed from its corners; that single picture gives every length in the figure in terms of the side. Choosing the base $DG$ parallel to a side of the bounding square makes the height fall out as a sum of known pieces, so no trigonometry is needed.

Coordinates from the same box: $A = (\sqrt2, 0)$, $D = (2 + 2\sqrt2,\ 2 + \sqrt2)$, $G = (0,\ 2 + \sqrt2)$. The base $DG$ is horizontal with length $2 + 2\sqrt2$ and the height is the $y$-coordinate $2 + \sqrt2$, giving the same product.

• Taking the cut-off corner triangles to have legs 2 (instead of 2/sqrt(2) = sqrt(2)), which inflates every length and yields a wrong area.
• Misidentifying $D$ and $G$ so that the chosen base is a slanted diagonal, which forces a much harder height computation.