2002 AMC 10B Problem 17

Question

A regular octagon

ABCDEFGH

has sides of length two. What is the area of

\triangle ADG?

4 + 2\sqrt{2}

6 + \sqrt{2}

4 + 3\sqrt{2}

3 + 4\sqrt{2}

8 + \sqrt{2}

Place the octagon on a grid; each slanted side moves

\sqrt2

horizontally and vertically

DG

is a horizontal chord; use it as the base and the vertical distance from

A

as the height

Set the octagon on coordinate axes with the axis-aligned sides of length

2

and each slanted side spanning

\sqrt2

horizontally and

\sqrt2

vertically. Then

\begin{aligned} A &= (\sqrt2, 0), \\ D &= (2 + 2\sqrt2, 2 + \sqrt2), \\ G &= (0, 2 + \sqrt2). \end{aligned}

Since

D

and

G

share the height

2 + \sqrt2,

segment

DG

is horizontal with length

2 + 2\sqrt2,

and the height from

A

up to that level is

2 + \sqrt2.

Therefore

\begin{aligned} [\triangle ADG] &= \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) \\ &= (1 + \sqrt2)(2 + \sqrt2) \\ &= 4 + 3\sqrt2. \end{aligned}

Thus, the correct answer is C.

Problem 17 in other years