Question

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

$4 + 2\sqrt{2}$

$6 + \sqrt{2}$

$4 + 3\sqrt{2}$

$3 + 4\sqrt{2}$

$8 + \sqrt{2}$

Solution:

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 $A = (\sqrt2, 0), \quad D = (2 + 2\sqrt2, 2 + \sqrt2), \quad G = (0, 2 + \sqrt2).$

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 $[\triangle ADG] = \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) = (1 + \sqrt2)(2 + \sqrt2) = 4 + 3\sqrt2.$

Thus, the correct answer is C.

2002 AMC 10B Problem 17

Solution:

Problem 17 in Other Years