2007 AMC 10A Problem 18

Question

Consider the

12

-sided polygon

ABCDEFGHIJKL,

as shown. Each of its sides has length

4,

and each two consecutive sides form a right angle. Suppose that

\overline{AG}

and

\overline{CH}

meet at

M.

What is the area of quadrilateral

ABCM?

\dfrac{44}{3}

16

\dfrac{88}{5}

20

\dfrac{62}{3}

Place the cross on coordinates using the side length

4

Find

M

as the intersection of lines

AG

and

CH,

then apply the shoelace formula to

ABCM

Put the figure on coordinates with

A = (-2, 6),

B = (2, 6),

C = (2, 2),

G = (2, -6),

and

H = (-2, -6).

Line

AG

is

y = -3x,

and line

CH

is

y = 2x - 2.

Their intersection is

M = \left(\tfrac25, -\tfrac65\right).

Applying the shoelace formula to

A, B, C, M

gives

\begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5}. \end{aligned}

Thus, the correct answer is C.

Problem 18 in other years