imc

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 AG and CH meet at M. What is the area of quadrilateral ABCM? FIGURE

Accepted Answer

FIGURE We can use coordinates to solve this. Let H=(0,0). Thus, we have A=(0,12), C=(4,8), G=(4,0). Therefore, AG has equation -3x+12=y and HC has equation 2x=y. Solving, we have M=( 125, 245). Using the Shoelace Theorem (or you could connect LC and solve for the resulting triangle + trapezoid areas), we find [ABCM]= 885.