1983 AMC 12 Problem 4

Question

In the adjoining plane figure, sides

AF

and

CD

are parallel, as are sides

AB

and

FE,

and sides

BC

and

ED.

Each side has length

1.

Also,

\angle FAB=\angle BCD=60^\circ.

The area of the figure is

\frac{\sqrt3}{2}

1

\frac32

\sqrt3

2

Place

AF

and

CD

vertically and use the

60^\circ

angles to assign coordinates

The six vertices can be written using horizontal changes of

\frac{\sqrt3}{2}

The parallel directions and

60^\circ

angles place the vertices, up to rigid motion, at

\begin{aligned} A&=(0,0),\\ B&=(\tfrac{\sqrt3}{2},-\tfrac12),\\ C&=(\sqrt3,0),\\ D&=(\sqrt3,-1),\\ E&=(\tfrac{\sqrt3}{2},-\tfrac32),\\ F&=(0,-1). \end{aligned}

Applying the shoelace formula to these six vertices gives area

\sqrt3.

Therefore, the correct answer is D.

Problem 4 in other years