The convex pentagon ABCDE has ∠ A=∠ B=120°, EA=AB=BC=2 and CD=DE=4. What is the area of ABCDE?

1993 AMC 12 Problem 14

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14.

The convex pentagon

ABCDE

has

\angle A=\angle B=120^\circ,

EA=AB=BC=2

and

CD=DE=4.

What is the area of

ABCDE?

10

7\sqrt3

15

9\sqrt3

12\sqrt5

Draw

\overline{CE}

to split the pentagon into a trapezoid and a triangle

The angle conditions give

CE=4,

\triangle CDE

is equilateral

Place

A=(0,0)

and

B=(2,0).

The

120^\circ

angles and side lengths

EA=BC=2

give

E=(-1,\sqrt3),\qquad C=(3,\sqrt3).

Thus

ABCE

is a trapezoid of height

\sqrt3

and bases

2

and

4,

so its area is

3\sqrt3.

Also

CE=4=CD=DE,

making

\triangle CDE

equilateral with area

4\sqrt3.

The total area is

7\sqrt3.

Thus the correct answer is B.