In rectangle ABCD, angle C is trisected by CF and CE, where E is on AB, F is on AD, BE=6, and AF=2. Which of the following is closest to the area of the rectangle ABCD?

1996 AMC 12 Problem 17

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17.

In rectangle

ABCD,

angle

C

is trisected by

\overline{CF}

and

\overline{CE},

where

E

is on

\overline{AB},

F

is on

\overline{AD},

BE=6,

and

AF=2.

Which of the following is closest to the area of the rectangle

ABCD?

110

120

130

140

150

Each of the three angles at

C

30^\circ

Use the two right triangles adjacent to sides

CB

and

CD

to determine the rectangle’s height and width

Let the rectangle have width

w

and height

h.

In right triangle

CBE,

\tan30^\circ=\frac{BE}{CB}=\frac{6}{h},

h=6\sqrt3.

In the triangle using

C,F,D,

the horizontal run is

w

and the vertical drop is

h-2,

\tan60^\circ=\frac{w}{h-2}.

Hence

w=\sqrt3(6\sqrt3-2)=18-2\sqrt3.

The area is

6\sqrt3(18-2\sqrt3)

=108\sqrt3-36\approx151.1,

closest to

150.

Thus the correct answer is E.