Problem:
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?
Answer Choices:
A. 110
B. 120
C. 130
D. 140
E. 150
Solution:
In the 30∘−60∘−90∘ triangle CEB,BC=63​. Therefore, FD=AD−AF=63​−2. In the 30∘−60∘−90∘ triangle CFD,CD=FD3​=18−23​. The area of rectangle ABCD is