20-16 As shown in the figure, $A B, C D$ are diameters of $\odot O$, and $A B \perp C D$, chord $D F$ intersects $A B$ at $E$. If $D E=6, E F=2$, then the area of the circle is 
(A) $23 \pi$
(B) $\frac{47}{2} \pi$.
(C) $24 \pi$.
(D) $\frac{49}{2} \pi$.
(E) $25 \pi$
(46th American High School Mathematics Examination, 1995)
[Solution] Connect $F$, and let the radius of $\odot O$ be $R$.
From $\triangle D O E \backsim \triangle D F C$, we have $\frac{O D}{D E}=\frac{F D}{D C}$, which is $\frac{R}{6} = \frac{(2+6)}{2 R}$, leading to $R^{2}=24$. Thus, $S_{\odot O}=\pi R^{2}=24 \pi$.
Therefore, the answer is $(C)$.