20.11 As shown in the figure, a rectangle intersects a circle, where $AB=4, BC=5, DE=3$, then $EF$ equals
(A) 6.
(B) 7.
(C) $\frac{20}{3}$.
(D) 8.
(E) 9.
(35th American High School Mathematics Examination, 1984)
[Solution 1] As shown in Figure (1), draw the perpendicular bisector of chord $EF$ intersecting the circle at $S$ and $BC$ at $P$, which must pass through the center $O$. Therefore, $PO \perp BC$, and it bisects $BC$.
Since $PQ \parallel AD$, we have $DQ = AP = 4 + \frac{5}{2} = \frac{13}{2}$.
Thus, $EF = 2EQ = 2(DQ - DE) = 7$.
Therefore, the answer is (B).
[Solution 2] As shown in Figure (2), draw perpendiculars from $B$ and $C$ to $AC$, denoted as $BR$ and $CS$, respectively, making quadrilaterals $BCSR$ and $ABRD$ rectangles.
Given $ER = 1$, and by symmetry $SF = 1$. Also, $RS = BC = 5$.
Therefore,
$$
EF = 1 + 5 + 1 = 7.
$$

Therefore, the answer is (B).