$20 \cdot 88$ As shown, the center of the circle is $O$, with a radius of $a$ inches. Chord $E F$ is parallel to chord $C D, O$, $G, H, J$ are on the same line, $G$ is the midpoint of $C D$, let the area of trapezoid $C D F E$ be $K$ square inches, and the area of rectangle $E L M F$ be $R$ square inches. Now, move $E F, C D$ upwards, keeping $J H=H G$, then as $O G$ approaches $a$, the ratio $K: R$ can approach arbitrarily close to
(A) 0.
(B) 1.
(C) $\sqrt{2}$.
(D) $\frac{1}{\sqrt{2}}+\frac{1}{2}$.
(E) $\frac{1}{\sqrt{2}}+1$.
［Solution］From the area formulas of trapezoid and rectangle, we have
$$
\frac{K}{R}=\frac{\frac{1}{2} H G(E F+C D)}{H G \cdot E F}=\frac{1}{2}\left(1+\frac{C D}{E F}\right) .
$$

In $\mathrm{Rt} \triangle O G D$ and $\mathrm{Rt} \triangle O H F$, by the Pythagorean theorem, we have
$$
\frac{C D}{E F}=\frac{2 G D}{2 H F}=\frac{G D}{H F}=\frac{\sqrt{O D^{2}-O G^{2}}}{\sqrt{O F^{2}-O H^{2}}}=\frac{\sqrt{a^{2}-O G^{2}}}{\sqrt{a^{2}-O H^{2}}} \text {. }
$$

Let $J H=z$, then $J G=2 z$, and $O H=a-z, O G=a-2 z$. Thus, we have
$$
\frac{C D}{E F}=\frac{\sqrt{a^{2}-(a-2 z)^{2}}}{\sqrt{a^{2}-(a-z)^{2}}}=\frac{\sqrt{4 a z-4 z^{2}}}{\sqrt{2 a z-z^{2}}}=\frac{\sqrt{4 a-4 z}}{\sqrt{2 a-z}} .
$$

When $O G \rightarrow a$, $z \rightarrow 0$, so $\frac{C D}{E F} \rightarrow \frac{\sqrt{4 a}}{\sqrt{2 a}}=\sqrt{2}$.
Therefore, when $O G \rightarrow a$, $\frac{K}{R}=\frac{1}{2}\left(1+\frac{C D}{E F}\right) \rightarrow \frac{1}{2}(1+\sqrt{2})$. Hence, the answer is $(D)$.