AHSME 1968 Problem 35

In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD$. $O$,$G$,$H$,$J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF.$ Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ becomes arbitrarily close to: $\text{(A)} 0 \text{(B)} 1 \text{(C)} \sqrt{2} \text{(D)} \frac{1}{\sqrt{2}}+\frac{1}{2} \text{(E)} \frac{1}{\sqrt{2}}+1$