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:
$\text{(A)} 0\quad\text{(B)} 1\quad\text{(C)} \sqrt{2}\quad\text{(D)} \frac{1}{\sqrt{2}}+\frac{1}{2}\quad\text{(E)} \frac{1}{\sqrt{2}}+1$
Let $OG = a - 2h$, where $h = JH = HG$. Since the areas of rectangle $EHGL$ and trapezoid $EHGC$ are both half of rectangle $LMFE$ and trapezoid $EFDC$, respectively, the ratios between their areas will remain the same, so let us consider rectangle $EHGL$ and trapezoid $EHGC$.
Draw radii $OC$ and $OE$, both of which obviously have length $a$. By the Pythagorean Theorem, the length of $EH$ is $\sqrt{a^2 - (OG + h)^2}$, and the length of $CG$ is $\sqrt{a^2 - OG^2}$. It follows that the area of rectangle $EHGL$ is \[EH\cdot HG = h\sqrt{a^2 - (OG + h)^2}\] while the area of trapezoid $EHGC$ is  \[\frac{HG}{2}(EH + CG)=\frac{h}{2}\left(\sqrt{a^2 - (OG + h)^2} + \sqrt{a^2 - OG^2}\right).\]
Now, we want to find the limit, as $OG$ approaches $a$, of $\frac{K}{R}$. Note that this is equivalent to finding the same limit as $h$ approaches $0$. Substituting $a - 2h$ into $OG$ yields that trapezoid $EHGC$ has area \[\frac{h}{2}\left(\sqrt{a^2 - (a - 2h + h)^2} + \sqrt{a^2 - (a - 2h)^2}\right) =\frac{h}{2}\left(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2}\right)\] and that rectangle $EHGL$ has area \[h\sqrt{a^2 - (a - 2h + h)^2} = h\left(\sqrt{2ah - h^2}\right).\] Our answer thus becomes
\begin{align*} \lim_{h\rightarrow 0}\dfrac{\frac{h}{2}\bigl(\sqrt{2ah - h^2} + \sqrt{4ah - 4h^2}\bigr)}{h\bigl(\sqrt{2ah - h^2}\bigr)} &= \lim_{h\rightarrow 0}\left[\dfrac{1}{2}\cdot\dfrac{\sqrt{h}\bigl(\sqrt{2a - h} + 2\sqrt{a - h}\bigr)}{\sqrt{h}\bigl(\sqrt{2a - h}\bigr)}\right] \\ \implies \frac{1}{2}\cdot\frac{\sqrt{2a} + 2\sqrt{a}}{\sqrt{2a}} &= \frac{1}{2}\left(1 + \frac{2}{\sqrt{2}}\right) = \boxed{\textbf{(D) }\frac{1}{2}+\frac{1}{\sqrt{2}}.} \end{align*}