3 circles in an iso. trap.

Let the lower left circle touches the line $\overline {AB}$ at $H$ , mid point of $\overline {AB}$ be $G$ , mid point of $\overline {CD}$ be $F$ and $\overline {AC}$ and $\overline {BD}$ meet at $E$ . Let $|\overline {AH}|=x$ and $|\overline {EF}|=y$ . Then $\dfrac{r}{x}=\dfrac{1}{2}$ , or $x=2r$ . Again $\dfrac{y+4r}{\dfrac{7r}{2}}=\dfrac{4}{3}$ , or $y=\dfrac{2r}{3}$ . Also, $\dfrac{y}{|\overline {CD}|}=\dfrac{y+4r}{7r}$ . Therefore $|\overline {CD}|=r$ , and the required ratio is $\dfrac{7r}{r}=\boxed 7$