jmc

Let $h$ be the height of the trapezoid. The height of the trapezoid is also a height of $ABC$ and of $ADC$. Specifically, we have $[ABC]=(AB)(h)/2$ and $[ADC]=(CD)(h)/2$, so $[ABC]:[ADC]=AB:CD$. Since we are given that this area ratio equals $7:3$, we know that $AB:CD=7:3$. Therefore, $AB=7x$ and $CD=3x$ for some value of $x$. Since $AB+CD=210$ cm, we have $7x+3x=210$, so $10x=210$ and $x=21$. Therefore, $AB=7\times 21=\boxed{147\text{ cm}}$.