From the figure, $QS \perp PS$ (right angle at $P$), $QP = 12$, and $QR = 20$.

Area of $\triangle PQS = \dfrac{1}{2} \times PS \times QP = 45$

$\dfrac{1}{2} \times PS \times 12 = 45 \Rightarrow PS = \dfrac{90}{12} = 7.5$

Now find $SR$: In right triangle $QPR$ (right angle at $P$), using $QP = 12$ and $QR = 20$:

$PR = \sqrt{QR^2 - QP^2} = \sqrt{400 - 144} = \sqrt{256} = 16$

$SR = PR - PS = 16 - 7.5 = 8.5$

Column A $= PS = 7.5$, Column B $= SR = 8.5$. Column B is greater.