Difference of dipole moments of dichloromethane and trichloromethane?
Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.  

Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$
where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.  

For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$
which will be $1A$, lesser than in the case of dichloromethane.  

To visualise how can you get those angles and cosines, see this page.