smc

Using Pythagoras' Theorem, $BD=\sqrt{3^2+4^2}=5$, and $AD=\sqrt{(13-3{)}^2+4^2}=\sqrt{116}$. Thus the sum is $5+\sqrt{116}$, and as $100<116<121$, $10<\sqrt{116}<11$, so that the sum is between $5+10$ and $5+11$, or $15$ and $16$, which is answer $\boxed{\text{between 15 and 16}}$.