jmc

Label the feet of the altitudes from $B$ and $C$ as $E$ and $F$ respectively. Considering right triangles $AEB$ and $DFC$, $AE=\sqrt{10^2-8^2}=\sqrt{36}=6\text{ cm}$, and $FD=\sqrt{17^2-8^2}=\sqrt{225}=15\text{ cm}$. So the area of $△AEB$ is $\frac{1}{2}(6)(8)=24{\text{ cm}}^2$, and the area of $△DFC$ is $\left(\frac{1}{2}\right)(15)(8)=60{\text{ cm}}^2$. Rectangle $BCFE$ has area $164-(24+60)=80{\text{ cm}}^2$. Because $BE=CF=8$ cm, it follows that $BC=\boxed{10\text{ cm}}$.