The area of trapezoid $ ABCD$ is $ 164 \text{cm}^2$.  The altitude is $ 8 \text{cm}$, $ AB$ is $ 10 \text{cm}$, and $ CD$ is $ 17 \text{cm}$.  What is $ BC$, in centimeters?
[asy]/* AMC8 2003 #21 Problem */
size(4inch,2inch);
draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);
draw((11,8)--(11,0), linetype("8 4"));
draw((11,1)--(12,1)--(12,0));
label("$A$", (0,0), SW);
label("$D$", (31,0), SE);
label("$B$", (6,8), NW);
label("$C$", (16,8), NE);
label("10", (3,5), W);
label("8", (11,4), E);
label("17", (22.5,5), E);[/asy]

$ \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$
1. **Identify the given values and the formula for the area of a trapezoid:**
   - The area of trapezoid \(ABCD\) is \(164 \text{ cm}^2\).
   - The altitude (height) is \(8 \text{ cm}\).
   - The lengths of the parallel sides \(AB\) and \(CD\) are \(10 \text{ cm}\) and \(17 \text{ cm}\) respectively.
   - We need to find the length of \(BC\).

2. **Use the area formula for a trapezoid:**
   \[
   \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
   \]
   Substituting the given values:
   \[
   164 = \frac{1}{2} \times (10 + 17) \times 8
   \]
   Simplify the equation:
   \[
   164 = \frac{1}{2} \times 27 \times 8
   \]
   \[
   164 = \frac{1}{2} \times 216
   \]
   \[
   164 = 108
   \]
   This confirms that the given area and dimensions are consistent.

3. **Draw perpendiculars from \(B\) and \(C\) to \(AD\):**
   - Let \(E\) be the foot of the perpendicular from \(B\) to \(AD\).
   - Let \(F\) be the foot of the perpendicular from \(C\) to \(AD\).

4. **Calculate the lengths \(AE\) and \(DF\) using the Pythagorean theorem:**
   - In \(\triangle ABE\):
     \[
     AB = 10 \text{ cm}, \quad BE = 8 \text{ cm}
     \]
     \[
     AE^2 + BE^2 = AB^2
     \]
     \[
     AE^2 + 8^2 = 10^2
     \]
     \[
     AE^2 + 64 = 100
     \]
     \[
     AE^2 = 36
     \]
     \[
     AE = 6 \text{ cm}
     \]

   - In \(\triangle CFD\):
     \[
     CD = 17 \text{ cm}, \quad CF = 8 \text{ cm}
     \]
     \[
     DF^2 + CF^2 = CD^2
     \]
     \[
     DF^2 + 8^2 = 17^2
     \]
     \[
     DF^2 + 64 = 289
     \]
     \[
     DF^2 = 225
     \]
     \[
     DF = 15 \text{ cm}
     \]

5. **Calculate the length of \(EF\):**
   - Since \(AD = AE + EF + DF\):
     \[
     AD = 6 + EF + 15
     \]
     \[
     AD = 21 + EF
     \]
     Since \(AD\) is the total length of the base of the trapezoid, and we know \(AD = 27 \text{ cm}\):
     \[
     27 = 21 + EF
     \]
     \[
     EF = 6 \text{ cm}
     \]

6. **Calculate the length of \(BC\):**
   - Since \(BC\) is the horizontal distance between \(B\) and \(C\), and \(EF\) is the horizontal distance between the feet of the perpendiculars from \(B\) and \(C\):
     \[
     BC = EF = 10 \text{ cm}
     \]

The final answer is \(\boxed{10}\).