In `DeltaABC,` point M is the midpoint of BC,`AB^(2)+AC^(2)=290cm,`AM=8cm,find BC.

Answer

To solve the problem step by step, we will use Apollonius's theorem, which relates the sides of a triangle to the median drawn from one vertex to the midpoint of the opposite side.

### Step-by-Step Solution:

1. **Identify Given Information:**
- \( AB^2 + AC^2 = 290 \, \text{cm}^2 \)
- \( AM = 8 \, \text{cm} \) (where M is the midpoint of BC)

2. **Apply Apollonius's Theorem:**
Apollonius's theorem states:
\[ AB^2 + AC^2 = 2AM^2 + \frac{BC^2}{4} \]
Here, \( AM \) is the median, and \( BC \) is the side where the median is drawn.

3. **Substitute the Known Values:**
Substitute \( AB^2 + AC^2 = 290 \) and \( AM = 8 \) into the equation:
\[ 290 = 2(8^2) + \frac{BC^2}{4} \]

4. **Calculate \( 2(8^2) \):**
\[ 8^2 = 64 \implies 2(8^2) = 2 \times 64 = 128 \]
So, we can rewrite the equation as:
\[ 290 = 128 + \frac{BC^2}{4} \]

5. **Isolate \( \frac{BC^2}{4} \):**
Subtract 128 from both sides:
\[ 290 - 128 = \frac{BC^2}{4} \]
\[ 162 = \frac{BC^2}{4} \]

6. **Multiply Both Sides by 4:**
\[ 162 \times 4 = BC^2 \]
\[ 648 = BC^2 \]

7. **Take the Square Root:**
\[ BC = \sqrt{648} \]

8. **Simplify \( \sqrt{648} \):**
Factor 648:
\[ 648 = 81 \times 8 = 9^2 \times 8 \]
Thus,
\[ BC = 9\sqrt{8} = 9 \times 2\sqrt{2} = 18\sqrt{2} \, \text{cm} \]

### Final Answer:
\[ BC = 18\sqrt{2} \, \text{cm} \]