To determine the length of $AC$, we note that $\triangle ABC$ is isosceles with $AB = BC = 10 \, \text{cm}$ and $BD$ is perpendicular to $AC$, splitting it into two equal parts, $AD = DC$.

Step-by-step solution:

1. Use the Pythagorean theorem in $\triangle ABD$: $AB^2 = AD^2 + BD^2$ Substitute $AB = 10$ and $BD = 8$: $10^2 = AD^2 + 8^2$ $100 = AD^2 + 64$ $AD^2 = 36$ $AD = 6 \, \text{cm}$

2. Calculate $AC$: Since $AD = DC$ in $\triangle ABC$, and $AD = 6$: $AC = AD + DC = 6 + 6 = 12 \, \text{cm}$

Final Answer:

The length of $AC$ is 12 centimeters.

Would you like further clarification or additional examples? Here are five related questions for practice:

1. How would you calculate the length of $BD$ if it was not given but $AC$ and $AB$ were provided?
2. Can you determine the area of $\triangle ABC$?
3. What happens to $AC$ if $AB$ is doubled but $BD$ remains the same?
4. What is the relationship between the height and base in an isosceles triangle?
5. How do you prove that $\triangle ABD$ and $\triangle CBD$ are congruent?

Tip: In isosceles triangles with a perpendicular dropped from the vertex, always check for symmetry to simplify calculations.