Answer: The distance is 8 cm.
Let \(\triangle ABC\) have AB = AC = 13 cm and BC = 10 cm, and let AD be the median to the base. In an isosceles triangle this median is also perpendicular to the base, and BD = 5 cm.
\[AD^2 = 13^2 - 5^2 = 169 - 25 = 144\] \[AD = 12\ \text{cm}\]The centroid G divides the median in the ratio 2 : 1 from the vertex, so
\[AG = \frac{2}{3} \times 12 = 8\ \text{cm}\]