In an equilateral triangle ABC, G is the centroid and AB = 10 cm. What is the length of AG (in cm)?

Step 1: Find median of equilateral triangle

Side $a = 10$ cm

Median (height) of equilateral triangle:
$\frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}$

Step 2: Centroid division
$AG = \frac{2}{3} \times \text{median}$So,
$AG = \frac{2}{3} \times 5\sqrt{3} = \frac{10\sqrt{3}}{3}$

Final Answer: $\frac{10\sqrt{3}}{3}$