$\triangle A B C \sim \triangle P Q R$. The areas of $\triangle A B C$ and $\triangle P Q R$ are $64 \mathrm{~cm}^2$ and $81 \mathrm{~cm}^2$, respectively and $A D$ and $P T$ are the medians of $\triangle A B C$ and $\triangle P Q R$, respectively. If $P T=10.8 \mathrm{~cm}$, then $A D=$ ?

9.6 cm

\(\frac{area\;of\;ABC}{area\;of\;PQR}\) = (\(\frac{AD}{PT}\)) x (\(\frac{AD}{PT}\))

= (\(\frac{AD}{PT}\)) x (\(\frac{AD}{PT}\)) = \(\frac{64}{81}\)

= \(\frac{AD}{PT}\) = \(\frac{8}{9}\)

= \(\frac{AD}{10.8}\) = \(\frac{8}{9}\)

= AD = \(\frac{8\;×\;10.5}{9}\) = 9.6 cm

Therefore, AD is 9.6 cm.