$\triangle A B C$ is an equilateral triangle with side $18 \mathrm{~cm} . D$ is a point on $B C$ such that $B D=\frac{1}{3} B C$. Then length (in $\mathrm{cm}$) of $A D$ is:

$6 \sqrt{7}$

In \(\Delta \)ABD

cos B = (\( { AB}^{2 } \) + \( { BD}^{2 } \) - \( { AD}^{2 } \))/(2 x AB x BD)

= cos 60 = (\( { 18}^{2 } \) + \( { 6}^{2 } \) - \( { AD}^{2 } \))/(2 x 18 x 6)

= \(\frac{1}{2}\) = (324 + 36 - \( { AD}^{2 } \))/216

= 216 = 648 + 72 - 2\( { AD}^{2 } \)

= 2\( { AD}^{2 } \) = 648 + 72 - 216

= \( { AD}^{2 } \) = 252

= AD = 6\(\sqrt {7 }\)

Therefore, AD is 6\(\sqrt {7 }\).