The base of a right pyramid is an equilateral triangle with area 16√3 cm². If the area of one of its lateral faces is 30 cm², then its height (in cm) is:

$$\eqalign{ & {\text{Area of }}\Delta ABC = \frac{{\sqrt 3 }}{4}{a^2} \cr & 16\sqrt 3 = \frac{{\sqrt 3 }}{4}{a^2} \cr & a = 8 \cr & {\text{In }}\Delta ABD, \cr & \frac{1}{2} \times 8 \times MD = 30 \cr & MD = \frac{{15}}{2} \cr & {\text{Now, in }}\Delta MOD, \cr & MO = \frac{a}{{2\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \cr & M{D^2} = M{O^2} + O{D^2} \cr & \frac{{225}}{4} = {\left( {\frac{4}{{\sqrt 3 }}} \right)^2} + O{D^2} \cr & OD = \sqrt {\frac{{225}}{4} + \frac{{16}}{3}} \cr & OD = \sqrt {\frac{{611}}{{12}}} \cr} $$