The base of a right pyramid is an equilateral triangle with area $$16\sqrt{3}cm^2$$. If the area of one of its lateral facesis 30 $$cm^2$$, then its height (in cm) is:

Base area = $$16\sqrt{3}cm^2$$
$$\frac{\sqrt{3}a^2}{4} = 16\sqrt{3}cm^2$$
a = 8 cm
In radius(r) = $$\frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$$
Lateral face area = $$\frac{1}{2} \times B \times h$$
30 = $$\frac{1}{2} \times 8 \times h$$
h = l = $$\frac{15}{2}$$
h = $$\sqrt{(\frac{15}{2})^2 - (\frac{4}{\sqrt{3}})^2}$$
= $$\sqrt{(\frac{225}{4})^2 - (\frac{16}{3})^2}$$
= $$\sqrt{\frac{675 - 64}{12}} = \sqrt{\frac{611}{12}}$$