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:

Answer: Option C

Solution

Answer: Option C

Solution:

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