Question

The frustum of a regular triangular pyramid has equilateral triangles for its bases. The lower and upper base edges are $9m$ and $3m$, respectively. If the volume is $118.2cu.m$, how far apart (m) are the base?

• A. $9$
• B. $8$
• C. $7$
• D. $10$

Answer 1

The correct option is C $7$

Given : Volume $=118.2cu.m$

Upper base edge $=9m$, lower base edge $=3m$

We know that,

Volume $=\frac{h}{3}(A_1+A_2+\sqrt{A_1{A}_2})$ ....... $\left(1\right)$, where $A_1,A_2$ are area of upper and lower bases.

$A_1=\frac{\sqrt{3}}{4}\times 9^2=35.074$

$A_2=\frac{\sqrt{3}}{4}\times 3^2=3.897$

From $\left(1\right)$ we get,

$118.2=\frac{h}{3}(35.074+3.897+\sqrt{35.074\times 3.897})$

$⟹118.2=\frac{h}{3}(38.971+11.6911)$

$⟹118.2\times 3=h\left(50.6621\right)$

$⟹h=7m$

Hence, the bases are $7m$ far from each other.