Question

The volume of the frustum of a regular triangular pyramid is $135cu.m$. The lower base is an equilateral triangle with an edge of $9m$. The upper base is $8m$ above the lower base. What is the upper base edge in meters?

Answer 1

Given : Volume$\left(V\right)=135cu.m$ Edge of lower base $=9m$, height$\left(h\right)=8m$

We know that, Volume$\left(V\right)=\frac{h}{3}(A_1+A_2+\sqrt{B_1{B}_2})$ ....... (1),

Where $A_1$ and $A_2$ are the areas of lower base and upper base

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times sid{e}^2$

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

$=35.074m^2$

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

$=0.433x^2{m}^2$

From $\left(1\right)$

$135cu.m=\frac{8}{3}(35.074+0.433x^2+\sqrt{35.074\times 0.433x^2})$

$⟹50.625=35.074+0.433x^2+3.897x$

$⟹x^2+9x-36=0$ .......... (Dividing by $0.433$)

$⟹(x+12)(x-3)=0$

$⟹x=3m$ ..... (Since, length can't be negative)

Hence, upper base edge is $3m$.