Answer to Question #145386 in Geometry for bjt

Question

The volume of a frustum of a right circular cone is 52π cm*. Its altitude is
3cm and the measure of its lower radius is three times the measure of its
upper radius.
1. What is the radius of the upper base?
2. What is the slant height of the frustum?

Answer 1

$\displaystyle\textsf{Joining a smaller cicular cone on the top,}\\ \textsf{we have a complete triangle}\\ \frac{h_1}{r_1} = \frac{h_1 + h_2}{r_2}\\ \textsf{By the law of similar traingles}\\ h_1 \,\, \textsf{is the height of the smaller cone}\\ h_2\,\, \textsf{is the height of the larger cone}\\ r_2 = 3 r_1 \\ \frac{h_1}{r_1} = \frac{h_1 + h_2}{3 r_1}\\ \frac{h_1}{r_1} = \frac{h_1 + 3}{3 r_1}\\ h_1 = \frac{h_1 + 3}{3}\\ 3h_1 = h_1 + 3 \\ 2h_1 = 3, h_1 = \frac{3}{2}\\ \begin{aligned} V &= \pi \left({r_2}^2\left(\frac{3}{2} + 3\right) - {r_1}^2\times\frac{3}{2}\right) = 52\pi\\ \\&=\pi \left(9{r_1}^2\left(\frac{3}{2} + 3\right) - {r_1}^2\times\frac{3}{2}\right) = 52\pi\\ \end{aligned}\\ 9{r_1}^2\left(\frac{3}{2} + 3\right) - {r_1}^2\times\frac{3}{2} = 52\\ 9{r_1}^2\cdot\frac{9}{2} - {r_1}^2\times\frac{3}{2} = 52\\ \frac{81{r_1}^2}{2} - \frac{3{r_1}^2}{2} = 52\\ \frac{78{r_1}^2}{2} = 52\\ \frac{78{r_1}^2}{2} = 52\\ 39{r_1}^2 = 52\\ {r_1}^2 = \frac{4}{3}\\ r_1 = \frac{2}{\sqrt{3}}\\ \therefore\textsf{The upper radius is}\,\, \frac{2}{\sqrt{3}}\,\textsf{cm}\\ \textsf{Slant height of the larger cone is}\, =\sqrt{{r_2}^2 + (h_1 + h_2)^2}\\ \textsf{Slant height of the smaller cone is}\, =\sqrt{{r_1}^2 + {h_1}^2}\\ \begin{aligned} \textsf{Slant height of the larger cone is}\, &= \sqrt{\left(3\times\frac{2}{\sqrt{3}}\right)^2 + \left(\frac{9}{2}\right)^2}\\ \\&=\sqrt{12 + \left(\frac{9}{2}\right)^2} = \sqrt{12 + \frac{81}{4}} \\&= \sqrt{\frac{48 + 81}{4}} = \sqrt{\frac{129}{4}} = \frac{\sqrt{129}}{2} \end{aligned}\\ \begin{aligned} \textsf{Slant height of the smaller cone is}\, &= \sqrt{\frac{4}{3} + \left(\frac{3}{2}\right)^2}\\ \\& = \sqrt{\frac{4}{3} + \frac{9}{4}} = \sqrt{\frac{16 + 27}{12}} \\&= \sqrt{\frac{43}{12}} = \frac{\sqrt{43}}{2\sqrt{3}} \end{aligned}\\ \begin{aligned} \therefore \textsf{The slant height of the frustum is}\, &= \frac{\sqrt{129}}{2} - \frac{\sqrt{43}}{2\sqrt{3}} \\&= \frac{\sqrt{129}}{2} - \frac{\sqrt{129}}{6} = \frac{\sqrt{129}}{3} \end{aligned}\\ \therefore \textsf{The slant height of the frustum is}\,\, \frac{\sqrt{129}}{3}\,\textsf{cm}$