Question #145386
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?
1
Expert's answer
2020-11-19T18:14:27-0500

Joining a smaller cicular cone on the top,we have a complete triangleh1r1=h1+h2r2By the law of similar trainglesh1is the height of the smaller coneh2is the height of the larger coner2=3r1h1r1=h1+h23r1h1r1=h1+33r1h1=h1+333h1=h1+32h1=3,h1=32V=π(r22(32+3)r12×32)=52π=π(9r12(32+3)r12×32)=52π9r12(32+3)r12×32=529r1292r12×32=5281r1223r122=5278r122=5278r122=5239r12=52r12=43r1=23The upper radius is23cmSlant height of the larger cone is=r22+(h1+h2)2Slant height of the smaller cone is=r12+h12Slant height of the larger cone is=(3×23)2+(92)2=12+(92)2=12+814=48+814=1294=1292Slant height of the smaller cone is=43+(32)2=43+94=16+2712=4312=4323The slant height of the frustum is=12924323=12921296=1293The slant height of the frustum is1293cm\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}


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