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A cyclindrical water tank has a radius of 6 feet and a height of 20 feet. A larger tank with the same height has a volume that is 3 times the volume of the smaller tank. What is the best approximation of the diameter of the larger tank?

Respuesta :

[tex]\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\ -----\\ r=6\\ h=20 \end{cases}\implies V=\pi 6^2(20)\implies \stackrel{\textit{smaller cylinder}}{V=720\pi }[/tex]

since the smaller tank has a volume of 720π, then the larger tank has a volume of 3 times that much or 3*720π, or 2160π, and its height is the same as the smaller one's, 20.

[tex]\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\ -----\\ V=2160\pi \\ h=20 \end{cases}\implies 2160\pi =\pi r^2(20)\implies \cfrac{2160\pi }{20\pi }=r^2 \\\\\\ 108=r^2\implies \sqrt{108}=r\implies 6\sqrt{3}=r \\\\\\ \stackrel{\textit{and since the \underline{d}iameter is twice the radius}}{d = 2r\implies d=12\sqrt{3}}[/tex]
Luv2Teach
We need to find the volume of the smaller cylinder.  [tex]V= \pi r^2h[/tex].  For us, that looks like  [tex]V= \pi (6)^2(20)[/tex].  So the volume is [tex]V=720 \pi [/tex].  3 times that is the volume of the larger at [tex]V=2160 \pi [/tex].  Using that volume in the formula to solve for the radius we have  [tex]2160 \pi = \pi (r^2)(20)[/tex].  Divide both sides by 20pi to get  [tex]r^2=108[/tex].  Taking the square root of both sides gives us that r = 10.3923  or, in radical form, r = [tex]6 \sqrt{3} [/tex].  Diameter is twice that at 12sqrt(3)

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