A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is $39$ $39$ and the height of the cylinder is $17$ $17$ . If the volume of the solid is $120 π$ $120 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-05-10T22:12:40

$π r^{2} = 40 π$ $pir^2 = 40pi$

This question is surprisingly simple, despite what look like awkward numbers.

$Vol cone = \frac{1}{3} π r^{2} h$ $"Vol cone" = 1/3pir^2h$ and $Vol cylinder = π r^{2} H$ $"Vol cylinder" = pir^2H$

Total volume = $\frac{1}{3} π r^{2} h + π r^{2} H = 120 π$ $1/3pir^2h + pir^2H = 120pi$

$\frac{1}{3} π r^{2} 39 + π r^{2} 17 = 120 π \Rightarrow$ $1/3pir^2 39 + pir^2 17 = 120pirArr$ $1/cancel3pir^2cancel39^13 + pir^2 17 = 120pi$

$13pir^2 + 17 pir^2 = 120pi$

$30 pir^2 = 120pi" divide both sides by 30"$

$pir^2 = 40pi$

Nothing more is needed, because the area of the circular base is given by $pi r^2$ .

A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is 39 3939 and the height of the cylinder is 17 1717 . If the volume of the solid is 120 pi120π120 pi, what is the area of the base of the cylinder?