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 $18$ $18$ and the height of the cylinder is $36$ $36$ . If the volume of the solid is $420 π$ $420 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2018-07-30T02:22:16

$radius of the cone = r = \sqrt{10}, units$ $color(green)("radius of the cone " = r = sqrt 10, " units"$

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$Given h_{1} = 18, h_{2} = 36, V = 420 π$ $"Given " h_1 = 18, h_2 = 36, V = 420 pi$

$Volume of the solid = V = V_{c o \neq} + V_{c y l}$ $"Volume of the solid " = V = V_(cone) + V_(cyl)$

$V_{c o \neq} = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} 18 = 6 π r^{2}$ $V_(cone) = 1/3 pi r^2 h = 1/3 pi r^2 18 = 6 pi r^2$

$V_{c y l} = π r^{2} h = π r^{2} 36 = 36 π r^{2}$ $V_(cyl) = pi r^2 h = pi r^2 36 = 36 pi r^2$

$V = 6 π r^{2} + 36 π r^{2} = 420 π$ $V = 6 pi r^2 + 36 pi r^2 = 420 pi$

$42 π r^{2} = 420 π$ $42 pi r^2 = 420 pi$

$r^2 = cancel(420 pi)^color(red)(10) / cancel(42 pi) = 10$

$color(green)("radius of the cone " = r = sqrt 10, " units"$

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 18 1818 and the height of the cylinder is 36 3636 . If the volume of the solid is 420 pi420π420 pi, what is the area of the base of the cylinder?