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 $12$ and the height of the cylinder is $24$ . If the volume of the solid is $42 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2018-08-04T01:37:26

$color(maroon)("Cylinder base area " = A = 3/2 pi " sq units"$

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$"Volume of cone " V_(cone) = 1/3 pi r^2 h_1$

$"Volume of cylinder " V_(cyl) = pi r^2 h_2$

$"Volume of solid " V = 1/3 pi r^2h_1 + pi r^2 h_2 = pi r^2 (1/3 h_1 + h_2)$

$"Area of cylinder base " = A = pi r^2$

$"Given " V = 42 pi, h_1 = 12, h_2 = 24, pi r^2 = ?$

$pi r^2 (1/3 h_1 + h_2) = 42 pi$

$pi r^2 = (42 pi) / (1/3 h_1 + h_2)$

$A = pi r^2 = (42 pi) / (1/3 * 12 + 24) = (42 pi) / 28$

$color(maroon)(A = 3/2 pi " sq units"$

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 12 12 and the height of the cylinder is 24 24 . If the volume of the solid is 42 pi42 pi, what is the area of the base of the cylinder?