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 $9$ $9$ and the height of the cylinder is $6$ $6$ . If the volume of the solid is $45 π$ $45 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-05-05T14:40:46

Volume of a cone is given by $V = \frac{1}{3} π r^{2} h$ $V = 1/3pir^2h$

Volume of a cylinder is given by $V = π r^{2} h$ $V = pir^2h$

$\frac{1}{3} π {(r)}^{2} (9) + r^{2} (6) (π) = 45 π$ $1/3pi(r)^2(9) + r^2(6)(pi) = 45pi$

$3 r^{2} π + 6 r^{2} π = 45 π$ $3r^2pi + 6r^2pi = 45pi$

$9 r^{2} π = 45 π$ $9r^2pi = 45pi$

$9 r^{2} = \frac{45 π}{π}$ $9r^2 = (45pi)/pi$

$9 r^{2} = 45$ $9r^2 = 45$

$r^{2} = 5$ $r^2 = 5$

$r = \sqrt{5}$ $r = sqrt(5)$

The area of a circle is given by $r^{2} π$ $r^2pi$

${(\sqrt{5})}^{2} π$ $(sqrt(5))^2pi$

= $5 π$ $5pi$

The area of the base of the cylinder is $5 π$ $5pi$

Hopefully this helps!

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 9 99 and the height of the cylinder is 6 66 . If the volume of the solid is 45 pi45π45 pi, what is the area of the base of the cylinder?