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 $66$ $66$ and the height of the cylinder is $5$ $5$ . If the volume of the solid is $64 π$ $64 pi$ , what is the area of the base of the cylinder?

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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 $66$ $66$ and the height of the cylinder is $5$ $5$ . If the volume of the solid is $64 π$ $64 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2018-06-06T07:06:00

$A = \frac{64}{27} π$ $A=64/27pi$ $u^{2}$ $u^2$

Explanation:

The volume of the cone is given by: $v = \frac{1}{3} π r^{2} h$ $v=1/3pir^2h$
Since the height of the cone is 66, then $h = 66$ $h=66$
So, $v = \frac{1}{3} π r^{2} \times 66 = 22 π r^{2}$ $v=1/3pir^2times66=22pir^2$

The volume of a cylinder is given by: $v = π r^{2} h$ $v=pir^2h$
Since the height of the cylinder is 5, then $h = 5$ $h=5$
So, $v = π r^{2} h = π r^{2} \times 5 = 5 π r^{2}$ $v=pir^2h = pir^2times5=5pir^2$

The total volume of the solid is $64 π$ $64pi$
Therefore, $22 π r^{2} + 5 π r^{2} = 64 π$ $22pir^2+5pir^2=64pi$

$27 π r^{2} = 64 π$ $27pir^2=64pi$
$r^{2} = \frac{64 π}{27 π}$ $r^2=(64pi)/(27pi)$
$r^{2} = \frac{64}{27}$ $r^2=64/27$
$r = \pm \frac{8}{3 \sqrt{3}}$ $r=+-8/(3sqrt3)$
Since r is the radius, it must be have the restriction: $r > 0$ $r>0$
Therefore, $r = \frac{8}{3 \sqrt{3}}$ $r=8/(3sqrt3)$ units

To find the base of the cylinder, we need to know that the base is a circle. The area of a circle is given by $A = π r^{2} = π \times {(\frac{8}{3 \sqrt{3}})}^{2} = \frac{64}{27} π$ $A=pir^2 = pitimes(8/(3sqrt3))^2=64/27pi$ $u^{2}$ $u^2$

Answer 2 · 2018-06-06T07:35:29

The area of the base of the cylinder is: $A = π r^{2} = \frac{64 π}{27}$ $A=pir^2=(64pi)/27$

Explanation:

The area of the base we need to find is: $A = π r^{2}$ $A=pir^2$ , where $r$ $r$ is the radius of the cylinder.

The volume of the cylinder is: $π r^{2} \cdot h_{1}$ $pir^2*h_1$
where $h_{1}$ $h_1$ is the height of the cylinder.
The volume of the cone is $π r^{2} \cdot \frac{h_{2}}{3}$ $pir^2*h_2/3$
where $h_{2}$ $h_2$ is the height of the cone.

The volume of the solid is the sum of those two volumes: $V = π r^{2} \cdot h_{1} + π r^{2} \cdot \frac{h_{2}}{3}$ $V=pir^2*h_1 + pir^2*h_2/3$
Factoring $π r^{2}$ $pir^2$ :
$V = π r^{2} (h_{1} + \frac{h_{2}}{3})$ $V= pir^2(h_1+h_2/3)$
$64 π = π r^{2} (5 + \frac{66}{3}) = π r^{2} (5 + 22) = π r^{2} (27)$ $64pi = pir^2(5+66/3) = pir^2(5+22) = pir^2(27)$

$64 π = 27 π r^{2}$ $64pi = 27pir^2$

$π r^{2} = \frac{64 π}{27}$ $pir^2 = (64pi)/27$
And that is the area of the base: $A = π r^{2} = \frac{64 π}{27}$ $A=pir^2=(64pi)/27$

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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 $66$ $66$ and the height of the cylinder is $5$ $5$ . If the volume of the solid is $64 π$ $64 pi$ , what is the area of the base of the cylinder?

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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 6666 and the height of the cylinder is 55 . If the volume of the solid is 64π64 pi, what is the area of the base of the cylinder?

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Explanation:

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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 $66$ $66$ and the height of the cylinder is $5$ $5$ . If the volume of the solid is $64 π$ $64 pi$ , what is the area of the base of the cylinder?