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 $33$ and the height of the cylinder is $17$ . If the volume of the solid is $168 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-04-10T04:52:24

$6pi$

Consider the diagram

enter image source here

Note: $color(brown)(pi=22/7,V=volume,h=height$

Remember the formulas

$color(purple)(V_(con)=1/3pir^2h$

$color(purple)(V_(cyl)=pir^2h$

$color(purple)(Area_(base)=pir^2$

And $r$ is the radius of the base (circle)

We know that the volume of the whole solid is $168pi$

Our aim is to find Area of base ( $pir^2$ )

And we could see that $pir^2$ is present in both of the volume formulas

So, consider $pir^2$ as $a$

$rarr1/3*a*33+a*17=168pi$

$rarr1/cancel3^1*a*cancel33^11+a*17=168pi$

$rarra*33+a*17=168pi$

$rarr28a=168pi$

$rarra=168/28pi$

$color(green)(rArra=6pi$

Source of image: paint (my drawing)

And,if you don't like this drawing and want to change it, your changes are welcome

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 33 33 and the height of the cylinder is 17 17 . If the volume of the solid is 168 pi168 pi, what is the area of the base of the cylinder?