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A toy consists of a cylinder of diametre 6 cm 'sandwiched' between a hemisphere and a cone of the same diameter. If the cone is of height 8cm and a cylinder is of height 10cm, what is the volume of the toy.

Jeffreymars16  Jun 1, 2017
edited by Jeffreymars16  Jun 1, 2017
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radius = diameter / 2

r   =   6/2   =   3

 

volume of toy = volume of hemisphere + volume of cylinder + volume of cone

 

volume of toy = \(\frac12*\frac43*\pi*r^3\) ) + ( \(\pi*r^2*\text{cylinder height}\) ) + ( \(\frac13*\pi*r^2*\text{cone height}\) )

 

volume of toy = \(\frac12*\frac43*\pi*3^3\) ) + ( \(\pi*3^2*10\) ) + ( \(\frac13*\pi*3^2*8\) )

 

volume of toy = \(\frac12*\frac43*\pi*27\) ) + ( \(\pi*9*10\) ) + ( \(\frac13*\pi*9*8\) )

 

volume of toy = \(18\pi\) ) + ( \(90\pi\) ) + ( \(24\pi\) )

 

volume of toy = 132π   cubic centimeters               ≈ 414.69 cubic cm

hectictar  Jun 1, 2017

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