+0  
 
+8
346
1
avatar+132 

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
 #1
avatar+7324 
+1

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

23 Online Users

avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.