Posted by CJ on Monday, February 14, 2011 at 5:42am.
it looks like you just added .2 cm to the height and .2 cm to (1/2) the width, then used the formula for volume of the cone.
For all practical purposes, this will be ok.
But, ...
There is a rather complicated calcululation needed for the extensions at the vertex of the cone and the base of the cone
Draw a cross-section of the cone with r= 3.8 and h = 15.2
Now draw a larger cross-section around it so that the distance between the two triangles is .2 cm
From the top of the original triangle draw a perpendicular to the new side and extend the height to meet the new triangle.
The extension of the height will be hypotenuse of a small right-angled triangle, where you know one of the sides to be .2
You will need the angle at the top.
That angle will be equal to the angle formed by the original height and the side of the original cone.
let that angle be Ø
tanØ = 3.8/15.2 = ....
you can find Ø
now in the little triangle, let the hypotenuse be h
sinØ = 0.2/h
h = 0.2/sinØ
Of course at the base we can just add 0.2 to the origianal height.
So new height = 15.2 + h + 0.2
you will have to do a similar calculation at the base of the cone.
remember that the angle there will be 90-Ø
Good luck.
BTW, the answers you gave should have been in cm^3 , not m^3
For you second part, you are only concerned with the volume of the original cone, the chocolate cover does not affect how much the cone can hold.