Volume of a frustum; Y11 maths hw

[Jux]

I actually have no idea what this question is even asking really.

There is a diagram of a cone, with the frustum indicated about halfway down, and r is indicated at that line. h is indicated alongside the frustum and the top cone. There are no actual measurements. It does not say that the height of the full cone is 2h so I am guessing that one h is the same as the other h.

The question is to show that the volume of the frustum is

7/3 pi r sq h

How on earth can that be, if the volume of a cone is 1/3 pi r sq h?

Oh wait, it's to do with r and R? Too tired on a Sun evening to pee about with it (we had friends over for a very long lunch and wine has been consumed. In quantity). Please help.

Just asked DD, who is the family maths genius, she feels there is something missing, but would need to see the diagram to confirm.

I've just learnt that a Frustrum is a cone with its top cut off :)

Looking at the large cone, it's height is 2h and its radius at the base is twice the radius half way up, ie 2r. (Draw a triangular cross section to see the large and small triangles are similar)
So the volume of the large cone is pi (2r)^2 * 2h /3
This simplifies to 8/3 * pi r^2
The small top cone is 1/3 pi r^2
subtracting this from the whole cone gives 7/3 pi r^2.

Missed out *h. Sorry, it's getting late, but hope this makes sense.

Thank you JustRichmal. I thought that r would be 1/2 R but didn't actually know.

Another way of looking at this is that if lengths double, the volume goes up by 2^3, or 8. So taking the original volume away will leave a volume 7 times as big.