A level Maths

Drop a height line down from the top of the cone to the base of a cylinder, and a radius R out from the bottom of this line.

You then have a pair of similar triangles to work with.

So would rR euqal to p as well, as hH is equal to P. Due to the fact that they are similar triangles.

Not quite. What is the height of the smaller triangle?

I'm not really sure, would it be H-h?

Yes.

You know that h/H = p

You also know that (from the similarity of the triangles) (H-h)/r = H/R.
You can rearrange this to find r/R

Would someone mind explaining this further? I understand about the similar triangles, but how do I go from knowing that (H-h)/r = H/R and the fact that h/H is p to finding out r/R, I am confused about how to take the next step to rearrange

You're replying to a 7-month old thread but first of all can you rearrange (H-h)/r = H/R to get r/R on one side?

Hahahah sorry I realised it was old but I think they must have been doing the same maths course as I am now! I couldn't get r/R for some reason, I think I'm over complicating it ://///

That's OK - have you got it now?

nope haha I can't work it out

OK so you've got $\dfrac{H - h}{r} = \dfrac{H}{R}$

What can you do first to get r onto the RHS?

Multiply by r so Hr/R, How do I get rid of the H though ?

So now you have H - h = rH/R

Divide both sides by H and what do you get?

oh actually, r/R=1-h

Sorry to ask, I understood all of the rearranging of this formula to make r/R the subject of the formula, but I am very confused about how you got the formula in the first place. How did you know (H-h)/r = H/R ? What law of similar shapes do I have to know to arrive at this equation here?