Sahba7413
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Figure 2 shows a hollow cone with radius R and vertical height H and a cylinder with radius r and height h . The cone sits on top of the cylinder so that their bases are level and the top of the cylinder is in contact with the inside of the cone as shown. The ratio of the heights hH is equal to p .

Express the ratio of the radii rR in terms of p .


I would appreciate any hints.
Last edited by Sahba7413; 8 months ago
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dextrous63
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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.
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Sahba7413
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(Original post by dextrous63)
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.
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dextrous63
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(Original post by Sahba7413)
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?
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Sahba7413
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(Original post by dextrous63)
Not quite. What is the height of the smaller triangle?
I'm not really sure, would it be H-h?
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dextrous63
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(Original post by Sahba7413)
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
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Sahba7413
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(Original post by dextrous63)
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
Thanks a lot.
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jasmine_GCSE
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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
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davros
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(Original post by jasmine_GCSE)
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?
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jasmine_GCSE
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(Original post by davros)
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 ://///
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davros
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(Original post by jasmine_GCSE)
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?
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jasmine_GCSE
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(Original post by davros)
That's OK - have you got it now?
nope haha I can't work it out
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davros
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(Original post by jasmine_GCSE)
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?
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jasmine_GCSE
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(Original post by davros)
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 ?
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davros
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(Original post by jasmine_GCSE)
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?
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jasmine_GCSE
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ahhhh, r/R=-h, still don't know how to get to the last bit
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jasmine_GCSE
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(Original post by davros)
So now you have H - h = rH/R

Divide both sides by H and what do you get?
oh actually, r/R=1-h
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davros
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(Original post by jasmine_GCSE)
oh actually, r/R=1-h
Nearly - the '1' is correct but the 'h' should be something else
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ila.1
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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?
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davros
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(Original post by ila.1)
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?
Again, this is an old thread, but if you refer to the initial diagram and the instructions in the posts following this, it tells you how to get similar triangles to compare. You're using the fact that the ratios of corresponding sides of similar triangles are equal.
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