# right circular cylinder is inscribed in a right circular cone

• August 21st 2010, 07:42 PM
cutiemike1
right circular cylinder is inscribed in a right circular cone
A right circular cylinder is inscribed in a right circular cone of altitude h and radius of base x, as shown in the figure. Find the radius of the cylinder if its lateral area is equal to the lateral area of the small cone which surmounts the cylinder.

http://i104.photobucket.com/albums/m...necylinder.jpg

This is my effort and i don't know if my effort is really an effort.

First i solved for the lateral area of the cone:
A= xhpi(sqrt of (x^2 + h^2))

Second i solved for the lateral area of the right cylinder:
A = 2pirh

And lastly i equated them and getting the result of the radius since the lateral area of the cone is equal to the lateral area of the right cylinder:
xhpi(sqrt of (x^2 + h^2)) = 2pirh
r = x(sqrt of (x^2 + h^2))/2

THE CORRECT ANSWER stated on the book is:
(2xh)/(2h+(sqrt of (x^2 + h^2)))

I want to know where in my solution is wrong..Thanks!
• August 21st 2010, 11:18 PM
Vlasev
I think you are using the wrong formulas. Can you please tell me which is which, because I don't know what you mean by all the variables. In the drawing you are given r and h are the radius and height of the big cone.
• August 22nd 2010, 12:33 AM
earboth
Quote:

Originally Posted by cutiemike1
A right circular cylinder is inscribed in a right circular cone of altitude h and radius of base x, as shown in the figure. Find the radius of the cylinder if its lateral area is equal to the lateral area of the small cone which surmounts the cylinder.

...Thanks!

1. I've modified your sketch a little bit. (see attachment)

2. As far as I understand the question you have 2 constant values (r, h) and you are looking for the radius x of the cylinder which satisfies the given conditions(?).

If so:

3. Use proportions:

$\frac tx = \frac hr~\implies~t = \frac{h x}r$

4. Lateral surface of the cylinder: $A_{cyl} = 2\cdot \pi \cdot x \cdot h$

Lateral surface of the small cone on top: $A_{topcon} = \pi\cdot x \cdot \sqrt{x^2+t^2}$

Both areas are equal:

$2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}$

5. This can be simplified to:

$x^2+t^2 = 4h^2$. Substitute t from 3.:

$x^2+\left(\frac{h x}r\right)^2 = 4h^2$ and solve for x.
• August 22nd 2010, 01:17 AM
cutiemike1
OH!(Headbang) I AM REALLY SORRY!!! I messed up my question.

Ok.. h refers to the height of the cone and r is supposed to be x as the base radius of the cone..

@earboth.. where and how did you get x^2 + t^2 = 4h^2??
• August 22nd 2010, 04:34 AM
earboth
I wrote

Quote:

Originally Posted by earboth
...
Both areas are equal:

$2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}$

5. This can be simplified to:

$x^2+t^2 = 4h^2$.

Quote:

Originally Posted by cutiemike1
OH!(Headbang) I AM REALLY SORRY!!! I messed up my question.

Ok.. h refers to the height of the cone and r is supposed to be x as the base radius of the cone..

@earboth.. where and how did you get x^2 + t^2 = 4h^2??

I'll show you the steps of transformation:

$2\cdot \pi \cdot x \cdot h = \pi\cdot x \cdot \sqrt{x^2+t^2}$

This equation implies that x = 0 which isn't very plausible with your question. So divide the equation through by $\pi x$ which yields:

$2\cdot h = \sqrt{x^2+t^2}$

To get rid of the square-root square both sides of the equation:

$4h^2 = x^2+t^2$
• August 22nd 2010, 04:49 AM
cutiemike1
got it!...but it is different from the answer written in the book? how's that?
• August 22nd 2010, 04:55 AM
earboth
Quote:

Originally Posted by cutiemike1
got it!...but it is different from the answer written in the book? how's that?

Well ... that's easy to explain:

Quote:

Originally Posted by cutiemike1
OH!(Headbang) I AM REALLY SORRY!!! I messed up my question.

...

• August 22nd 2010, 05:24 AM
cutiemike1
oh sorry sir..it really is my bad. (Doh)
• September 7th 2010, 05:37 AM
ronnieryan
so whats the final answer to this?