# A cylinder inscribed in a cone [closed]

A cylinder inscribed in a cone with cone H = 8 and cylinder h = 4

a) Find the ratio of cone-volume / inscribed-volume. Solution:

b) Find the ratio of cone-area / inscribed-area ?

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## closed as off-topic by Goos, Michael Hoppe, Davide Giraudo, M Turgeon, egregFeb 27 at 14:22

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Please tell us more about what you want help with and what you've tried so far. –  naslundx Feb 27 at 13:01
I can't found the cone area, cuz I can't found s (length of slant) –  whyguy Feb 27 at 13:03
In your solution you've written $H-h = 4$, but should it not be $3$? –  naslundx Feb 27 at 13:24
my mistake, h = 4* –  whyguy Feb 27 at 13:56

If you know the radius of the base of the cone is $R$ and height $H=8$, by the pythagorean theorem the slant $S = \sqrt{R^2 + 8^2}$.

By the formulae for surface area of cylinder and cone, we have

$$A_{cone} = \pi R(R+S) = \pi R\left(R + \sqrt{R^2 + 64}\right)$$

$$A_{cylinder} = 2 \pi r(r + h) = 2 \pi \frac{R}{2}\left(\frac{R}{2} + 4\right) = \pi R \left(\frac{R}{2} + 4\right)$$

Divide the two expressions and simplify to get the ratio of the two quantities:

$$\frac{A_{cone}}{A_{cylinder}}=\frac{\pi R\left(R + \sqrt{R^2 + 64}\right)}{\pi R \left(\frac{R}{2} + 4\right)}=\frac{R + \sqrt{R^2 + 64}}{\frac{R}{2} + 4}$$

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I am in the same answer, but how to simplify ?? –  whyguy Feb 27 at 14:07
That's why it's good to tell us more about how far you've managed yourself. I added the final step, does it make sense? –  naslundx Feb 27 at 14:11
It's not a answer, cuz we dont know R. The answer will be sth like the a) (volume ratio) –  whyguy Feb 27 at 14:16
Since we only know the heights, there is not enough information to calculate $R$. –  naslundx Feb 27 at 14:21
No needed to calculate R, just to simplify it :/ –  whyguy Feb 27 at 14:23