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

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

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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, egreg Feb 27 at 14:22

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Goos, Michael Hoppe, Davide Giraudo, M Turgeon, egreg
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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
In your solution you've written $H-h = 4$, but should it not be $3$? –  naslundx Feb 27 at 13:24

1 Answer 1

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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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
Since we only know the heights, there is not enough information to calculate $R$. –  naslundx Feb 27 at 14:21
The expression is not constant so it cannot be simplified to an expression without the variable $R$. I would say the area ratio is dependant on the radius $R$. –  naslundx Feb 27 at 14:27
That's great, thank u :)) –  whyguy Feb 27 at 14:34

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