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How to calculate the volume of this shape?

enter image description here

I've tried to do $2.5\cdot 2.5\cdot 3.14\cdot 12$ and then divide that with $3$ since the first part is a cone. But I only get $78.5$ by doing that. Am I missing a part or what? I'm confused.

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  • $\begingroup$ The volume of a cone is $1/3\cdot \text{Area(base)} \cdot \text{height}$ $\endgroup$
    – David P
    Dec 27, 2014 at 22:22
  • $\begingroup$ Your calculation looks correct to me. I can't find any mistakes, assuming that the shape is a cone. $\endgroup$
    – user141592
    Dec 27, 2014 at 22:22
  • $\begingroup$ Well I know that, I said that I tried dividing with three.. Am I doing something wrong with my current calculation? EDIT to Johanna: That's strange.. $\endgroup$
    – didnotcomeuptosomething
    Dec 27, 2014 at 22:23
  • $\begingroup$ Is $12 cm$ the height of the cone or the slant height? $\endgroup$
    – JimmyK4542
    Dec 27, 2014 at 22:23
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    $\begingroup$ Bot: This is a perfectly fine question. OP's tried to do it, showed us his/her work, and asked for help. This is a legitimate use of Math.SE. $\endgroup$
    – John Hughes
    Dec 27, 2014 at 22:33

1 Answer 1

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It must be an "ice-cream cone"

$$\underbrace{\dfrac{1}{3}\pi(2.5)^2\cdot\underbrace{\sqrt{12^2-2.5^2}}_{\text{height of cone}}}_{\text{cone part}} + \underbrace{\dfrac{1}{2}\cdot \dfrac{4}{3}\pi(2.5)^3}_{\text{semisphere}} \approx 109.54141$$

I only now noticed the new picture. In this picture the height of the cone is given as $12cm$. In this case it works out to about $111cm$. Did the question specify to round to the nearest $10cm$?

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  • $\begingroup$ I actually just noticed that trying to calculate this and no, the question didn't specify to round to the nearest 10cm. I got the answer to actually be 111.2 cm^3 .. it's pretty strange. $\endgroup$
    – didnotcomeuptosomething
    Dec 27, 2014 at 22:49
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    $\begingroup$ However checking at the answers for the previous questions, it does seem like the answers are fairly rounded down or up so I think that's the case here too. Thank you for all your insightful help! $\endgroup$
    – didnotcomeuptosomething
    Dec 27, 2014 at 22:56
  • $\begingroup$ This is assuming the height given (12) is the slant height. I don't know what's supposed to be implied by the picture. $\endgroup$
    – GFauxPas
    Dec 27, 2014 at 23:14

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