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A math class is working with similar figures. Mr. Kole told the class that the surface area of the cones are 180 and 320 square units, and the volume of the smaller cone is 151 cubic units. He challenged his class to find the volume of the bigger cone. What is the volume of the bigger cone?

I tried to use the proportion 320/180 = x/151 to solve the problem. But my answer did not match the correct answer, which is 358 cubic units.

How would you solve this problem?

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  • $\begingroup$ Surface areas of similar figures are in proportion to the squares of corresponding lines; volumes are in proportion to the cubes. $\endgroup$
    – saulspatz
    Oct 3, 2019 at 19:27
  • $\begingroup$ Are cones open bases or closed? $\endgroup$
    – Mohammad Riazi-Kermani
    Oct 3, 2019 at 19:27

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Since the cones are similar the ratio of their areas is equal to the square of corresponding lines. (For example, the height or the radius of the base.) The ration of those lines is $$a=\sqrt{{320\over180}}=\frac43$$ The ratio of the volume is $a^3$ so the volume of the large cone is $$\left(\frac43\right)^3\cdot151$$

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Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that.

The area scales as a square of the radius $A\propto r^2$ and the volume as the cube $V\propto r^3$. In particular $A^3\propto r^6,\, V^2\propto r^6$, then the quantity $$ \frac{A^3}{V^2} $$ is an invariant for similar cones (or any other figure for that mater). And we have $$ \frac{320^3}{x^2} = \frac{180^3}{151^2} $$

This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?

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  • $\begingroup$ Yes. Your answer 358 is right. I made a typo. $\endgroup$
    – orangebull
    Oct 4, 2019 at 3:02
  • $\begingroup$ But why did you put $\frac{A^3}{V^2}$? Shouldn't it have been $\frac{A^2}{V^3}$? $\endgroup$
    – orangebull
    Oct 4, 2019 at 3:04
  • $\begingroup$ Ah sorry, I went a little fast there. $A^3\propto r^6$ and $V^2\propto r^6$ then the ratio of this two quantities doesn't depend on the radius. $\frac{A^3}{V^2}\sim \frac{r^6}{r^6}\sim \text{constant}$. Look that @saulspatz answer is exactly the same argument. $\endgroup$
    – David Jaramillo
    Oct 4, 2019 at 8:23
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No, surface area is not directly proportional to volume.

I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.

Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.

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  • $\begingroup$ All you need is the ratio of the areas. The shape of the cone doesn't matter. $\endgroup$
    – saulspatz
    Oct 3, 2019 at 19:30
  • $\begingroup$ @saulspatz Really? Aren't there infinitely many cones with surface area $320$ but different volumes, depending on their heights and radii? And only one of these will be similar to the smaller cone? $\endgroup$
    – 79037662
    Oct 3, 2019 at 19:34
  • $\begingroup$ I guess I assumed the cones being similar was relevant information. $\endgroup$
    – 79037662
    Oct 3, 2019 at 19:35
  • $\begingroup$ It is relevant. That why all you need is the ratio of the areas. $\endgroup$
    – saulspatz
    Oct 3, 2019 at 19:36
  • $\begingroup$ @saulspatz At the first glance I think the ratio alone is not sufficient. $\endgroup$
    – callculus42
    Oct 3, 2019 at 19:36
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Volumes are proportional to cube and areas to square of linear dimension in similitude.

$$ V = k_1 L^3,\, A = k_2 L^2,\, V = k_3 A^{\frac32} $$

Accordingly the larger Volume is

$$ V_2 = V_1 \big(\frac{320}{180}\big)^ {1.5} =151\times 64/ 27 \approx 357.926 $$

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The similarity of two cones imply: $$\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{h_1}{h_2} \Rightarrow \frac{R_1}{R_2}=\frac{R_1+l_1}{R_2+l_2}$$ The areas of cones: $$S_1=\pi R_1l_1+\pi R_1^2=180;\\ S_2=\pi R_2l_2+\pi R_2^2=320;\\ \frac{S_1}{S_2}=\frac{\pi R_1(l_1+R_1)}{\pi R_2(l_2+R_2)}=\left(\frac{R_1}{R_2}\right)^2=\frac{180}{320} \Rightarrow \left(\frac{R_1}{R_2}\right)^3=\left(\frac{180}{320}\right)^{3/2}$$ The volumes of cones: $$V_1=\pi R_1^2h_1=151;\\ V_2=\pi R_2^2h_2;\\ \frac{V_1}{V_2}=\frac{\pi R_1^2h_1}{\pi R_2^2h_2}=\left(\frac{R_1}{R_2}\right)^3=\left(\frac{180}{320}\right)^{3/2}=\frac{151}{V_2} \Rightarrow V_2\approx 357.93.$$

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