Question 242496
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In the first place, you didn't provide the actual value of the height.  However,  the problem can be solved in general terms if (and only if) the meaning of "similar" as used in your question is the technical meaning of "similar" as defined for similar solids.


The volume of a right circular cone, inverted or otherwise, is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \frac{\pi r^2h}{3}]


So, if we let *[tex \Large r_1] and *[tex \Large h_1] be the base radius and height of the smaller cone, and if we let *[tex \Large r_2] and *[tex \Large h_2] be the base radius and height of the larger cone, and furthermore if we let *[tex \Large x] and *[tex \Large y] be the ratio of the smaller radius to the larger radius and the smaller height to the larger height respectively.


Then, the smaller cone has a volume of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_s\ =\ \frac{\pi r_1^2h_1}{3}]


and the larger cone has a volume of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_l\ =\ \frac{\pi r_2^2h_2}{3}]


But since we have defined the ratios we can say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1x\ =\ r_2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h_1y\ =\ h_2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_l\ =\ \frac{\pi (r_1x)^2h_1y}{3}]


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_l\ =\ 3V_s], we can see that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2y\ =\ 3]


However, since the radii and heights of the two cones must be in the same proportion in order that the cones be similar, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r_1}{r_2}\ =\ \frac{h_1}{h_2}]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r_1}{r_1x}\ =\ \frac{h_1}{h_1y}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1h_1y\ =\ r_1h_1x]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2x\ =\ x^3\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \sqrt[3]{3}]


If the dimensions of the original cone are *[tex \Large r_1] and *[tex \Large h_1], then the dimensions of a similar cone with 3 times the volume are *[tex \Large r_1\sqrt[3]{3}] and *[tex \Large h_1\sqrt[3]{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt[3]{3}\ \approx\ 1.4422495703074083823216383107801]


But 1.44 should suffice for your purposes.  Round any numerical approximation answer to two decimal places -- that is the least precision in your input measurements.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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