Cone and cube

About what percentage hase cone with base radius r larger volume than same high cube with square base with edge length r?

Result

p =  4.7 %

Solution:

$p = 100 \dfrac{V_2-V_1}{V_1} = 100 (\dfrac{ \dfrac{\pi}{3}r^2v}{r^2v} - 1) = 100 ( \dfrac{ \pi}{3}-1) = 4.7 \%$

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