Cone and cube

By approximately what percentage does the volume of a cone with base radius r exceed that of a cube with the same height and square base edge r?

Final Answer:

p =  4.7 %

Step-by-step explanation:

$$\begin{gathered} p=100\cdot \frac{V_2-V_1}{V_1} \\ {V}_2=\frac{1}{3}\pi r^{2}h \\ {V}_1=r^{2}h \\ p=100\cdot \frac{V_2-V_1}{V_1} \\ p=100\cdot (\frac{1}{3}\cdot \pi -1) \\ p=100\cdot (\pi /3-1)=100\cdot (3.1416/3-1)=4.7\% \end{gathered}$$

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