Eight

Eight small Christmas balls with a radius of 1 cm have the same volume as one large Christmas ball. What has a bigger surface: eight small balls, or one big ball?

Correct result:

x =  8

Solution:

$$\begin{gathered} n=8 \\ r=1\text{ cm } \\ {V}_1=n\cdot {\displaystyle \frac{4}{3}}\cdot \pi \cdot r^3=8\cdot {\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 1^3\doteq 33.5103{\text{cm}}^3 \\ {V}_2=V_1=33.5103\doteq 33.5103{\text{cm}}^3 \\ {V}_2={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_2^3 \\ {r}_2=\sqrt[3]{3\cdot V_2\mathrm{/}4\mathrm{/}\pi }=\sqrt[3]{3\cdot 33.5103\mathrm{/}4\mathrm{/}3.1416}=2\text{ cm } \\ {S}_1=n\cdot 4\pi \cdot r^2=8\cdot 4\cdot 3.1416\cdot 1^2\doteq 100.531{\text{cm}}^2 \\ {S}_2=4\pi \cdot r_2^2=4\cdot 3.1416\cdot 2^2\doteq 50.2655{\text{cm}}^2 \\ {S}_1>S_2 \\ x=8 \end{gathered}$$

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