Cube in ball

Cube is inscribed into the sphere of radius 181 dm.

How many percent is the volume of
cube of the volume of the sphere?

Correct answer:

p =  36.7553 %

Step-by-step explanation:

$$\begin{gathered} r=6\text{ dm } \\ D=2\cdot r=2\cdot 6=12\text{ dm } \\ D=\sqrt{3}a \\ a=D\mathrm{/}\sqrt{3}=12\mathrm{/}\sqrt{3}=4\sqrt{3}\text{ dm}\doteq 6.9282\text{ dm } \\ {V}_1=a^3=6.9282^3=192\sqrt{3}{\text{cm}}^3\doteq 332.5538{\text{cm}}^3 \\ {V}_2={\displaystyle \frac{4}{3}}\cdot \pi \cdot r^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 6^3\doteq 904.7787{\text{dm}}^3 \\ p=100\cdot {\displaystyle \frac{V_1}{V_2}}=100\cdot {\displaystyle \frac{332.5538}{904.7787}}\doteq 36.7553 \\ p={\displaystyle \frac{200}{\sqrt{3}\cdot \pi }}\mathrm{\%}=36.7553\mathrm{\%} \end{gathered}$$

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