# Cube in ball

Cube is inscribed into sphere of radius 181 dm.

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

Correct result:

p =  36.755 %

#### Solution:

$r=6 \ \text{dm} \ \\ D=2 \cdot \ r=2 \cdot \ 6=12 \ \text{dm} \ \\ \ \\ D=\sqrt{ 3 } a \ \\ a=D / \sqrt{ 3 }=12 / \sqrt{ 3 } \doteq 4 \ \sqrt{ 3 } \ \text{dm} \doteq 6.9282 \ \text{dm} \ \\ \ \\ V_{1}=a^3=6.9282^3 \doteq 192 \ \sqrt{ 3 } \ \text{cm}^3 \doteq 332.5538 \ \text{cm}^3 \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 6^3 \doteq 904.7787 \ \text{dm}^3 \ \\ \ \\ p=100 \cdot \ \dfrac{ V_{1} }{ V_{2} }=100 \cdot \ \dfrac{ 332.5538 }{ 904.7787 } \doteq 36.7553 \ \\ \ \\ p=\dfrac{ 200 }{ \sqrt{ 3 } \cdot \ \pi }=36.755 \%$

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