Inscribed sphere

How many percents of the cube volume takes the sphere inscribed into it?

Result

p =  52.36 %

Solution:

$r=a/2 \ \\ V_{1}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3=\dfrac{ 4 }{ 3 \cdot \ 2^3 } \cdot \ \pi \cdot \ a^3 \ \\ V_{2}=a^3 \ \\ p=100 \cdot \ \dfrac{ V_{1} }{ V_{2} } \ \\ p=100 \cdot \ \dfrac{ 4 }{ 3 \cdot \ 2^3 } \cdot \ \pi=100 \cdot \ \dfrac{ 4 }{ 3 \cdot \ 2^3 } \cdot \ 3.1416 \doteq 52.3599 \doteq 52.36 \%$

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