# Inscribed sphere

How many % of the volume of the cube whose edge is 6 meters long is a volume of a sphere inscribed in that cube?

Correct result:

p =  52.36 %

#### Solution:

$a=6 \ \text{m} \ \\ r=a/2=6/2=3 \ \text{m} \ \\ \ \\ V_{1}=4/3 \pi \cdot \ r^3=4/3 \cdot \ 3.1416 \cdot \ 3^3 \doteq 113.0973 \ \text{m}^3 \ \\ V_{2}=a^3=6^3=216 \ \text{m}^3 \ \\ \ \\ p=100 \cdot \ V_{1}/V_{2}=100 \cdot \ 113.0973/216=52.36 \%$

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