# The ball

The ball has a radius of 2m. What percentage of the surface and volume is another sphere whose radius is 20% larger?

Result

p1 =  44 %
p2 =  72.8 %

#### Solution:

$r_{1}=2 \ \text{m} \ \\ r_{2}=(1+20/100) \cdot \ r_{1}=(1+20/100) \cdot \ 2=\dfrac{ 12 }{ 5 }=2.4 \ \\ \ \\ S_{1}=4 \pi \cdot \ r_{1}^2=4 \cdot \ 3.1416 \cdot \ 2^2 \doteq 50.2655 \ \text{m}^2 \ \\ S_{2}=4 \pi \cdot \ r_{2}^2=4 \cdot \ 3.1416 \cdot \ 2.4^2 \doteq 72.3823 \ \text{m}^2 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ S_{2}-S_{1} }{ S_{1} }=100 \cdot \ \dfrac{ 72.3823-50.2655 }{ 50.2655 }=44 \ \\ \ \\ \text{ Correctness test: } \ \\ q=1.2 \ \\ p_{11}=100 \cdot \ (q^2 - 1)=100 \cdot \ (1.2^2 - 1)=44 \ \% \ \\ p_{1}=p_{11}=44 \%$
$V_{1}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{1}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 2^3 \doteq 33.5103 \ \text{m}^3 \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{2}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 2.4^3 \doteq 57.9058 \ \text{m}^3 \ \\ \ \\ p_{2}=100 \cdot \ \dfrac{ V_{2}-V_{1} }{ V_{1} }=100 \cdot \ \dfrac{ 57.9058-33.5103 }{ 33.5103 }=\dfrac{ 364 }{ 5 }=72.8 \ \\ \ \\ \text{ Correctness test: } \ \\ p_{21}=100 \cdot \ (q^3 - 1)=100 \cdot \ (1.2^3 - 1)=\dfrac{ 364 }{ 5 }=72.8 \ \% \ \\ p_{2}=p_{21}=\dfrac{ 364 }{ 5 }=72.8 \%$

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