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:

r1=2 m r2=(1+20/100)⋅ r1=(1+20/100)⋅ 2=125=2.4  S1=4π⋅ r12=4⋅ 3.1416⋅ 22≐50.2655 m2 S2=4π⋅ r22=4⋅ 3.1416⋅ 2.42≐72.3823 m2  p1=100⋅ S2−S1S1=100⋅ 72.3823−50.265550.2655=44   Correctness test:  q=1.2 p11=100⋅ (q2−1)=100⋅ (1.22−1)=44 % p1=p11=44%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 \%
V1=43⋅ π⋅ r13=43⋅ 3.1416⋅ 23≐33.5103 m3 V2=43⋅ π⋅ r23=43⋅ 3.1416⋅ 2.43≐57.9058 m3  p2=100⋅ V2−V1V1=100⋅ 57.9058−33.510333.5103=3645=72.8   Correctness test:  p21=100⋅ (q3−1)=100⋅ (1.23−1)=3645=72.8 % p2=p21=3645=72.8%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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