Square - increased perimeter

How many times is increased perimeter of the square, where its sides increases by 150%?

If the perimeter of square will increase twice, how much% increases the content area of the square?

Result

n =  0
p2 =  300 %

Solution:

a1=1 a2=a1+1.50 a1=1+1.50 1=52=2.5 o1=4 a1=4 1=4 o2=4 a2=4 2.5=10 n=o2/o1=10/4=52=2.5=0a_{1}=1 \ \\ a_{2}=a_{1}+1.50 \cdot \ a_{1}=1+1.50 \cdot \ 1=\dfrac{ 5 }{ 2 }=2.5 \ \\ o_{1}=4 \cdot \ a_{1}=4 \cdot \ 1=4 \ \\ o_{2}=4 \cdot \ a_{2}=4 \cdot \ 2.5=10 \ \\ n=o_{2}/o_{1}=10/4=\dfrac{ 5 }{ 2 }=2.5=0
p2=(2212)/12 100=300%p_{2}=(2^2-1^2)/1^2 \cdot \ 100=300 \%



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