Percentage and rectangle

About what percentage increases perimeter and area of a rectangle if both the sides 12 cm and 10 cm long we increase by 20%?

Result

p1 =  20 %
p2 =  44 %

Solution:

a=12 b=10 S=a⋅ b=12⋅ 10=120 o=2⋅ (a+b)=2⋅ (12+10)=44 a1=a⋅ 1.20=12⋅ 1.20=725=14.4 b1=b⋅ 1.20=10⋅ 1.20=12 S1=a1⋅ b1=14.4⋅ 12=8645=172.8 o1=2⋅ (a1+b1)=2⋅ (14.4+12)=2645=52.8 p1=p1=100⋅ (o1−o)/o=100⋅ (52.8−44)/44=20=20%a = 12 \ \\ b = 10 \ \\ S = a \cdot \ b = 12 \cdot \ 10 = 120 \ \\ o = 2 \cdot \ (a+b) = 2 \cdot \ (12+10) = 44 \ \\ a_{ 1 } = a \cdot \ 1.20 = 12 \cdot \ 1.20 = \dfrac{ 72 }{ 5 } = 14.4 \ \\ b_{ 1 } = b \cdot \ 1.20 = 10 \cdot \ 1.20 = 12 \ \\ S_{ 1 } = a_{ 1 } \cdot \ b_{ 1 } = 14.4 \cdot \ 12 = \dfrac{ 864 }{ 5 } = 172.8 \ \\ o_{ 1 } = 2 \cdot \ (a_{ 1 }+b_{ 1 }) = 2 \cdot \ (14.4+12) = \dfrac{ 264 }{ 5 } = 52.8 \ \\ p_1=p_{ 1 } = 100 \cdot \ (o_{ 1 }-o)/o = 100 \cdot \ (52.8-44)/44 = 20 = 20 \%
p2=p2=100⋅ (S1−S)/S=100⋅ (172.8−120)/120=44=44%p_2=p_{ 2 } = 100 \cdot \ (S_{ 1 }-S)/S = 100 \cdot \ (172.8-120)/120 = 44 = 44 \%

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