Two squares

Two squares with sides in the ratio 3:7 have a sum of their perimeters 58 cm. Calculate the sum of the area of these two squares.

Correct answer:

s = 121.945 cm²

Step-by-step explanation:

a_{1}:a_{2} = 3:7\ = \ o_{1}:o_{2} \ \\ o = 58 \ \text{cm} \ \\ \ \\ o\ = \ o_{1}\ +\ o_{2}\ = \ 4(\ a_{1}\ +\ a_{2}\ ) \ \\ \ \\ o\ = \ 4(a_{1}+7/3 \cdot \ a_{1}) \ \\ \ \\ a_{1} = \dfrac{ 3 \cdot \ o }{ 4 \cdot \ (3+7) } = \dfrac{ 3 \cdot \ 58 }{ 4 \cdot \ (3+7) } = \dfrac{ 87 }{ 20 } = 4.35 \ \text{cm} \ \\ \ \\ a_{2} = \dfrac{ 7 }{ 3 } \cdot \ a_{1} = \dfrac{ 7 }{ 3 } \cdot \ 4.35 = \dfrac{ 203 }{ 20 } = 10.15 \ \text{cm} \ \\ \ \\ S_{1} = a_{1}^2 = 4.35^2 = \dfrac{ 7569 }{ 400 } = 18.9225 \ \text{cm}^2 \ \\ S_{2} = a_{2}^2 = 10.15^2 = 103.0225 \ \text{cm}^2 \ \\ \ \\ s = S_{1} + S_{2} = 18.9225 + 103.0225 = 121.945 = 121.945 \ \text{cm}^2 \ \\ \ \\ \text{ Verifying Solution: } \ \\ r = a_{1}:a_{2} = 4.35:10.15 = \dfrac{ 3 }{ 7 } \doteq 0.4286 = 3:7 \ \\ \ \\ o_{1} = 4 \cdot \ a_{1} = 4 \cdot \ 4.35 = \dfrac{ 87 }{ 5 } = 17.4 \ \text{cm} \ \\ o_{2} = 4 \cdot \ a_{2} = 4 \cdot \ 10.15 = \dfrac{ 203 }{ 5 } = 40.6 \ \text{cm} \ \\ O = o_{1}+o_{2} = 17.4+40.6 = 58 \ \text{cm}

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