Question

the sum of the area of two squares is 468 m^2 and the difference of perimeters are 24 m. tell the sides of the squares?

Answer 1

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Answer 2

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$Let, \\ \\The \: sides \: of \: the \: two \: squares \: are:-\\ \: a\: and \:b \\ \\ Sum \: of \: their \: areas \: \\= a^2 + b^2 = 468 \\ \\ Difference \: of \: their \: perimeters \: \: \\=>4a - 4b = 24 \\ => a - b = 6 \\ => a = b + 6 \\ So, we \: get \: the \: equation \\ (b + 6)^2 + b^2 = 468 \\ => 2b^2 + 12b + 36 = 468 \\ => b^2 + 6b - 216 = 0 \\ =>b^2+18b-12b-216=0\\ =>b (b+18)-12(b+18)=0\\ =>(b+18)(b-12)\\\\ => b = 12\: or\: b\:=\:-18 \\ we\: take\: b\:=12 \:positive\: only\\ =>a=b+6\\ => a = 18 \\ The \: sides \: of \: the \: two \: squares \: are\\ \: 12m \: and \: 18m$