Question

Sum of areas of two square is 640. If difference in there perimeter is 64 find there sides

Answer 1

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Here is your answer
Let the Side of First Square is x

Area of First Square is = x^2
Perimeter of First Square is = 4x

Let Side of Second Square is y

Area of Second square is y^ 2
Perimeter of Second Square is = 4y

ATQ
$x {}^{2} + {y}^{2} = 640$
AND
$4x + 4y = 64 \\ 4(x + y) = 64 \\ x + y = \frac{64}{4} \\ x + y = 16 \\ x = 16 - y$
Now put the value of x in above we get

$(16 - y) {}^{2} + y {}^{2} = 640 \\ 256 + y {}^{2} - 32y + {y}^{2} = 640 \\ 2 {y}^{2} - 32y + 256 - 640 = 0 \\ 2 {y}^{2} - 32y - 384 = 0 \\ {y}^{2} - 16y - 192 = 0 \\ {y}^{2} - 24y + 8y - 192 = 0 \\ y(y - 24) + 8(y - 24) = 0 \\ (y - 24)(y + 8) = 0 \\ y = 24 \: \: and \: - 8 \\ as \: side \: is \: not \: negative \: so \\ y = 24$
Hence put it we get
${x}^{2} + (24) {}^{2} = 640 \\ {x}^{2} + 576 = 640 \\ {x}^{2} = 640 - 576 \\ {x}^{2} = 64 \\ x = \sqrt{64} \\ x = 8$

Therefore
Side of First Square is 8
Side of Second Square is 24

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