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The sum of the areas of two squares is `640m^2` . If the difference in their perimeter be 64m, find the sides of the two square

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#### Solution

Let the length of the side of the first and the second square be x and y. respectively. According to the question:

`x^2+y^2=640` ................(1)

Also,

`4x-4y=64`

⇒`x-y=16`

⇒`x=16+y`

Putting the value of x in (1), we get:

`x^2+y^2=640`

⇒`(16+y)^2+y^2=640`

⇒`256+32y+y^2+y^2=640`

⇒`2y^2+32y-384=0`

⇒`y^2+16y-192=0`

⇒`y^2+(24-8)y-192=0`

⇒`y^2+24y-8y-192=0`

⇒`y(y+24)-8(y+24)=0`

⇒`(y+24)(y-8)=0`

⇒`y=-24 or y=8`

∴ y=8 (∵ Side cannot be negativ)

∴` x=16+y=16+8=24m`

Thus, the sides of the squares are 8 m and 24 m.

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