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Question
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
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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