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Sum of the areas of two squares is 468 m2 . If the difference of their perimeters is 24 m, find the sides of the two squares.

Sum of the areas of two squares is 468 m2 . If the difference of their perimeters is 24 m, find the sides of the two squares.

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student

Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y .
And area of the squares will be x^2 and y^2 .
Given,
4x- 4y= 24
x-y=6
x=y+6
Also,x^2+y^2= 468
⇒(6+y)^2+y^2=468
⇒36+y^2+12y+y^2=468
⇒2y^2+ 12y+ 432 = 0
⇒y^2+ 6y - 216 = 0
⇒y^2+18y-12y-216=0
⇒y(y+18) -12(y+ 18) = 0
⇒(y+ 18)(y- 12) = 0
⇒y= -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Regards

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