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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.

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3 months ago Anand Kumar Pandey
2937 Points
```							Dear StudentLet 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= 24x-y=6x=y+6Also,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, 12As 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
```
3 months ago
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