the side of a square exceeds the side of another square by 3 cm and the sum of the areas of the two squares is 549 cm ^2.find the perimeters of the square .
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The sides will be x of first then x+3 of there
X^2+(x+3)^2 =549
X2 + x2+9 +6x = 549
2X^2 +6x - 540 =0
X^2 + 3x - 275 =0
Now solve it
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