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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. Q.11

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How i solve the best question of class 10th of quadratic equations of question number 11. Sir please help me for solving this question in easy method 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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1. Let the sides of the two squares beÂ xÂ m andÂ yÂ m.

Therefore, their perimeter will be 4xÂ and 4yÂ respectively

And area of the squares will beÂ x2Â andÂ y2Â respectively.

Given,

4xÂ â€“ 4yÂ = 24

xÂ â€“Â yÂ = 6

xÂ =Â yÂ + 6

Also,Â x2Â +Â y2Â = 468

â‡’ (6Â +Â y2)Â +Â y2Â = 468

â‡’ 36Â +Â y2Â + 12yÂ +Â y2Â = 468

â‡’ 2y2Â + 12yÂ + 432 = 0

â‡’Â y2Â + 6y â€“ 216 = 0

â‡’Â y2Â + 18yÂ â€“ 12yÂ â€“ 216 = 0

â‡’Â y(yÂ +18) -12(yÂ + 18) = 0

â‡’ (yÂ + 18)(yÂ â€“ 12) = 0

â‡’Â yÂ = -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.

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