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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# 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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Let the sides of the two squares beÂ

xÂ m andÂyÂ m.Therefore, their perimeter will be 4

xÂ and 4yÂ respectivelyAnd area of the squares will beÂ

x^{2}Â andÂy^{2}Â respectively.Given,

4

xÂ â€“ 4yÂ = 24xÂ â€“ÂyÂ = 6xÂ =ÂyÂ + 6Also,Â

x^{2Â }+^{Â }y^{2}Â = 468â‡’ (6Â +Â

y^{2})Â +Ây^{2}Â = 468â‡’ 36Â +Â

y^{2}Â + 12yÂ +Ây^{2}Â = 468â‡’ 2

y^{2}Â + 12yÂ + 432 = 0â‡’Â

y^{2}Â + 6y â€“ 216 = 0â‡’Â

y^{2}Â + 18yÂ â€“ 12yÂ â€“ 216 = 0â‡’Â

y(yÂ+18) -12(yÂ + 18) = 0â‡’ (

yÂ + 18)(yÂ â€“ 12) = 0â‡’Â

yÂ = -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.