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

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

Solution) Let the sides of the two squares be x m and y m.

Thus, their perimeters will be 4x and 4y and areas will be x² and y².

Hence, according to question –

=4x-4y=24

⇒x-y=6

⇒x=y+6

And x²+y² = 468

Substituting value of x-

(y+6)²+y²=468

⇒36+y²+12y+y²=468

⇒2y²+12y-432=0

⇒y²+6y-216=0

⇒y²+18y-12y-216=0

⇒y(y+18)-12(y+18)=0

⇒(y+18)(y-12)=0

⇒y=-18 or 12


Since, side cannot be negative.

Therefore, the sides of the square are 12 m and (12+6)m i.e 18 m.

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