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The sum of areas of two squares is 468 sq.m. If the difference between their perimeter is 24 m, then find sides of both squares.

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Let the side of a square is x m.

Perimeter of that square = 4x m

Difference in perimeter is 24 m

Perimeter of second square 

= 4x + 24 m

Then, side of second square 

= (4x + 24)/4 = 4(x + 6)/4 = (x + 6)m

Area of first square = x2 sq m

Area of second square 

= (x + 6)2 sq m = x2 + 12x + 36 sq m

Sum of areas of both squares = 468 sq m

x2 + (x2 + 12x + 36) = 468

⇒ 2x2 + 12x + 36 – 468 = 0

⇒ 2x2+ 12x – 432 =0

⇒ 2(x2 + 6x – 216) = 0

⇒ x2 + 6x – 216 = 0

⇒ x2 + 18x – 12x – 216 = 0

⇒ x(x + 18) – 12(x + 18) = 0

⇒ (x + 18)(x – 12) = 0

when x + 18 = 0, then x = -18 (not possible)

or x – 12 = 0, then x = 12

∴ x = 12

Side of smaller square = 12 m

and side of larger square 

= x + 6 = 12 + 6 

= 18 m

Thus sides of both  squares are 12 m. and 18 m respatively

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