Question

HOLA FRIENDS♥️♥️♥️♥️♥️

Sum of the areas of two squares is
$468m {?}^{2}$
. if the difference of their perimeters is 24m, find the sides of the two squares.

Answer 1

Let the side of the first square be x and the another square be y.

(1)

Given that Sum of areas of two squares is 468m^2.

We know that Area of square = (S)^2

= > x^2 + y^2 = 468 ------- (1)

(2)

Given that Difference of their parameters is 24m.

We know that perimeter of the squares = 4S.

= > 4x - 4y = 24

= > x - y = 6

= > x = 6 + y ------- (2)

Substitute (2) in (1), we get

= > (6 + y)^2 + y^2 = 468

= > 36 + y^2 + 12y + y^2 = 468

= > 2y^2 + 12y + 36 = 468

= > 2y^2 + 12y - 432 = 0

= > y^2 + 6y - 216 = 0

= > y^2 - 12y + 18y - 216 = 0

= > y(y - 12) + 18(y - 12) = 0

= > (y + 18)(y - 12) = 0

= > y = 12, -18

Since, y cannot be negative. so y = 12.

Substitute y = 12 in (1), we get

= > x^2 + y^2 = 468

= > x^2 + 144 = 468

= > x^2 = 468 - 144

= > x^2 = 324

= > x = 18

Therefore, the sides of two squares are 18cm and 12cm.

Hope this helps!

Answer 2

hello@

here is u r answer ##

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this attached file help u