Maths probl! Cant solve it can anyone helpppp me!!!!!!!!!!

A piece of wire 44cm long is cut into two parts and each part is bent to form a square. If the total area of the two squares is 65cm(sqr) , find the perimeter of the two squares????

The sides of the squares are S and T

4*S + 4*T = 44...length of wire [1]

S^2 + T^2 = 65...sum of areas [2]

S + T = 11 from [1]

S = 11 - T

(11 - T)^2 + T^2 = 65...substitute for S in [2]

121 - 22*T + 2*T^2 = 65

56 - 22*T + 2*T^2 = 0

T^2 - 11*T + 28 = 0

(T - 7)*(T - 4) = 0

T = 7 or T = 4

So the sides of the squares are 7 and 4 cm

The perimeters of the squares are 4*4 and 4*7 ie 16 and 28 cm <<<

let x and y be the two sides of two squers x^2+y^2=65, 4x+4y=44 , so x+y=11, y=11-x

x^2+y^2=65 ----(1) put y=11-x we get x^2-11x+28=0 , x=7 or x=4 if x=7 then y=4 and if x=4 then y=7

perimeter of 1 st square is 16cm perimeter of 2 nd square is 26 cm

(x/4)^2 + (11-x/4)^2 = 65
x^2/16+121+x^2/16-22x/4=65
x^2-44x+448 = 0
(x-28)(x-16) = 0
Lengths cut = 28 and 16 cm
Sides of squares are 7 and 4 cm
Perimeters are 28 and 16 cm

Square 1 :4 x 4 = 16

Square 2 : 7 x 7 = 49 16 + 49 = 65

Square 1 : 4 + 4 + 4 + 4 = 16 cm Perimeter

Square 2 : 7 + 7 + 7 + 7 = 28 cm Perimeter 16 + 28 = 44 cm .... Your starting length

Answer :-

Total perimeter = 44 cm;
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Total area = 60.5 sq.cm;

5.7cm