First find the areas of the two separate rectangles where:
Area of rectangle = length x width
Area of the small rectangle = x(x–4)
Area of small rectangle = [tex]x^{2} -4x[/tex]
Area of big rectangle = (x–2)(2x+x)
Area of big rectangle = (x–2)(3x)
Area of big rectangle = [tex]3x^{2} -6x[/tex]
The total area of the compound shape is 36, therefore:
[tex]3x^{2} -6x +x^{2} -4x=36[/tex]
[tex]4x^{2} -10x-36=0[/tex] (collecting like-terms and by bringing the +36 to the other side)
By dividing this by 2, you get:
[tex]2(2x^{2} -5x-18)=0[/tex]
[tex]2x^{2} -5x-18=0[/tex]
To find the length of AB (x), solve the quadratic by either factorising, completing the square or by using the quadratic formula, etc to solve for x.
[tex]2x^{2} -5x-18=0[/tex]
[tex](2x-9)(x+2)=0[/tex]
[tex]x=\frac{9}{2}[/tex]
[tex]x=-2[/tex]
However, a length cannot be negative, so x must be [tex]\frac{9}{2}[/tex] or 4.5cm, so length AB = 4.5cm.
Hope this helps :)