The area of the rectangle is given by:
[tex]A_{rectangle}=2x(x-1)=2x^2-2x[/tex]
The area of the trapezium is given by:
[tex]A_{trapezium}=\frac{x\lbrack(2x+2)+(x-1)\rbrack}{2}=\frac{3x^2+x}{2}[/tex]
Then we have:
[tex]\begin{gathered} A_{rectangle}+3=A_{trapezium} \\ 2x^2-2x+3=\frac{3x^{2}+x}{2} \\ 4x^2-4x+6=3x^2+x \\ x^2-5x+6=0 \\ x=\frac{5\pm\sqrt{(-5)^2-4\cdot6}}{2} \\ x_+=3 \\ x_-=2 \end{gathered}[/tex]
Then, the two possible values for the area of the rectangle are:
[tex]\begin{gathered} 2\cdot3^2-2\cdot3=12\text{ cm}^2 \\ 2\cdot2^2-2\cdot2=4\text{ cm}^2 \end{gathered}[/tex]
