We will solve this problem thus:
The Area of rectangle A is:
![\begin{gathered} \text{Area of A} \\ =2x*2x \\ =4x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/im9b29lkhyg4o52mneg187rgr5wzcb2kpn.png)
The Area of rectangle B is:
![\begin{gathered} =2x* x \\ =2x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qk0h1znwdkjpjizm3p9k695lo98htt3a1r.png)
The area of the largest rectangle is:
![\begin{gathered} =2x*(2x+x) \\ =2x*(3x) \\ =6x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2ab9v2wzzko9ov8l0f9yywn0wxls57ajn3.png)
To find the fraction of rectangle A in relation to the largest rectangle, we will proceed thus:
![\begin{gathered} =\frac{\text{Area of rectangle A}}{Area\text{ of largest rectangle}} \\ =(4x^2)/(6x^2) \\ =(4)/(6) \\ =(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d3g95z87uv937ay683atjetyu3vwavkua4.png)
So we can say that the area of rectangle A is 2/3 times the area of the largest rectangle.