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Given:
The length of the floor is 7 feet greater than its width w.
Therefore, length, l=7+w.
Since the dimensions are increased by 2 feet,
The new width, w'=w+2.
The new length, l'=l+2=7+w+2=9+w.
Now, the new area N of the floor of the rectangular cage is,
[tex]\begin{gathered} N=l^{\prime}w^{\prime} \\ =(9+w)(w+2) \\ =9(w+2)+w(w+2) \\ =9w+9\times2+w^2+2w \\ =11w+18+w^2 \\ =w^2+11w+18 \end{gathered}[/tex]Therefore, the new area of the floor of the rectangular cage is,
[tex]w^2+11w+18[/tex]