The two classrooms are identical in length and width. On the other hand, the dimensions of the storage closet are
[tex](40-34)\times(36-30)=6\times6[/tex]
The shape of both classrooms and the storage closet is rectangular; therefore, their areas are
[tex]\begin{gathered} A_{\text{rectangle}}=l\cdot w\to length\cdot width_{} \\ \Rightarrow A_{\text{Friedman}}=40\cdot36 \\ _{}A_{\text{Elliot}}=40\cdot36 \\ A_{storage}=6\cdot6 \\ \end{gathered}[/tex]
Simplifying,
[tex]\begin{gathered} \Rightarrow A_{\text{storage}}=36ft^2 \\ \Rightarrow A_{\text{Friedman}}=A_{\text{Elliot}}=1440ft^2 \end{gathered}[/tex]
Finally, the total area of the compound is
[tex]\begin{gathered} A_{\text{total}}=A_{\text{Friedman}}+A_{\text{Elliot}}-A_{\text{storage}} \\ \Rightarrow A_{\text{total}}=2\cdot1440-36=2844 \end{gathered}[/tex]
Thus, the total area of the two classrooms plus the closet is 2844ft^2
Then,
