Note that the side of each square is unity.
Along the horizontal, it has 3 complete squares, so the length of Henry's rectangle is 3 units,
[tex]\begin{gathered} L=1+1+1 \\ L=3 \end{gathered}[/tex]Along the vertical, it has 2 complete squares while 1 half heighted rectangle, so the width of Henry's rectangle is,
[tex]\begin{gathered} W=1+1+\frac{1}{2} \\ W=1+1+0.5 \\ W=2.5 \end{gathered}[/tex]Solve for the area of the rectangle as,
[tex]\begin{gathered} \text{Area}=\text{Length}\times\text{ Width} \\ \text{Area}=3\times2.5 \\ \text{Area}=7.5 \end{gathered}[/tex]Thus, the Henry's rectangle is 3 units long, 2.5 units wide, and the area of the rectangle is 7.5 square units.