Solution:
Given the figure below:
The above figure, when closed, results into a cuboid.
This can be proven in the diagram below:
where the cuboid has
[tex]\begin{gathered} \text{length}=7\text{ units} \\ \text{width}=5\text{ units} \\ \text{height}=2\text{ units} \end{gathered}[/tex]
The surface area of a cuboid is expressed as
[tex]\begin{gathered} \text{Area = 2(L}\times W)+2(L\times H)+2(H\times W) \\ \text{where} \\ L\Rightarrow\text{length} \\ W\Rightarrow\text{width} \\ H\Rightarrow\text{height} \end{gathered}[/tex]
Thus, the surface area of the cuboid is evaluated as
[tex]\begin{gathered} \text{Area = 2(L}\times W)+2(L\times H)+2(H\times W) \\ =2(7\text{ units}\times5\text{ units)+2(7 units}\times2\text{ units)+2(2 units}\times5\text{ units)} \\ =2(35\text{ square units)+2(14 square units})+2(10\text{ square units)} \\ =(70+28+20)\text{ square units} \\ =118\text{ square units} \end{gathered}[/tex]
Hence, the surface area of the box is
[tex]118\text{ square units}[/tex]
The correct option is D.
