Answer:
[tex]132\ in^{2}[/tex] is closest to the total surface area of the box
Step-by-step explanation:
we know that
The surface area of the box is equal to
[tex]SA=2B+PH[/tex]
where
B is the area of the base
P is the perimeter of the base
H is the height of the box
we have
[tex]L=7\frac{3}{4}\ in=\frac{7*4+3}{4}=\frac{31}{4}\ in[/tex]
[tex]W=3\frac{1}{4}\ in=\frac{3*4+1}{4}=\frac{13}{4}\ in[/tex]
[tex]H=3\frac{7}{8}\ in=\frac{3*8+7}{8}=\frac{31}{8}\ in[/tex]
Find the area of the base B
[tex]B=LW[/tex]
[tex]B=(\frac{31}{4})(\frac{13}{4})=\frac{403}{16}\ in^{2}[/tex]
Find the perimeter of the base P
[tex]P=2(L+W)[/tex]
[tex]P=2(\frac{31}{4}+\frac{13}{4}))[/tex]
[tex]P=2(\frac{44}{4})=22\ in[/tex]
Find the surface area
[tex]SA=2(\frac{403}{16})+22(\frac{31}{8})[/tex]
[tex]SA=(\frac{403}{8})+(\frac{682}{8})[/tex]
[tex]SA=\frac{1,085}{8}=135.625\ in^{2}[/tex]