Answer:
[tex]s=8.1\ in[/tex]
Step-by-step explanation:
we know that
The area of a square is
[tex]A=s^{2}[/tex]
Solve for s
[tex]s=\sqrt{A}[/tex]
where
s is the length side of the square
A is the area of the square
In this problem we have
[tex]A=65\ in^2[/tex]
substitute in the formula
[tex]s=\sqrt{65}[/tex]
[tex]s=8.06\ in[/tex]
Round to the nearest tenth of a inch
[tex]s=8.1\ in[/tex]