Find the area of a regular hexagon with the given measurement
48-inch perimeter
A = sq. in.

AnswerArea of a regular hexagon is given by:[tex]A = \frac{3\sqrt{3}}{2}a^2[/tex]            .....[1]where, A is the area of a regular hexagona is the side of the hexagon.As per the statement:Perimeter of a regular hexagon is 48 inch.Perimeter(P) of a regular hexagon is given by:[tex]P=6a[/tex]Substitute the given values we have;[tex]48 = 6a[/tex]Divide both sides by 6 we have;[tex]8 = a[/tex]ora = 8 inch.Substitute the value of a = 8 inches  in [1] we have;[tex]A = \frac{3\sqrt{3}}{2}(8)^2[/tex] ⇒[tex]A = 3\sqrt{3} \cdot 32 = 96\sqrt{3}[/tex]Therefore, the area of a regular hexagon is[tex]A = 96\sqrt{3}[/tex] sq. in

[tex]Area = \frac{3 \sqrt{3} }{2} s^{2}
\\ \\ \frac{48}{6} = 8=s
\\ \\ Area =\frac{3 \sqrt{3} }{2} 64
\\ \\ Area =166.28[/tex]