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

Sagot :

MathG33k
[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]

Answer

Area 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 hexagon

a 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]

or

a = 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

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