contestada

The regular hexagon has a radius of 4 in.



What is the approximate area of the hexagon?

24 in.2
42 in.2
48 in.2
84 in.2

Respuesta :

[tex]\bf \textit{area of a regular polygon}=\cfrac{1}{2}nR^2\cdot sin\left(\frac{360}{n} \right) \\ \quad \\ \quad \\ \begin{cases} n=\textit{number of sides in the polygon}\\ R=\textit{radius of the polygon} \end{cases}[/tex]

so, sides= n= 6
radius = R = 4
[tex]\bf \textit{area of a regular polygon}=\cfrac{1}{2}nR^2\cdot sin\left(\frac{360}{n} \right) \\ \quad \\ \quad \\ \begin{cases} n=\textit{number of sides in the polygon}\\ R=\textit{radius of the polygon} \end{cases} \\ \quad \\ \quad \\ \cfrac{1}{2}6\cdot 4^2\cdot sin\left(\frac{360}{6} \right)\implies \cfrac{1}{2}\cdot 96\cdot sin\left(60^o \right) \\ \quad \\ 48\cdot sin\left(60^o \right)[/tex]
calculista

we know that

The Area of a Regular Polygon is equal to the formula

[tex] A=\cfrac{1}{2}nr^2\cdot sin\left(\frac{360}{n} \right) [/tex]

where

n is number of sides in the polygon

r is the radius of the polygon

In this problem

the regular polygon is a hexagon with radius of [tex] 4 [/tex] in

So

[tex] n=6\\ r=4 in [/tex]

Substitute the value of n and r in the formula above

[tex] A=\cfrac{1}{2}*6*4^2\cdot sin\left(\frac{360}{6} \right)\\ \\A=\cfrac{96}{2}* sin(60) \\ \\ A=\frac{96}{2} *\frac{\sqrt{3}}{2} \\ \\ A=24\sqrt{3} \\ \\ A=41.57 in^{2} [/tex]

therefore

the answer is the option

[tex] 42 in.2 [/tex]

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