Answered

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

Verified Answer :

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

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

calculista

we know that

The Area of a Regular Polygon is equal to the formula

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

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 \(4\) in

So

\(n=6\\ r=4 in\)

Substitute the value of n and r in the formula above

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

therefore

the answer is the option

\(42 in.2\)

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