To begin, we first find the measure of the internal angles of the regular hexagon, using the formula below:
[tex]\frac{n-2}{n}\times180[/tex]
Since a hexagon is six-sided, we have that n = 6. So:
[tex]\frac{6-2}{6}\times180=\frac{4}{6}\times180=4\times30=120[/tex][tex]\Rightarrow120^o[/tex]
Thus, we have each internal angle to be 120 degrees.
Now, we consider the triangle ABC formed in the figure shown in the question.
- We know that
- Also, we know that all the sides of a regular ploygon are always equal and so AB = BC = 4 in.
Thus, we can redraw the figure, as shown below:
Now, we simply apply the Cosine rule to triangle ABC in order to obtain x, which is the length of side AC.
By cosine rule, we have:
[tex]x^2=4^2+4^2-2(4)\cdot(4)\cdot\cos 120^o[/tex]
Simplifying gives:
[tex]x^2=16^{}+16^{}-32\cdot\cos 120^o[/tex][tex]x^2=32^{}-32\cdot(-\frac{1}{2})[/tex][tex]x^2=32^{}-(-16)[/tex][tex]\begin{gathered} x^2=32^{}+16=48 \\ \Rightarrow x^2=48 \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow x=\sqrt[\square]{48} \\ \Rightarrow x=\sqrt[]{16\times3}=\sqrt[\square]{16}\times\sqrt[\square]{3}=4\times\sqrt[\square]{3} \\ \Rightarrow x=4\sqrt[]{3}\text{ in.} \end{gathered}[/tex]
Therefore, the length of the diagonal AC is:
[tex]4\sqrt[]{3}\text{ in.}[/tex]
Correct answer: option A
