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Sagot :

Answer:

[tex]162\sqrt{3}[/tex]

Step-by-step explanation:

The equation for finding the area of a hexagon with a side length is

[tex](s^23\sqrt{3} )/2[/tex]

insert [tex]6\sqrt{3}[/tex]

and you are left with 162\sqrt{3}

Hope that helps :)

Answer:

[tex] \displaystyle E)\: 162 \sqrt{3} [/tex]

Step-by-step explanation:

we are given a side a polygon

and said to figure out the area

recall the formula of regular polygon

[tex] \displaystyle \: \frac{ {na}^{2} }{4} \cot \left( \frac{ {180}^{ \circ} }{n} \right) [/tex]

where a represents the length of a side

and n represents the number of sides

the given shape has 6 sides

and has a length of [tex]\displaystyle 6\sqrt{3}[/tex]

so our n is 6 and a is 6√3

substitute the value of n and a:

[tex] \displaystyle \: \frac{ {6 \cdot \:( 6 \sqrt{3} })^{2} }{4} \cot \left( \frac{ {180}^{ \circ} }{6} \right) [/tex]

reduce fraction:

[tex] \displaystyle \: \frac{ {6 \cdot \:( 6 \sqrt{3} })^{2} }{4} \cot \left( \frac{ { \cancel{180}^{ \circ}} ^{ {30}^{ \circ} } }{ \cancel{6 \: } } \right) [/tex]

[tex] \displaystyle \: \frac{6 \cdot \: (6 \sqrt{ {3} } {)}^{2} }{4} \cot( {30}^{ \circ} ) [/tex]

simplify square:

[tex] \displaystyle \: \frac{6 \cdot \: 36 \cdot \: 3 }{4} \cot( {30}^{ \circ} ) [/tex]

reduce fraction:

[tex] \displaystyle \: \frac{6 \cdot \: \cancel{36} \: ^{9} \cdot \: 3 }{ \cancel{ 4 \: } } \cot( {30}^{ \circ} ) [/tex]

[tex] \displaystyle \: 6 \cdot \: 9 \cdot \: 3 \cot( {30}^{ \circ} ) [/tex]

simplify multiplication:

[tex] \displaystyle \: 162\cot( {30}^{ \circ} ) [/tex]

recall unit circle:

[tex] \displaystyle \: 162 \sqrt{3} [/tex]

hence, our answer is E