Math Help???! Please?

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The base of a regular pyramid is a hexagon.

What is the area of the base of the pyramid?

Express your answer in radical form.

2 Answers
Mar 27, 2018

#216sqrt3 " cm"^2#

Explanation:

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A regular hexagon can be divided into six congruent triangles, #DeltaABC# is one of them, as shown in the figure,
#=> AC=AB=BC=12# cm
#CD=a=12cos30=12*sqrt3/2=6sqrt3# cm
area of #DeltaABC=A_1=1/2*AB*CD=1/2*12*6sqrt3=36sqrt3 " cm"^2#
Area of the regular hexagon #A_H=6*A_1=6*36sqrt3=216sqrt3 " cm"^2#

Side notes : formula for a regular hexagon is #A_H=(3sqrt3)/2*x^2#, where #x# is the length of one side of the hexagon.
given that #AC=12, => AB=AC=12# cm
#=> A_H=(3sqrt3)/2*AB^2=(3sqrt3)/2*12^2=216sqrt3 " cm"^2#

Mar 27, 2018

#Area = C^2 sqrt(27/4)#
Where, c: distance from center to hexagon corner = 12
#Area = (12)^2 sqrt(27/4)# = #216 sqrt(3)#

Explanation:

Finding the rectangle high "a" from hexagon center to center of one hexagon side "c/2"

#a = sqrt ((c^2) -(c/2)^2)#
Or,

#a = c sqrt (3/4)#

#"Area of an equilateral triangle"# = #"c . c"#__#sqrt(27/4)#

The hexagon area will be 6 times the equilateral triangle:

#Area = C^2 6/4 sqrt(3)#

Finally,

#Area = C^2 sqrt(27/4)#

Where, c: distance from center to hexagon corner

If c = 12

Then
#Area = (12)^2 sqrt(27/4)# = #216sqrt(3)#