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"^22163 cm2

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)