Question

Math Help???! Please?

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.

Answer 1

`216sqrt3 " cm"^2`

Explanation:

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`

Answer 2

`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)`