Google - find the surface area of a hexagonal pyramid?

As I understand this, there is a straight pyramid, with it's top 4 length units u above the plane of the base. The connection line from the top through the centre of the regular hexagonal base is perpendicular to the plane of that base.

The radius r of the base (?) Well, given that the bottom of each triangular side which is also a side b of the hexagonal base is 2u, r=1.7u seems to be too rough an approximation of the incircle radius. We will derive the exact value of this radius, as we will need it
1stly to derive the formula for the area of the hexagon A_h, and
2ndly as one of the legs of a triangle to obtain the height of the isosceles triangular side h_ti of the pyramid.
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The base is a regular hexagon. Drawing the three diameters from three adjacent corners to their respective opposite corner via the centre, we get six congruent equilateral triangles. The sum of the areas of these triangles is the area of the hexagon.

The area of a triangle is 1/2 * b * h, with height h and base b.
Height of our equilateral triangle = h_te
(btw the radius of the incircle). Pythagoras tells us that
h_te = sqrt(b^2 - (1/2*b)^2) = sqrt(4/4 * b^2 - 1/4 * b^2) = b * sqrt(3) / 2.

The area of the equilateral triangle is then
A_te = 1/2 * b * b * sqrt(3) / 2 = b^2 * sqrt(3) / 4.

For b = 2u this means,
A_te = 1/2 *2 * sqrt(3) = sqrt(3)u^2 ;;; ≈1.732u^2.

The area of the hexagon A_hg is then
A_hg = 6 * sqrt(3)u^2.
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The mantle or envelope of the pyramid consists of 6 isosceles triangles ti. To determine the area A_ti of one of these, at first we have to calculate its height h_ti as the hypotenuse of a right triangle.
One leg of this triangle is the radius of the incircle again, which for b = 2 is equal to the height
h_te = sqrt(3). The other leg is the height of the pyramid h_p = 4. Thus
h_ti = sqrt( (sqrt(3))^2 + 4^2) = sqrt(19).
The area of one isosceles triangle side of the pyramid is
A_ti = 1/2 * 2 * sqrt(19) = sqrt(19)u^2.

The area of the mantle of the pyramid is then
6*A_ti = 6 * sqrt(19)u^2
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Including the base, the whole surface of the pyramid is

6 * ( sqrt(3) + sqrt(19) ) ≈ 6 * ( 1.732 + 4.359 ) ≈ 36.5457 u².