Given:
Slant = 6
a = side lenght = 4
Apply : area of an hexagon
[tex]A=\frac{3\sqrt[\placeholder{⬚}]{3}}{2}a^2[/tex]
Where:
A = area
Replacing:
[tex]A=\frac{3\sqrt[\placeholder{⬚}]{3}}{2}(4)^2=24\sqrt[\placeholder{⬚}]{3}\text{ square units}[/tex]
Now, the lateral face of the pyramid is a triangle , base = 4 , height = 6
An Hexagon has 6 triangular faces.
Area of a triangle = 1/2 bh
LA = 6 x 1/2 x 4 x 6 = 72 square units
Total area (TA) = A + LA = 24 √3 + 72 square units
Answer: 72 + 24 √3
