To find the area of the triangle shown we have to find the length of a and the base.
a can be found using the sine function. Recall sin = (opp)/(hyp)
Here [tex]sin 60 = \frac{a}{8}
[/tex]
[tex] \frac{ \sqrt{3} }{2}= \frac{a}{8}
[/tex]
[tex]2a=8 \sqrt{3} [/tex]
[tex]a=4 \sqrt{3} [/tex]
The base can be found using the sin of 30 degrees as this is a 30-60-90 right triangle.
[tex]sin30= \frac{b}{8} [/tex]
[tex]\ \frac{1}{2} = \frac{b}{8} [/tex]
[tex]b=4[/tex]
The area of the triangle in the picture is given by [tex]A=( \frac{1}{2} )bh= (\frac{1}{2})(4)( \frac{ \sqrt{3} }{2} )= \sqrt{3} [/tex].
The base of the pyramid is a hexagon which can be divided into 6 triangles each of which is double the area of the one we just found. So the area of the base = 2(area of the triangle we found)(6) = [tex]12 \sqrt{3} [/tex]
