Answer:
The total area is [tex]16\sqrt{3}\ units^{2}[/tex]
Step-by-step explanation:
we know that
The surface area of the regular pyramid is equal to the area of its four triangular faces
Each face is an equilateral triangle
so
Applying the law of sines
The surface area is equal to
[tex]SA=4[\frac{1}{2}b^{2}sin(60\°)][/tex]
we have
[tex]b=4\ units[/tex]
[tex]sin(60\°)=\frac{\sqrt{3}}{2}[/tex]
[tex]SA=4[\frac{1}{2}(4)^{2}\frac{\sqrt{3}}{2}][/tex]
[tex]SA=16\sqrt{3}\ units^{2}[/tex]
