This is a regular tetrahedron in which all four faces are equilateral triangles.
The area of an equilater triangle of base b, is
[tex]A=\frac{\sqrt[]{3}}{4}b^2[/tex]Since, the tetrahedron is 4 faces, then its surface area is sqrt(3)*b^2
[tex]\begin{gathered} SA=4\cdot A \\ =4\cdot\frac{\sqrt[]{3}}{4}\cdot b^2 \\ =\sqrt[]{3}\cdot b^2 \end{gathered}[/tex]