5. If the three views (front view, side view, top view) of a tetrahedron are all isosceles triangles with a base of 2 and a height of 2, then the surface area of this geometric body is
5. $2+\sqrt{5}+\sqrt{21}$.

The base $ABC$ of the tetrahedron $P-ABC$ is an isosceles triangle with a base of 2 and a height of 2; the front side $PBC$ is an isosceles triangle with a base of 2 and a height of $\sqrt{5}$; the other two sides are non-isosceles triangles with a base of $\sqrt{5}$ and a height of $\sqrt{\frac{21}{5}}$. Therefore, the surface area of the tetrahedron is
$$
\begin{aligned}
S & =\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 2 \times \sqrt{5}+2 \times \frac{1}{2} \times \sqrt{5} \times \sqrt{\frac{21}{5}} \\
& =2+\sqrt{5}+\sqrt{21} .
\end{aligned}
$$