4. As shown in Figure 2, the radii of $\odot O_{1}$ and $\odot O_{2}$ are 2 and 4 respectively, and $O_{1} O_{2}=10$. Then the area of $\triangle M N P$ formed by the two inner common tangents and one outer common tangent of the two circles is $\qquad$ .
4. $\frac{32}{3}$.

As shown in Figure 6, the external common tangent $AB = 4 \sqrt{6}$, the internal common tangents $CD = EF = 8$, $O_{1}P = \frac{10}{3}$, $O_{2}P = \frac{20}{3}$. By the Pythagorean theorem, we get
$$
PC = PF = \frac{8}{3}, PD = PE = \frac{16}{3}.
$$

Let $MA = MF = x$, $ND = NB = y$.
By the tangent segment theorem, we have $MB = ME$.
Thus, $AB - x = 8 + x$. Therefore, $x = 2 \sqrt{6} - 4$.
Similarly, we get $y = 2 \sqrt{6} - 4$.
At this point, we calculate
$S_{\text{side } s A O_{1} m U} = 2(2 \sqrt{6} - 4)$,
$S_{\text{side } B O_{2} D \mathrm{DN}} = 4(2 \sqrt{6} - 4)$,
$S_{\triangle O_{1} F P} = \frac{8}{3}, S_{\triangle O_{2} O P} = \frac{32}{3}$. $=\frac{32}{3}$.