3. As shown in the left figure, it is a regular pentagram, and $S_{\triangle A F J}$ $=10$. Then the area of the entire pentagram is $\qquad$

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3. $50+10 \sqrt{5}$.

Connect $G J, G I$. By the property of the golden ratio, we get $\frac{F G}{A F}=$ $\frac{\sqrt{5}-1}{2}=\frac{S_{\triangle G F J}}{S_{\triangle A F J}}$, then $S_{\triangle G F J}:=5(\sqrt{5}-1)$, $S_{\triangle G H I}=5(\sqrt{5}-1)$, and $\frac{S_{\triangle G I I}}{S_{\triangle G J A}}=\frac{J I}{J A}=$ $\frac{\sqrt{5}-1}{2}$, then
$$
\begin{array}{l}
S_{\triangle G J I}=\frac{\sqrt{5}-1}{2}[10+5(\sqrt{5}-1)], \\
\therefore S_{\text {irregular polygon }}=50+10(\sqrt{5}-1) \\
+\frac{\sqrt{5}-1}{2}[10+5(\sqrt{5}-1)] \\
=50+10 \sqrt{5} . \\
\end{array}
$$