181 Given a regular quadrilateral pyramid with edge lengths and base side lengths both equal to 1. The radius $r=$ $\qquad$ of the escribed sphere that is tangent to the extensions of the four side faces and the base.
$181 \frac{\sqrt{6}+\sqrt{2}}{4} . \quad S_{\text {side }}=\frac{\sqrt{3}}{4}, S_{\text {base }}=1$, volume
$$
\begin{array}{c}
V=\frac{1}{3} \times 1 \times \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{1}{3 \sqrt{2}} \\
4 \times \frac{1}{3} r \times S_{\text {side }}-\frac{1}{3} r \times S_{\text {base }}=V .
\end{array}
$$

That is
$$
\begin{array}{c}
4 \times r \times \frac{\sqrt{3}}{4}-r=\frac{1}{\sqrt{2}} . \\
r=\frac{1}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{6}+\sqrt{2}}{4} .
\end{array}
$$