10.073. In a square with a side of 12 cm, the midpoints of its adjacent sides are connected to each other and to the opposite side of the square. Find the radius of the circle inscribed in the resulting triangle.
Solution.

Let's find the area of triangle $NMD$ (Fig. 10.71); $S_{\triangle NMD}=$ $=S_{ABCD}-2 S_{\triangle MCD}-S_{\triangle NBM} ; S_{\triangle NBM}=0.5 \cdot 6 \cdot 6=18\left(\mathrm{~cm}^{2}\right) ; S_{\triangle MCD}=0.5 \cdot 6 \cdot 12=$ $=36\left(\mathrm{~cm}^{2}\right) ; S_{\triangle NMD}=144-2 \cdot 36-18=54\left(\mathrm{~cm}^{2}\right)$. Since $MD=DN=$ $=\sqrt{12^{2}+6^{2}}=\sqrt{180}=6 \sqrt{5}(\mathrm{~cm}), MN=\sqrt{6^{2}+6^{2}}=6 \sqrt{2}(\mathrm{~cm})$, , $P_{\triangle NMD}=$ $=0.5(12 \sqrt{5}+6 \sqrt{2})=6 \sqrt{5}+3 \sqrt{2}$ (cm). Then, using formula (10.4), we get $r=\frac{54}{6 \sqrt{5}+3 \sqrt{2}}=\frac{18}{2 \sqrt{5}+\sqrt{2}}=2 \sqrt{5}-\sqrt{2}($ cm $)$.

Answer: $2 \sqrt{5}-\sqrt{2}$ cm.