6. The base of the pyramid $S A B C$ is an isosceles triangle $A B C$, with $A B = B C = 3$, and $A C = 5$. The height of the pyramid $S A B C$ is the segment $S O$, where $O$ is the intersection point of the line passing through vertex $B$ parallel to side $A C$ and the line passing through $C$ perpendicular to side $A C$. Find the distance from the center of the inscribed circle of triangle $A B C$ to the plane containing the lateral face $B S C$, if the height of the pyramid is $7 / 12$.
# Solution:

$$
A B=B C=b=3, \quad A C=a=5, d=\rho\left(O_{1}, B S C\right)
$$

$$
O B=a / 2, \quad O C=\sqrt{b^{2}-a^{2} / 4}, O K=\frac{O B \cdot O C}{B C}=\frac{a \sqrt{4 b^{2}-a^{2}}}{4 b}
$$

$r=\frac{2 S_{A B C}}{P_{A B C}}=\frac{a \sqrt{4 b^{2}-a^{2}}}{2(a+2 b)}, L K=r$, in triangle OSK we have $\frac{O T}{d}=\frac{O K}{L K}, \quad d=\frac{O T \cdot 2 b}{a+2 b}$,

$S O=h, \quad h \cdot O K=O T \sqrt{h^{2}+O K^{2}}$,

$O T=\frac{h \cdot O K}{\sqrt{h^{2}+O K^{2}}}=\frac{h a \sqrt{4 b^{2}-a^{2}}}{\sqrt{16 h^{2} b^{2}+a^{2}\left(4 b^{2}-a^{2}\right)}}$, $d=\frac{2 h a b \sqrt{4 b^{2}-a^{2}}}{(a+2 b) \sqrt{16 h^{2} b^{2}+a^{2}\left(4 b^{2}-a^{2}\right)}}=\frac{35 \sqrt{11}}{396}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-8.jpg?height=980&width=531&top_left_y=521&top_left_x=1436)

Answer: $\frac{35 \sqrt{11}}{396}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-8.jpg?height=474&width=938&top_left_y=1558&top_left_x=980)