11.226 The base of the pyramid $SABC$ is a triangle $ABC$ such that $AB=AC=10$ cm and $BC=12$ cm. The face $SBC$ is perpendicular to the base and $SB=SC$. Calculate the radius of the sphere inscribed in the pyramid if the height of the pyramid is 1.4 cm.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-276.jpg?height=260&width=347&top_left_y=1431&top_left_x=90)

The height $S K$ of the given pyramid $S A B C$ lies in the plane of the face $S B C$, since $S B C \perp A B C$ (by the condition). Therefore, $K$ is the midpoint of $B C$ (triangle $S B C$ is isosceles) $\Rightarrow$ $\Rightarrow A K \perp B C$. Let $S M$ be the apothem of the face $S A B$, then $M K \perp A B$ (Theorem 12),

$$
A K=\sqrt{A B^{2}-B K^{2}}=8 . \Delta B K M \sim \Delta B K A \Rightarrow \frac{M K}{8}=\frac{B K}{A B} \Rightarrow M K=4.8, S M=5
$$

Let $x$ be the radius of the sphere inscribed in the given pyramid. Then

$V_{S A B C}=\frac{x}{3}\left(S_{A B C}+S_{S B C}+2 S_{S A B}\right)=\frac{x}{3}\left(\frac{1}{2} A K \cdot B C+\frac{1}{2} S K \cdot B C+S M \cdot A B\right)=\frac{532}{15} x$.

On the other hand, $V_{S A B C}=\frac{1}{3} S K \cdot S_{A B C}=\frac{112}{5} \Rightarrow x=\frac{12}{19}$.

Answer: $\frac{12}{19}$ cm.