10,11

In a triangular pyramid $SABC$, it is known that $AB=AC=10, BC=16$. The height of the pyramid, dropped from vertex $S$, passes through vertex $B$ and is equal to 4. Find the total surface area of the pyramid and the radius of the sphere inscribed in the pyramid.
Let $Q$ be the area of the total surface of the given pyramid, $V$ its volume, $r$ the radius of the sphere inscribed in the pyramid, $A K$ and $B M$ the heights of triangle $A B C$. Then $K$ is the midpoint of the base $B C$ of the isosceles triangle $A B C$. Therefore,

$$
\begin{gathered}
A K=\sqrt{A B^{2}-B K^{2}}=\sqrt{100-64}=6, \\
B M=\frac{B C \cdot A K}{A C}=\frac{16 \cdot 6}{10}=\frac{18}{5}.
\end{gathered}
$$

Since $S B$ is perpendicular to the plane of the base $A B C$, $B M$ is the orthogonal projection of the inclined line $S M$ on the plane of the base $A B C$. Since $B M \perp A C$, by the theorem of three perpendiculars, $S M \perp A C$, i.e., $S M$ is the height of triangle $A S C$. From the right triangle $S B M$, we find that

$$
S M=\sqrt{B S^{2}+B M^{2}}=\sqrt{4^{2}+\left(\frac{18}{5}\right)^{2}}=4 \sqrt{1+\frac{111}{25}}=4 \cdot \frac{13}{5}=\frac{52}{5}
$$

Therefore,

$$
\begin{gathered}
Q=S_{\triangle S B C}+S_{\triangle S A B}+S_{\triangle S A C}+S_{\triangle A B C}= \\
=\frac{1}{2} B C \cdot S B+\frac{1}{2} A B \cdot S B+\frac{1}{2} A C \cdot S M+\frac{1}{2} B C \cdot A K= \\
=\frac{1}{2} \cdot 16 \cdot 4+\frac{1}{2} \cdot 10 \cdot 4+\frac{1}{2} \cdot 10 \cdot \frac{52}{5}+\frac{1}{2} \cdot 16 \cdot 6=32+20+52+48=152
\end{gathered}
$$

From the equality $V=\frac{1}{3} Q \cdot r=\frac{1}{3} S_{\triangle A B C} \cdot S B$ it follows that

$$
r=\frac{3 V}{Q}=\frac{3 \cdot 96}{152}=\frac{21}{19}
$$

## Answer

$152 ; \frac{21}{19}$