The base of the pyramid is a triangle with sides 6, 5, and 5. The lateral faces of the pyramid form angles of $45^{\circ}$ with the plane of its base. Find the volume of the pyramid.
Let $ABCD$ be a given triangular pyramid with vertex $D$, $AB=AC=5$, $BC=6$, and $O$ be the base of the height dropped from vertex $D$. Drop perpendiculars $OA_1$, $OB_1$, and $OC_1$ to the lines $BC$, $AC$, and $AB$ respectively. By the theorem of three perpendiculars, $DA_1 \perp BC$, $DB_1 \perp AC$, and $DC_1 \perp AB$, so $DA_1O$, $DB_1O$, and $DC_1O$ are the linear angles of the dihedral angles formed by the lateral faces of the pyramid with the plane of its base. According to the problem,

$$
\angle DA_1O = \angle DB_1O = \angle DC_1O = 45^\circ
$$

Therefore, the right triangles $DA_1O$, $DB_1O$, and $DC_1O$ are equal by the leg and acute angle. Thus, $OA_1 = OB_1 = OC_1$, i.e., point $O$ is equidistant from the lines $BC$, $AC$, and $AB$. Consequently, point $O$ is either the center of the inscribed circle of triangle $ABC$ (Fig.1) or the center of one of the excircles of this triangle (Fig.2). Let $O$ be the center of the inscribed circle of triangle $ABC$, and $r$ be the radius of the circle. Then $r = \frac{S}{p}$, where $S$ is the area of the triangle, and $p$ is its semiperimeter. Since $AA_1$ is the height of triangle $ABC$, then

$$
AA_1 = \sqrt{AB^2 - BA_1^2} = \sqrt{25 - 9} = 4, \quad S = \frac{1}{2} BC \cdot AA_1 = \frac{1}{2} \cdot 6 \cdot 4 = 12, \quad r = \frac{S}{p} = \frac{12}{8} = \frac{3}{2}
$$

From the right triangle $DA_1O$, we find that

$$
DO = OA_1 = r = \frac{3}{2}
$$

Therefore,

$$
V_{ABCD} = \frac{1}{3} S \cdot DO = \frac{1}{3} \cdot 12 \cdot \frac{3}{2} = 6
$$

If $O_a$ is the center of the circle touching side $BC$ of triangle $ABC$ and the extensions of sides $AB$ and $AC$, and $r_a$ is its radius, then

$$
r_a = \frac{S}{p - BC} = \frac{12}{8 - 6} = 6, \quad DO_a = OA_a = r_a = 6, \quad V_{ABCD} = \frac{1}{3} S \cdot DO_a = \frac{1}{3} \cdot 12 \cdot 6 = 24
$$

If $O_b$ is the center of the circle touching side $AC$ of triangle $ABC$ and the extensions of sides $AB$ and $BC$, and $r_b$ is its radius, then

$$
r_b = \frac{S}{p - AC} = \frac{12}{8 - 5} = 4, \quad DO_b = OA_a = r_b = 4, \quad V_{ABCD} = \frac{1}{3} S \cdot DO_b = \frac{1}{3} \cdot 12 \cdot 4 = 16
$$

If $O_c$ is the center of the circle touching side $AB$ of triangle $ABC$ and the extensions of sides $AC$ and $BC$, and $r_c$ is its radius, then

$$
r_c = r_b = 4, \quad V_{ABCD} = 16
$$

## Answer

$6; 24; 16; 16$.