Consider a triangular pyramid $ABCD$ with vertex $A$ and base $BCD$. Let $AM$ be the height of the isosceles triangle $ADC$. Then $M$ is the midpoint of $CD$ and

$$
AM = \sqrt{AD^2 - DM^2} = \sqrt{19 - 1} = 3\sqrt{2}
$$

Since the plane of the face $ADC$ is perpendicular to the plane of the base $BCD$, the line $AM$ is perpendicular to the plane of the base. Therefore, $AM$ is the height of the pyramid $ABCD$. The median $BM$ of the isosceles triangle $BCD$ is its height. The line $BM$ is the orthogonal projection of the inclined line $AB$ on the plane of the base of the pyramid. Since $BM \perp CD$, by the theorem of three perpendiculars, $AB \perp CD$. From the right triangle $BCM$, we find that

$$
BM = \sqrt{BC^2 - CM^2} = \sqrt{19 - 1} = 3\sqrt{2}.
$$

$$
AB = \sqrt{AM^2 + BM^2} = \sqrt{18 + 18} = 6
$$

Consider the section of the pyramid by a plane intersecting the edges $AC, BC, BD$, and $AD$ at points $P, Q, R$, and $T$ respectively. Suppose that $PQ \parallel RT$. Then through the parallel lines $PQ$ and $RT$ there are two planes $ABC$ and $ABD$, intersecting along the line $AB$. Therefore, the lines $PQ$ and $RT$ are parallel to the line $AB$. Similarly, if $PT \parallel QR$, then the lines $PT$ and $QR$ are parallel to the line $CD$. From the previously proven $AB \perp CD$. Therefore, if the plane intersecting the edges $DB, BC, AC$, and $AD$ forms a parallelogram in the section of the pyramid, then this parallelogram is a rectangle. Let $\frac{BK}{BD} = k$. Then

$$
RQ = kCD = 2k, \quad RT = (1-k)AB = 6(1-k)
$$

From the equation $2k = 6(1-k)$, we find that $k = \frac{3}{4}$. Therefore, if the specified section is a square, then $RQ = RT = \frac{3}{2}$. The area of such a section is $\frac{9}{4}$. Suppose now that the intersecting plane intersects the edges $DC, DB, AC$, and $AB$, and a parallelogram is formed in the section. Similarly to the previous case, we will prove that one pair of opposite sides of the parallelogram is parallel to the edge $AD$, and the other pair is parallel to the edge $BC$. Therefore, if a rectangle were formed in the section, then the opposite edges $AD$ and $BC$ of the pyramid would be perpendicular. We will prove that this is not the case. Suppose that $AD \perp BC$. Since the line $CD$ is the orthogonal projection of the inclined line $AD$ on the plane of the base of the pyramid, by the theorem of three perpendiculars, $CD \perp BC$, which is impossible ($\angle DCB < 90^\circ$ as the angle at the base of the isosceles triangle). Similarly, in the section of the pyramid $ABCD$ by a plane intersecting the edges $CD, BC, AD$, and $AB$, a square cannot be formed.

## Answer

$6; \frac{9}{4}$.
10,11 |

In a triangular pyramid, the lateral faces $D B C$ and $D C A$ are mutually perpendicular and are equal isosceles triangles with base $C D=2$ and lateral side equal to $\sqrt{19}$. Find the edge $A B$, as well as the areas of those sections of the pyramid that are squares.