Example 63. The base of a right prism, whose height is 1 cm, is a rhombus with sides equal to 2 cm and an acute angle of $30^{\circ}$. A cutting plane is drawn through the side of the base, the angle between which and the base plane is $60^{\circ}$. We need to find the area of the section.

Construction of the image. Let the figure $A B C D A_{1} B_{1} C_{1} D_{1}$ be the image of the given prism (Fig. 72). This is a complete image. Its parametric number $p=5$ (count it yourself).

Since the construction of the section in this example is a metric problem, and $p=5$, we cannot, by taking arbitrary points $M \in\left(D D_{1}\right)$ and $L \in\left(C C_{1}\right)$ (naturally, such that $\left.(M L) \|(A B)\right)$, claim that the quadrilateral $A M L B$ is the image of the given section. However, if we knew the position of point $M$ - the intersection point of the cutting plane with the edge $D D_{1}$, we could easily construct the image of the section.

If a perpendicular $D K$ is dropped from point $D$ to the edge $A B$, then in the resulting right triangle $A D K$ we will have $|A D|=2$ cm, $|D K|=1$ cm, and, consequently, $|A K|=\sqrt{3} \mathrm{~cm}$. Thus, to construct $[D K] \perp[A B]$, point $K$ should be chosen such that the equality $|A K|:|A B|=\sqrt{3}: 2$ is satisfied.

This construction can be performed, for example, as follows. Construct an auxiliary right triangle with the hypotenuse equal to $|A B|$ and one leg equal to $\frac{1}{2}|A B|$. Then the length of the second leg will be $\frac{\sqrt{3}}{2}|A B|$. Since $\frac{\sqrt{3}}{2}|A B|:|A B|=\sqrt{3}: 2$, $|A K|$ can be taken equal to the length of the second leg of the auxiliary triangle.

Further, since $\left[D D_{1}\right] \perp(A B C)$, for any point $M \in\left(D D_{1}\right)$, the segment $D K$ will serve as the projection of the segment $M K$ onto the plane $A B C$, i.e., $\angle M K D$ will be the linear angle of the dihedral angle at the edge $A B$.

If point $M$ is such that $\widehat{M K D}=60^{\circ}$, then in the right $\triangle M K D$, where $|D K|=1 \mathrm{cm},|M K|=2 \mathrm{cm},|M D|=\sqrt{3} \mathrm{cm}$. Thus, when constructing point $M$, it is necessary to take into account the equality $|M D|:\left|D D_{1}\right|=\sqrt{3}: 1$.

The construction of point $M$ can be performed, for example, as follows. Construct an auxiliary right triangle with one leg equal to $\left|D D_{1}\right|$ and the hypotenuse equal to $2\left|D D_{1}\right|$. Then the length of the second leg will be $\sqrt{3}\left|D D_{1}\right|$. Since $\sqrt{3}\left|D D_{1}\right|:\left|D D_{1}\right|=\sqrt{3}: 1$, $|M D|$ can be taken equal to the length of the second leg of the auxiliary triangle.

Having constructed point $M$, draw $(M L) \|(A B)$, where $L \in\left(C C_{1}\right)$. The construction of the line $A M$ allows finding the point $Q=(A M) \cap\left(A_{1} D_{1}\right)$, and the construction of the line $B L$ - the point $P=(B L) \cap\left(B_{1} C_{1}\right)$. The quadrilateral $A Q P B$ is the image of the given section.

Given: $A B C D A_{1} B_{1} C_{1} D_{1}$ is a prism, $A B C D$ is a rhombus, $\widehat{B A D}=$ $=30^{\circ},|A B|=2$ cm, $A Q P B$ is a section of the prism, $\left.((A Q P) \widehat{,(} A B C)\right)=$ $=60^{\circ}$.

To find: $S_{A Q P B}$.
Solution. It is clear that the quadrilateral $A Q P B$ is a parallelogram. Since $[M K] \perp [A B]$, then $[N K]$ is the height of this parallelogram, where $N=(M K) \cap (P Q)$. Thus, $S_{A Q P B} = |A B| \cdot |N K|$. Let's find $|N K|$.

In the right-angled $\triangle M D K$ we have: $|M K| = 2 \, \text{cm}, |M D| = \sqrt{3} \, \text{cm}$. It is not difficult to show that $(D K) \parallel (D_{1} N)$, i.e.,

$$
\frac{|M K|}{|N K|} = \frac{|M D|}{|D D_{1}|}, \text{ or } \frac{2}{|N K|} = \frac{\sqrt{3}}{1},
$$

from which $|N K| = \frac{2 \sqrt{3}}{3} \, \text{cm}$.

Thus, $S_{A Q P B} = \frac{4 \sqrt{3}}{3} \, \text{cm}^2$.

We omit the investigation.

Example 64. In a regular triangular pyramid, a section is made through the middle line of a lateral face, parallel to a lateral edge, and through the vertex of the pyramid not lying in this face. Determine the angle of inclination of the plane of the section to the plane of the base, given that the angle between the lateral edge of the pyramid and the plane of its base is $\alpha$.

Construction of the figure. Let the quadrilateral $S A B C$ with its diagonals be the image of the given pyramid (Fig. 73). This image is complete, $p=5$. Construct points $K$ and $E$ - the midpoints of edges $B C$ and $S B$, and then $\triangle A E K$ - the image of the given section of the pyramid (these constructions did not involve any parameter expenditure).

Given: $S A B C$ is a pyramid, $[S O]$ is its height, $|A B| = |B C| = |A C|, |B K| = |C K|, [E K]$ is the midline of $\triangle S B C, \triangle A E K$ is the section of the pyramid, $\widehat{S C O} = \alpha$.

Find: $((A E K), (A B C))$.

Solution. To determine the required dihedral angle, we need to construct its linear angle and find it. Since $p=5$, no arbitrary metric constructions can be performed on the existing image. We need to construct the images of two perpendiculars to the edge of the dihedral angle. We will perform the necessary construction as follows.

Construct $[O B]$ and in the plane $S O B$ construct $[E F] \parallel [S O]$. Then, since $[S O] \perp (A B C)$, it follows that $[E F] \perp (A B C)$, i.e., $[E F] \perp [O B]$. Since $[A K]$ is the median of the equilateral $\triangle A B C$, then $[B C] \perp [A K]$. Construct $(F L) \parallel [B C]$, then $(F L) \perp [A K]$. Finally, construct $[E L]$. Thus, $[A K] \perp [F L]$ and $[A K] \perp [E F]$. Therefore, $[A K] \perp [E L]$, i.e., $\angle E L F$ is the linear angle of the required dihedral angle. Let $\widehat{E L F} = x$. Noting that the given problem is of type $M$, introduce the auxiliary parameter $|A B| = a$. Now calculate $|E F|$ and $|F L|$.

Since $[E K]$ is the midline of $\triangle S B C$, then $|B E| = |S E|$. But by construction, $[E F] \parallel [S O]$. Therefore, $[E F]$ is the midline of $\triangle S O B$, so $|E F| = \frac{1}{2} |S O|$.

From the right-angled $\triangle S O C$ we find that $|S O| = |C O| \operatorname{tg} \alpha$, where $|C O| = \frac{2}{3} |C D| = \frac{a \sqrt{3}}{3}$.

Thus, $|E F| = \frac{a \sqrt{3}}{6} \operatorname{tg} \alpha$. Further, we easily find that $|F L| = \frac{1}{2} |B K| = \frac{1}{4} a$.

Thus, $\operatorname{tg} x = \frac{|E F|}{|F L|} = \frac{2 \sqrt{3}}{3} \operatorname{tg} \alpha$ and, therefore, $x = \operatorname{arctg} \left( \frac{2 \sqrt{3}}{3} \operatorname{tg} \alpha \right)$.

Investigation. By the meaning of the problem, $0^\circ < x < 90^\circ$, from which $\operatorname{tg} \alpha > 0$, and, therefore, $0^\circ < \alpha < 90^\circ$. But by the condition, $\alpha$ is the angle between the lateral edge of the pyramid and the plane of its base, i.e., the angle is acute. Thus, from the system of inequalities $0^\circ < x < 90^\circ$, no additional restrictions on $\alpha$ followed.

Remark. In the considered example, at first glance, it may seem that constructing the linear angle of the required dihedral angle is more expedient as follows (try this yourself): 1) $[O M] \parallel [B C]$, 2) $[S M]$, 3) $P = [S M] \cap [A E]$, 4) $[O P]$. Indeed, with such a construction, $\angle P O M$ is the linear angle of the dihedral angle formed by the planes $A E K$ and $A B C$. However, attempts to determine $\widehat{P O M}$ with such a construction encounter significant difficulties.

The need to construct a section sometimes arises even in problems where the section is not mentioned in the conditions (see Example 51, p. 82). In some cases, constructing the section is not necessary, but it can simplify the solution of the problem.