Example 62. At the base of the pyramid $SABC$ lies a right-angled triangle $ABC$. The edge $SA$ is perpendicular to the base plane, $|SA|=|AB|=|BC|=a$. Through the midpoint of the edge $AB$, a plane is drawn perpendicular to the edge $SC$, and we need to find the area of the resulting section.

Construction of the image. Let the quadrilateral $SABC$ with its diagonals $SB$ and $AC$ be the image of the given pyramid (Fig. 71). This image is complete. Its parametric number $p=5$. We will now proceed directly to the construction of the section.

1) $[AD]$ - the median of $\triangle SAB$,
2) point $M$ - the midpoint of the edge $AB$,
3) $[MN] \parallel [AD]$,
4) $[BP]$ - the median of $\triangle ABC$,
5) $[MK] \parallel [BP]$.

To construct $[KL] \perp [SC]$, we first construct $[AE] \perp [SC]$. This is a metric construction and cannot be performed arbitrarily. We will proceed as follows. Notice that in the right-angled

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-104.jpg?height=703&width=657&top_left_y=1656&top_left_x=1042)

Fig. 71
$\triangle ABC$, $|AB|=|BC|=a$ and therefore $|AC|=a\sqrt{2}$. Then, from $\triangle SAC$, where $|SA|=a$, we find that $|SC|=a\sqrt{3}$. Thus, for the segment $AE$ to be the image of the perpendicular to the edge $SC$, the following equality must hold:

$|SA|^2 - |SE|^2 = |AC|^2 - |CE|^2$, or $a^2 - |SE|^2 = 2a^2 - (a\sqrt{3} - |SE|)^2$, from which we find that $|SE| = \frac{a\sqrt{3}}{3}$, i.e., $|SE|:|SC|=1:3$.

Thus, we will complete the construction in the following order:

6) point $E$ such that $|SE|:|SC|=1:3$,
7) $[AE]$,
8) $[KL] \parallel [AE]$,
9) $[LN]$.

We will prove that the plane of the quadrilateral $MKLN$ is perpendicular to the edge $SC$. Indeed, $[SA] \perp (ABC)$, i.e., $[SA] \perp [MK]$, or $[MK] \perp [SA]$. Moreover, by construction, $[MK] \perp [AC]$. Then $[MK] \perp [LK]$ and $[MK] \perp [SC]$. Further, $[SC] \perp [LK]$ and $[SC] \perp [MK]$, i.e., $[SC] \perp (MKL)$. Thus, the section $MKLN$ meets the conditions of the problem and, therefore, is the desired one.

It is clear that since the cutting plane is perpendicular to a given line and passes through a given point belonging to the surface of the pyramid, the section of the pyramid defined by these conditions exists and is unique.

Thus, the construction of the image is complete, and we can proceed to the next stages of the solution.

Given: $SABC$ - pyramid, $S$ - vertex, $[SA] \perp (ABC)$, $[AB] \perp [BC]$, $|SA|=|AB|=|BC|=a$, $|AM|=|BM|$, $(MKLN)$ - section of the pyramid, $(MKL) \perp (SC)$.

Find: $S_{MKLN}$.
The problem is solved. To calculate the desired area, we first determine the type of quadrilateral $M K L N$.

From the right triangles $A B P$ and $S A B$, we have respectively:

$$
M K=\frac{1}{2}|B P| \text { and }|M N|=\frac{1}{2}|A D| .
$$

But $|B P|=|A D|=\frac{a \sqrt{2}}{2}$. Thus, $|M K|=|M N|=$ $=\frac{a \sqrt{2}}{4}$.

Since $[S A] \perp(A B C)$, then $[A B]$ is the projection of edge $S B$ onto the plane $A B C$. But $[A B] \perp[B C]$. Therefore, $[S B] \perp[B C]$.

From $\triangle S L N \sim \triangle S B C$

$$
\frac{|L N|}{|B C|}=\frac{|S N|}{|S C|}
$$

from which we find $|L N|=\frac{a \sqrt{6}}{4}$.

104

From $\triangle C L K \sim \triangle C A E$

$$
\frac{|L K|}{|A E|}=\frac{|C K|}{|A C|}=\frac{3}{4}, \quad \text { from which }|L K|=\frac{3}{4}|A E| \text {. }
$$

But $|A E| \cdot|S C|=|S A| \cdot|A C|$, i.e., $|A E|=\frac{|S A| \cdot|A C|}{|S C|}=\frac{a \sqrt{6}}{3}$, and then $|L K|=\frac{a \sqrt{6}}{4}$.

Thus, the quadrilateral $M K L N$ has the special property that $|M K|=|M N|,|L K|=|L N|$.

It is not difficult to see that $\triangle M K L \cong \triangle M N L$ and therefore $S_{M K L N}=2 S_{\triangle M K L}$. But $[M K] \perp[A C]$ and $[M K] \perp[S A]$, i.e., $[M K] \perp[L K]$.

Therefore, $S_{\triangle M K L}=\frac{1}{2}|M K| \cdot|L K|=\frac{a^{2} \sqrt{3}}{16}$,

and thus, $S_{M K L N}=\frac{a^{2} \sqrt{3}}{8}$.

Let's now consider problems of the second type.