At the base of the pyramid $S A B C$ lies an equilateral triangle $A B C$, and all lateral faces have equal areas. The edge $S A$ is equal to 2, and the edge $S B$ is equal to $\sqrt{2}$. A section of the pyramid is made through vertex $B$ perpendicular to the edge $S C$. Find the area of this section.
Since the lateral faces of the pyramid are equal, and their bases are equal to the sides of an equilateral triangle, the heights of the lateral faces, drawn from the common vertex $S$, are also equal to each other. Therefore, the height $S H$ of the pyramid passes either through the center of the inscribed circle or through the center of one of the exscribed circles of the base $A B C$. Since, by the condition of the problem, the pyramid is not regular ($S A > S B$), the height passes through the center $H$ of the exscribed circle, and this circle touches the side $B C$ and the extensions of the sides $A C$ and $A B$. Let $r$ be the radius of this circle, the side of the base $A B C$ be $a$, and $M$ be the midpoint of $B C$. Then

$$
H M=\frac{a \sqrt{3}}{2}, H A=H M+A M=a \sqrt{3}
$$

Drop a perpendicular $S P$ from the vertex of the pyramid to the line $A B$. By the theorem of three perpendiculars, $H P \perp A B$, so $H P$ is the radius of the considered exscribed circle. From the right triangles $S H A$, $S B P$, and $S H P$, we find that

$$
S H^2=S A^2-H A^2=4-3 a^2, S P^2=S B^2-B P^2=2-\frac{1}{4} a^2
$$

$$
S H^2=S P^2-H P^2=2-\frac{1}{4} a^2-\frac{3}{4} a^2=2-a^2
$$

From the equation $4-3a^2=2-a^2$, we find that $a=1$. Then $S H=2-a^2=1$. Let $B K$ be the height of the triangle $S B C$ with sides $S B=S C=\sqrt{2}, B C=1$. Then

$$
B K \cdot S C=B C \cdot S M, B K=\frac{B C \cdot S M}{S C}=\frac{1 \cdot \sqrt{2-\frac{1}{4}}}{\sqrt{2}}=\frac{\sqrt{7}}{2 \sqrt{2}}
$$

Since $S B > B C$, the triangle $S B C$ is acute, so the point $K$ lies on the segment $S C$. From the right triangle $S K B$, we find that

$$
S K=\sqrt{S B^2-B K^2}=\sqrt{2-\frac{7}{8}}=\frac{3}{2 \sqrt{2}}
$$

The cutting plane passes through the point $B$ perpendicular to the line $S C$, so it intersects the edge $S C$ at the point $K$, and since the angle $A C S$ is obtuse, the point $L$ of intersection of the cutting plane with the line $S A$ lies on the edge $S A$. Thus, the section mentioned in the problem is the triangle $B K L$. Let the segments $S M$ and $B K$ intersect at the point $E$. The line $B C$ is perpendicular to the plane $A M S$, so $L E \perp B C$, and since $L E \perp S K$, the line $L E$ is perpendicular to the plane $S B C$, so $L E \perp K B$, i.e., $L E$ is the height of the triangle section $B K L$. Denote $\angle A S M=\alpha$. By the cosine theorem

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-40.jpg?height=198&width=1001&top_left_y=1067&top_left_x=534)

The right triangles $S K E$ and $S M C$ are similar, so

$$
\frac{S E}{S C}=\frac{S K}{S M}, S E=\frac{S K \cdot S C}{S M}=\frac{\frac{3}{2 \sqrt{2}} \cdot \sqrt{2}}{\frac{\sqrt{7}}{2}}=\frac{3}{\sqrt{7}}
$$

By the theorem of three perpendiculars, $S E \perp L E$ ( $S K$ is the perpendicular to the cutting plane, $K E$ is the orthogonal projection of the inclined $S E$ ). From the right triangle $S E L$, we find that

$$
L E=S E \cdot \operatorname{tg} \alpha=\frac{3}{\sqrt{7}} \cdot \frac{\sqrt{3}}{5}=\frac{3 \sqrt{3}}{5 \sqrt{7}}
$$

Therefore,

$$
S_{\triangle B K L}=\frac{1}{2} B K \cdot L E=\frac{1}{2} \cdot \frac{\sqrt{7}}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{5 \sqrt{7}}=\frac{3 \sqrt{6}}{40}
$$

## Answer

$3 \sqrt{6}$

40.