A square-based right pyramid has a base edge and height of $40 \mathrm{~cm}$. We want to draw a line on the side faces from one vertex of the base to the opposite vertex of the base. What is the length of the shortest such line?
I. solution. The length of the sides of the base plate, which is also the height of the pyramid, is $a=m=40 \mathrm{~cm}$. The diagonal of the base plate is $d=a \sqrt{2}$, and let the slant height of the pyramid be $x$. The straight line of the pyramid's height bisects the base plate at the intersection of the diagonals, at point $O$. Let the vertices of the pyramid be $A, B, C, D, E$ as shown in Figure 1. In the triangle $EOC$, we apply the Pythagorean theorem:

$$
x^{2}=\left(\frac{a \sqrt{2}}{2}\right)^{2}+m^{2}
$$

Since $m=a$, we have $x=a \sqrt{\frac{3}{2}}$

![](https://cdn.mathpix.com/cropped/2024_05_02_845a87b07400e6f57356g-1.jpg?height=537&width=529&top_left_y=534&top_left_x=782)

Figure 1

Fold the triangular faces $ABE$ and $BCE$ around the edge $EB$ into a plane (Figure 2). From the equalities $AB=BC=a$, $AE=BE=CE=x$, and $\angle EAB = \angle ABE = \angle EBC = \angle ECB$, it follows that the quadrilateral $A'BC'E$ is a kite. The shortest distance between points $A'$ and $C'$ is the segment connecting them, which is the diagonal $A'C'$ of the kite. We need to calculate the length of this diagonal.

![](https://cdn.mathpix.com/cropped/2024_05_02_845a87b07400e6f57356g-1.jpg?height=439&width=631&top_left_y=1361&top_left_x=736)

Figure 2

In the triangle $A'BE$, the height $m_1$ to the side $A'B$; since $A'T=TB=\frac{a}{2}$, from the Pythagorean theorem:

$$
m_{1}=\sqrt{x^{2}-\left(\frac{a}{2}\right)^{2}}=\sqrt{a^{2} \frac{3}{2}-\frac{a^{2}}{4}}=a \sqrt{\frac{5}{4}}
$$

We know that the area of a kite is half the product of its diagonals. Write the area $T$ of the kite, which is nothing more than twice the area of the triangle $A'BE$, since $A'BE \triangle \cong BCE \triangle$.

The area of the triangle $A'BE$ is $\frac{a m_{1}}{2}$, so $T=a m_{1}$, and also $T=\frac{x \cdot A'C'}{2}$, thus

$$
a m_{1}=\frac{x \cdot A'C'}{2}
$$

From this,

$$
A'C'=\frac{2 a m_{1}}{x}
$$

Substitute the previously calculated values of $m_1$ and $x$, thus we get

$$
A'C'=2 a \frac{a \sqrt{\frac{5}{4}}}{a \sqrt{\frac{3}{2}}}=2 a \sqrt{\frac{5}{6}} \approx 73.03 \mathrm{~cm}
$$

II. solution. We know that the distance from a point to a line is the length of the perpendicular segment from the point to the line. Draw perpendiculars from $A$ and $C$ to the common edge $EB$ of the two triangles (Figure 3). From the congruence of triangles $AEB$ and $CEB$, it follows that the two perpendiculars intersect the edge $EB$ at the same point $P$. The distance $AP$ can be determined from the similarity of triangles $ETB$ and $APB$, both of which are right-angled and have the common angle $EBA$ (Figure 4). Thus,

$$
\frac{AP}{m_{1}}=\frac{a}{x},
$$

from which $AP=\frac{a m_{1}}{x}$, and twice this value is equal to the previously obtained value of $A'C'$. (Since $AP+PC$ is nothing more than the diagonal of the kite laid out in the plane in the previous steps.) From this, we can continue the calculation in the same way as in the first solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_845a87b07400e6f57356g-2.jpg?height=523&width=569&top_left_y=595&top_left_x=759)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_845a87b07400e6f57356g-2.jpg?height=401&width=336&top_left_y=1252&top_left_x=881)

Figure 4