A pyramid with a quadrilateral base $ABCD$ has a trapezoid as its base, where $AB=BC=CD=\frac{1}{2} AD$. The foot of the pyramid's height is at the intersection of the diagonals of the base, and the edges starting from vertices $A$ and $C$ form a right angle at the apex of the pyramid. What is the surface area and volume of the pyramid?
Let the length of side $AB$ of the trapezoid be denoted by $a$. The two bases cannot be of equal length, because in that case the legs would also be of equal length, although exactly 3 of the four sides are of equal length. Therefore, the bases of the trapezoid can only be $CB$ and $AD$, and the legs $AB$ and $CD$ are of equal length. Extend the trapezoid to form a triangle, and let $E$ be the intersection point of the legs (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_a0d6d3e839b26a9f3eb9g-1.jpg?height=488&width=528&top_left_y=325&top_left_x=777)

Figure 1

Since $CB \parallel AD$ and $CB = \frac{1}{2} AD$, the triangle $AED$ is equilateral with side length $2a$.

The side of the equilateral triangle $BEC$ is half the side of the triangle $ADE$, so its area is a quarter of that of $ADE$. Therefore, the area of the trapezoid is

$$
T_{ABCD} = \frac{3}{4} T_{ADE} = \frac{3a^2 \sqrt{3}}{4}
$$

Let the intersection point of the diagonals of the trapezoid be denoted by $M'$. Point $B$ bisects segment $AE$, and similarly, point $C$ bisects $DE$. Thus, $M'$ is the centroid of the equilateral triangle $ADE$, so

$$
AM' = \frac{2}{3} AC = \frac{2a}{\sqrt{3}}
$$

$$
CM' = \frac{1}{3} AC = \frac{a}{\sqrt{3}}
$$

Let the apex of the pyramid be denoted by $M$. According to the problem, the orthogonal projection of $M$ onto the plane $ABCD$ is exactly $M'$, so $MM'$ is the altitude of the triangle $AMC$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_a0d6d3e839b26a9f3eb9g-1.jpg?height=290&width=464&top_left_y=1704&top_left_x=814)

Figure 2

Also, according to the problem, $\angle AMC = 90^\circ$. Using the theorems for right triangles and (2),

$$
\begin{aligned}
MM' & = \sqrt{AM' \cdot M'C} = a \sqrt{\frac{2}{3}} \\
AM & = \sqrt{AM' \cdot AC} = a \sqrt{2} \\
CM & = \sqrt{CM' \cdot CA} = a
\end{aligned}
$$

The volume $V$ of the pyramid can now be calculated:

$$
V = \frac{T_{ABCD} \cdot MM'}{3} = \frac{a^3 \sqrt{2}}{4}
$$

The surface area of the pyramid is the sum of the areas of the bounding faces, one of which is the base area, which we already know.

![](https://cdn.mathpix.com/cropped/2024_05_02_a0d6d3e839b26a9f3eb9g-2.jpg?height=542&width=746&top_left_y=127&top_left_x=668)

The triangles $MM'B$ and $MM'C$, as well as $MM'A$ and $MM'D$, are congruent (Figure 3), so $MB = MC$, $MA = MD$. The triangle $BCM$ has all three sides equal to $a$, so its area is

$$
T_{BCM} = \frac{a^2 \sqrt{3}}{4}
$$

The triangle $AMD$ is isosceles, and since $MA^2 + MD^2 = 4a^2 = AD^2$, by the converse of the Pythagorean theorem, it is also right-angled. Thus,

$$
T_{AMD} = \frac{a \sqrt{2} \cdot a \sqrt{2}}{2} = a^2
$$

Finally, the triangles $MAB$ and $MCD$ are congruent and, because $MA = MD = a \sqrt{2}$, $AB = CD = a$, and $MB = MC = a$, they are also isosceles right triangles, and their combined area is $a^2$. Therefore, the surface area of the pyramid is

$$
F = 3 \frac{a^2 \sqrt{3}}{4} + a^2 + \frac{a^2 \sqrt{3}}{4} + a^2 = a^2 \sqrt{3} + 2a^2 = a^2 (\sqrt{3} + 2)
$$

Based on the work of László Dingó (Budapest XI., Petőfi S. Secondary School, 8th grade).