A truncated triangular pyramid has edges of one base measuring 148, 156, and $208 \mathrm{~mm}$. Each side face forms an angle of $112.62^{\circ}$ with this base. The height of the truncated pyramid is $27 \mathrm{~mm}$. What is the surface area and volume of the truncated pyramid?
The side faces form an obtuse angle with the given base, which is why the given base is the smaller base of the frustum, and thus the base of the complementary pyramid of the body. We can easily determine the height $h$ of this pyramid from the equality of the inclination angles. With this, we also know the ratio of the two bases. The area $t$ of the smaller base can be calculated using Heron's formula; for the larger area $T$, we have $T / t=(m+h)^{2} / h^{2}, T=\frac{(m+h)^{2}}{h^{2}} t^{2}$.

The volume:

$$
V=\frac{m}{3}(T+\sqrt{T t}+t)=\frac{m t}{3}\left[\left(\frac{m+h}{h}\right)^{2}+\frac{m+h}{h}+1\right]
$$

To calculate this, we need to determine the height $h$ of the complementary pyramid in addition to $t$.

The side faces of this pyramid are inclined to the base at the complementary angle of the given angle, $\delta=67.38^{\circ}$. This implies that the slant heights of the complementary pyramid are equal, as well as their projections on the base; indeed, we can measure the inclination angle of the side faces between these segments, and in the resulting three right triangles, the common leg is the height of the body, and the angles opposite to it are the equal dihedral angles, so the triangles are congruent.

![](https://cdn.mathpix.com/cropped/2024_05_02_355205a2cc006df2e57eg-1.jpg?height=640&width=960&top_left_y=726&top_left_x=580)

Accordingly, the projection $M^{\prime}$ of the apex $M$ of the complementary pyramid on the base is equidistant from the sides of the base triangle, and it lies within the triangle, thus coinciding with the center of the inscribed circle of the base triangle, and the common length of these distances is equal to the radius $\varrho$ of the inscribed circle. $\varrho$ can be calculated from the sides in a known way, from which we can then obtain the height $h$ of the complementary pyramid. Numerically:

$$
\begin{gathered}
\varrho=\frac{t}{s}=\frac{1}{s} \sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{256} \sqrt{256 \cdot 108 \cdot 100 \cdot 48}= \\
=\frac{11520}{256}=45 \mathrm{~mm} \\
h=\varrho \operatorname{tg} \delta \approx 45 \cdot 2.400=108.0 \mathrm{~mm}
\end{gathered}
$$

Thus,

$$
\begin{gathered}
(m+h) / h \approx 135 / 108=5 / 4, \text { and the volume } \\
V \approx \frac{27 \cdot 11520}{3}\left(\frac{25}{16}+\frac{5}{4}+1\right)=\frac{27 \cdot 11520 \cdot 61}{3 \cdot 16}= \\
=395280 \mathrm{~mm}^{3} \approx 395.3 \mathrm{~cm}^{3}
\end{gathered}
$$

(At most 4 significant digits of the value can be reliable, as the value of $h$ is only this precise.)

It is obvious that the height $m^{\prime}$ of the trapezoidal side faces is also equal, its value is $m^{\prime}=m / \sin \delta \approx 29.25 \mathrm{~mm}$; thus, we can calculate the total area of the 3 trapezoidal side faces, i.e., the lateral surface area $P$, in one step. Denoting half the perimeter of the larger base by $S$, from the similarity already used, we have $S: s=(m+h): m$, so

$$
\begin{gathered}
P=(s+S) m^{\prime}=\left(1+\frac{m+h}{h}\right) s m^{\prime} \approx \frac{9}{4} \cdot 256 \cdot 29.25= \\
=16848 \mathrm{~mm}^{2} \approx 168.5 \mathrm{~cm}^{2}
\end{gathered}
$$

Finally, the surface area of the frustum is

$$
F=t+T+P \approx\left(1+\frac{25}{16}\right) 11520+16850=46370 \mathrm{~mm}^{2}=463.7 \mathrm{~cm}^{2}
$$