I'm dealing with the following frustum:
Which is similar to a truncated pyramid but the lateral faces may have different inclinations (i.e. $\alpha \ne \beta$). How can I evaluate its volume and lateral surface area? Thank you in advance!
P.S.: Sorry I couldn't provide a 3D sketch, but I hope it's clear from the views above.
For the lateral area, use the following property:
In 3D space, if a planar polygon (it can be a more general shape in fact, but we don't need it) belonging to a plane $(P)$ with area $A$ is orth. projected onto a plane $(P')$, making an angle $\alpha$ with initial plane $(P)$, the resulting polygon has area
$$A'=A \cos \alpha \tag{1}$$
The lateral area is obtained by using (1) in a reverse way (= "unprojecting"):
$$\text{lateral area} = \text{area}(R)+2 \ \text{area}(T_1)/\sin(\alpha)+2 \ \text{area}(T_2)/\sin(\beta)$$
where $R$ is the area of the rectangle and $T_1$ and $T_2$ the two projected trapezoids (the area of a isosceles trapezoid being $\frac12(b_1+b_2)\times h$: average base length times height).