Let $m$ be the length of the diagonal of a regular four-sided prism which closes an angle $\alpha$ with the side of the prism

Let $$m$$ be the length of the diagonal of a regular four-sided prism which closes an angle $$\alpha$$ with the side of the prism. Calculate the surface of the perimeter.

I guessed the angle is the diagonal of the side of the prism and the diagonal of the whole prism. I then just guessed it was a right angle and got everything I need with simple trigonometry formulas and Pythagoreans theory and I got the correct answer. But the thing is that I just guessed and I don't really know why the angle is right or why that is the angle in the first place. If somebody can please help explain how and why it is like that I would really appreciate it.

• What is a regular four-sided prism? Is it a prism where the top and bottom are squares and the sides are parallelograms?
– Jens
Oct 19, 2018 at 18:46
• @Jens the bases are squares and the edges are all equal. Oct 20, 2018 at 19:34

Let $$a$$ be the edge of the base and $$h$$ the height of the prism. The diagonal of the prism is also a diagonal of the rectangle having as sides the diagonal of the base (length $$\sqrt2a$$) and a lateral edge (length $$h$$). Hence: $$h=m\cos\alpha,\quad \sqrt2 a=m\sin\alpha,$$ and the surface of the pyramid is: $$S =2a^2+4ah=m^2\sin^2\alpha+2\sqrt2 m^2 \sin\alpha\cos\alpha.$$