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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.

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

1 Answer 1

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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: $$ S =2a^2+4ah=m^2\sin^2\alpha+2\sqrt2 m^2 \sin\alpha\cos\alpha. $$

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