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Let there be a geometric shape $\Omega$ of area $S$ lying in a plane $B$. Let the horizontal plane (the plane $xy$) be $A$. Let the angle between the planes $A$ and $B$ be $\theta$.

It could be easily proved that $AB \cos \theta$ is the length of the orthographic projection of any segment $AB \in \Omega$.

How can it be proved that $S' = S \cos \theta$ is the area of the orthographic projection of the shape $\Omega$ onto the plane $A$?

marmistrz
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    What "can be easily proved" is wrong: The scaling factor for segments depends on their direction. Segments $\sigma\subset B$ orthogonal to $g:=A\wedge B$ are scaled by a factor $\cos\theta$, and segments parallel to $g$ keep their length under projection. Now find out how the area of suitable rectangles is scaled. It is a nontrivial theorem that all areas are scaled by the same factor. – Christian Blatter Apr 05 '15 at 15:49
  • Yes, indeed it's wrong. But using the fact mentioned by you, we could divide the shape into infinitesimal, rectangular shapes of area $dS$ and one segment perpendicular to $g$, one parallel. The area of the projection is thus $dS' = dS \cos\theta$. If we sum all $dS'$ we'll get $S' = S \cos\theta$ – marmistrz Apr 06 '15 at 10:35
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    Now you have it right! – Christian Blatter Apr 06 '15 at 14:26

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