This Question is in The History of Mathematics by John Stillwell : Show that the area of two triangles with the same base and height can be approximated arbitrarily closely by the same set of rectangles, differently stacked.

I came up with my own proof, roughly saying that any set of rectangles contained and approximating triangle 1 must also be contained and approximate triangle 2, situated differently, since the triangles have the same area (But different angles). If this were not the case, I reached a contradiction on the fact that the triangles had the same area.

However, this argument (potentially incorrect) does not appeal to Cavlieri's Principle, which I read can be used to solve this problem:

Cavalieri's Principle: Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

Here I am stuck. The triangles are surely contained between two parallel lines and have the same area. But, how can I use this to prove that they can be approximated by the same rectangles, differently stacked? Thank you in advance.


You don't need Cavalieri's principle. Dispose the triangles with their bases on the same line (see diagram below) and draw $n$ lines parallel to the bases so that they divide the altitudes into $n$ equal parts. Those lines intersect the triangles at $n$ couples of segments and it is easy to prove (by similar triangles) that segments on the same line are equal. On those segments, taken as bases, you can then construct $n$ couples of rectangles, all having the same height $h/n$. They are the approximating rectangles you are looking for.

enter image description here

  • $\begingroup$ Wow, thank you, this is a very elegant solution. I appreciate this. $\endgroup$ – wesssg Feb 25 '17 at 20:39

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