Equivalent triangles (Cavalieri 2D)

Problem

2D version of Cavalieri: show that two triangles having the same base and the same height have the same area.

Solution

We arrange the two triangles with their base on the same line $r$ and the opposite vertex on a line $s$ parallel to $r$ at distance $h$ (the height). A line parallel to $r$ at distance $t$ from the base ($0\le t\le h$) cuts each triangle in a segment whose length, by similarity, is $\ell(t) = b\left(1-\frac{t}{h}\right),$ where $b$ is the base. Since both triangles have the same base $b$ and the same height $h$, at every height $t$ the two segments have the same length. By Cavalieri’s principle (2D version) the two figures have the same area: $\boxed{A_1 = A_2 = \tfrac{1}{2}b h.}$

Links

Topics: Integral
Concepts: Cavalieri’s principle
Methods: Cavalieri’s principle
Skills: Prove · Synthetic geometry
Exercise types: Proof