## Equidecomposition of a Triangle and a Rectangle

What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe applet purports to illustrate a simple fact:

A triangle can be cut into three pieces which, after translation and rotation, combine into a rectangle of equal area.

It is said that the triangle and rectangle are *equidecomposable*. In general two shapes are *equidecomposable*, provided one could be cut into pieces, which can be rearranged into the second shape. The decomposition of a triangle into a rectangle is pretty tame, but there are rather weird ones. Much depends on what kind of a cut is allowed. The one we are concerned with here may be called *decomposition by dissection*. It is characterized by the fact that, as sets, any two adjacent pieces both contain their common boundary. The decomposition in the Tarski-Banach paradox is different. The pieces there, adjacent or not, have an empty intersection.

Note that the above demonstration does not prove that any triangle and rectangle of equal areas are equidecomposable, although they in fact are.

This demonstration does teach us serendipitously that the area of a triangle is the product of the base times half its altitude to the base:

AreaOfTriangle = base × altitude/2.

(The approach that is more readily interpreted as

AreaOfTriangle = (base × altitude) / 2

is considered elsewhere.)

Finally note that any rectangle is equidecomposable with a square. Thus we can also claim that any triangle is equidecomposable with a square.

### Equidecomposition by Dissection

- Carpet With a Hole
- Equidecomposition of a Rectangle and a Square
- Equidecomposition of Two Parallelograms
- Equidecomposition of Two Rectangles
- Equidecomposition of a Triangle and a Rectangle
- Equidecomposition of a Triangle and a Rectangle II
- Perigal's Proof of the Pythagorean Theorem
- Two Symmetric Triangles Are Directly Equidecomposable
- Wallace-Bolyai-Gerwien Theorem

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Copyright © 1996-2018 Alexander Bogomolny