Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Tangram and Pythagoras
Replies: 2   Last Post: Sep 27, 1994 2:03 PM

 Messages: [ Previous | Next ]
 Chih-Han sah Posts: 75 Registered: 12/3/04
Tangram and Pythagoras
Posted: Jun 2, 1994 4:21 PM

While testing out the geometer's sketchpad, I came across a
Tangram proof of the theorem of Pythagoras. I assume that it must be one
of umpteen thousand proofs. However, not being a geometer, I have
no idea if it is in fact wellknown.

Let me draw the picture:

B' C A'
_______ __________
| | C, C' are right angle vertices
| |
| |
A |
| |
| B
| |
D' |_____ ___________| D

C'

ACB is the given right triangle with C the right angle vertex.
AB'C and CA'B as well as AC'B are 45-45 right triangles by
construction. AD'C' and C'DB are right triangles which can be
checked to be congruent and similar to ACB with ratio 1:sqr(2)

B'A'DD' is therefore a square with sides of lengths:

(a + b)/sqr(2)

I need the special case of the Pythagoras theorem that the hypoteneuse
of a 45-45 right triangle is sqr(2) times the length of one leg. This
can be given a Tangram proof. Namely two of these form a square with
the hypoteneuse glued to each other. However, two of these may be
glued along a common leg to form a 45-45 right triangle so that 4 of
them form a square whose side is equal to the original hypoteneuse.
An area calculation for squares gives us the desired special case.

For right triangles, the area is half the product of the two perpendicular
legs. Now, the displayed square has area:

(a + b)^2 * (1/2)

It is the sum of 6 right triangles. Three are 45-45 right triangles
with hypoteneuses a, b, and c by contruction. Of the remaining 3,
one is the original one with area ab/2 while the other two are
congruent to each other and simiar to the given one with ratio
sqr(2):1, thus have areas equal to ab/4. Adding up, we get a total
area of:

(a^2 + b^2 + c^2)/2 + ab/2 + ab/4 + ab/4

Equating the two different calculations, multiplying by 2 on both
sides and simplify, we get the theorem of Pythagoras:

a^2 + b^2 = c^2

Note: Coxeter (Intro. to Geometry) gives a proof
that involves dropping the perpendicular from C to AB and uses
proportionality principle for similar triangles and their areas.
It is compact and neat, but the idea of a square is hidden.
The Tangram proof does not involve squares, but it involves a
number of half squares (from a half-wit (!)).

Han

Date Subject Author
6/2/94 Chih-Han sah
6/2/94 John Conway
9/27/94 Dan Bennett