Geometry Forum - Problem of the Week
Solutions - Pythagorean Theorem - March 25-29, 1996
Annie says:Maybe I am being too hard with my grading this week, but by my count, only two folks got this one right! I received 48 incorrect solutions, some of which got only the extra part right; this may partly be a case of my having an answer in mind and not really asking the right questions, so I'll try to do be more explicit next time.
A number of people said no, you don't have to use squares, and suggested that other shapes such as regular polygons would work. A couple of people showed that you could use hexagons, and proved it, but their answers weren't general enough - the key here is that you can use any shape you want, as long as the shapes on the three sides are similar ! Thomas and Brian both got that idea in; Thomas actually said similar shapes, and Brian said shapes with proportional lengths. So you could really say the proof says that, given similar shapes on all the edges, the area of the shape on one leg plus the area of the shape on the other leg equals the area of the shape on the hypotenuse. You can check this out in Sketchpad or something - it is pretty cool!
A lot of people said that squares are used because they are easiest, and that's certainly true - the area of a square looks just like the terms we are used to seeing in the Pythagorean Theorem, so this makes it really convenient.
If you come across a problem like this, and you determine that other shapes will work, try to figure out what those other shapes might look like, or what limitations there might be. While you might find a couple that work, see if you can find any that don't, and hence narrow things down a bit.
From: Thomas S. Kuo
Email: ssusd2@owens.ridgecrest.ca.us
School: Murray Junior High School, Ridgecrest, California
Grade: 7th
(1) We don't have to use squares to prove the Pythagorean Theorem.
(2) Hexagons or other shapes could be used too.
For a regular hexagon with side a:
*
* * * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * * * * * a * * c
* * * * * *
* * * * * *
* * * * * *
* * * * * * * * * * *
a b
The hexagon is formed by six congruent equilateral triangles. Each
triangle has side a and height (sqrt(3)/2) * a. Then the area of a
triangle is (1/2)*(a)*(sqrt(3)*a/2) = (a^2) * sqrt(3)/4. The area of the
hexagon is 6 * (a^2) * sqrt(3)/4 = (a^2) * 3 * sqrt(3) / 2.
Let a, b and c are three side of a right triangle as shown. When we
form three hexagons with side a, b, and c , their areas will be
(a^2) * 3 * sqrt(3) / 2, (b^2) * 3 * sqrt(3) / 2, and
(c^2) * 3 * sqrt(3) / 2, respectively. Now I have to check out if the
sum of areas of the first two is equal to that of the third one. After we
cancel out the common factor, we have only a^2, b^2 and c^2 left and
the relationship a^2 + b^2 = c^2 still holds.
The conclusion is that we can prove the Pythagorean Theorem by using
hexagons. It is also true if other shapes are used. The fact is that
as long as we use similar shapes with side a, b, and c, their areas will
have ratio of a^2, b^2, and c^2. After cancelling out the common factor,
it can be simplified to a^2 + b^2 = c^2.
(3) I believe that the reason to use squares in the proof is due to the fact
that it is the most direct and easiest method.
*****************************************************
From: Brian Gordon
Grade: 1992
School: Dartmouth
I have an answer for the bonus. It's James Garfield, right? I
believe the proof is like this:
|\
b | \ c
|__\
a
Paste two copies of this triangle together and connect the other
acute vertices to make a trapezoid like this:
a
____
| /*
b | /c *
|/ *
|`-c_ *
a |____\*
b
Now....
the area of the trapezoid= 1/2 * height * sum of bases
= 1/2 * (a+b) * (a+b)
The areas of the triangles are:
1. 1/2 * a * b
2. 1/2 * a * b
3. 1/2 * c * c because they meet at a right angle. This can be
shown by noting that the straight angle that is the left side
of the trapezoid is made up of two complementary angles (from
the two congruent right triangles) and the angle of the two
sides of length c (which is then 180 - 90 = 90).
Equate the trapezoid formula with the sum of the triangles:
Double everything to eliminate those pesty 1/2's:
(a+b)(a+b) = ab + ab + cc do the FOIL...
a^2 + 2ab + b^2 = 2ab + c^2 subtract the 2ab....
a^2 + b^2 = c^2
I like this one much more than the proof I originally learned,
which was based on the altitude-on-hypotenuse and similar
triangles.
I never actually learned the squares-on-the-sides proof.
As for other shapes, I think you could do it, if all the
constructions are done right. That's because the areas of the
shapes would be proportional to the square of the lengths
of the sides.
I have a formula for a regular n-gon with side s. It's
A=.25 * n * s^2 * cotangent(180/n). Could you guys patent that
for me? :)
--bri
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