Visual Representation of (a+b)(a-b)

Date: 03/22/2003 at 15:03:12
From: Amber
Subject: Is there a visual representation of (a+b)(a-b)?

Is there a way to model (a+b) * (a-b)?

I can use a rectangle array to model (a+b) times (a+b) but haven't 
been able to model a+b times a-b.

Date: 03/22/2003 at 15:34:18
From: Doctor Rick
Subject: Re: Is there a visual representation of (a+b)(a-b)?

Hi, Amber.

You can start with this figure:

  |           a             |
- +-----------------+-------+
  |/////////////////|       |
  |/////////////////|       |
  |/////////////////|       |
  |/////////////////|       |
  |/////////////////|       |
a |/////////////////|       |
  |/////////////////|       |
  |/////////////////|       |
  +-----------------+-------+
  |/////////////////|   b   |
  |/////////////////|b      |
  |/////////////////|       |
- +-----------------+-------+
  |/////////////////|
b |/////////////////|
  |/////////////////|
- +-----------------+
          a-b

The shaded area is (a+b)(a-b). Now compare it to this:

  |           a             |
- +-----------------+-------+
  |/////////////////|///////|
  |/////////////////|///////|
  |/////////////////|///////|
  |/////////////////|///////|
  |/////////////////|///////|
a |/////////////////|///////|
  |/////////////////|///////|
  |/////////////////|///////|
  +-----------------+-------+
  |/////////////////|   b   |
  |/////////////////|b      |
  |/////////////////|       |
- +-----------------+-------+

This area is a^2 - b^2: the area of the big square, with the area of 
the little (b by b) square removed. Compare the two: they are equal, 
because the rectangle at the right of the second figure is congruent 
to the rectangle at the bottom of the first figure (it's just rotated 
90 degrees). The remainder of the shaded region is the same in both 
figures.

If I could work with colors or more subtle shading, I could have 
combined the two figures into one, so the relation would be more 
obvious.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/