Remember the Math Adventurer’s Rule: Figure it out for yourself! Whenever I give a problem in an Alexandria Jones story, I will try to post the answer soon afterward. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. So if you haven’t worked these problems yet, go back to the original post. Figure them out for yourself — and then check the answers just to prove that you got them right.

Euclid’s Geometric Algebra

## Puzzle #1

$ab + ac = a \left(b + c\right)$

This is like the first example given in Euclid’s Geometric Algebra, showing the Distributive Property.

• Draw a rectangle with one side a.
• Make the other side (b + c).
• The area of the large rectangle equals the sum of the two smaller rectangles.

## Puzzle #2

$\left(a + b\right) \left(c + d\right) = ac + ad + bc + bd$

This is a double Distributive Property diagram.

• Draw a rectangle with sides (a + b) and (c + d).
• Split it into four smaller rectangles.
• The small rectangles have areas ac, ad, etc.

## Puzzle #3

${\left(a + b\right)}^2 = a^2 + 2ab + b^2$

• Draw a square with sides (a + b).
• Then cut it into pieces.
• You get a square with sides of a and a square with sides of b, along with two matching rectangles ab.

## Puzzle #4

$\left(a + b\right) \left(a - b\right) = a^2 - b^2$

This is the hardest diagram to figure out, so if you got this one, you understand geometric algebra very well.

• Draw a square with sides of a (the bold lines on this diagram below).
• Cut off a slice of width b (area ab).
• Lay this slice below the square. This new position is shown as dashed lines on the diagram.
• Cut off the section marked $b^2$.
• Throw this square away, leaving only the shaded rectangle with length (a + b) and width (ab).
• We started with $a^2$, took away $b^2$, and ended up with the rectangle we were looking for: $\left(a + b\right) \left(a - b\right)$.

## To Be Continued…

Read all the posts from the May/June 1999 issue of my Mathematical Adventures of Alexandria Jones newsletter.