Last week, in Fundamental Understanding of Mathematics LXV we took another look at inductive proofs, showing that the sum of all the odd numbers (starting at one) is N squared.
This week, I'm going to take a look at some geometry.
A while ago, we talked about the square unit: a new kind of measurement that had both length and width, which was so different from length that it needed a new name: area.
We defined it, if I recall correctly, as the space swept out by a line one unit long, moving one unit to one side or the other.
We can connect this geometry to arithmetic my noticing that, for rectangles, it is directly related to multiplication.
If we take length three, and sweep it to the right for width 4, we cross an area of 12 square units. I've shown the area cut into unit squares, and we can also see that the area is three unit squares (length 3) added to three more, then three more, then three more, as we move to the right. Repeated addition is multiplication. 3 + 3 + 3 + 3 = 3 x 4. By definition.
So our definition for area, (notice that it includes the idea of sweeping at right angles, or perpendicular, to the original line) matches our definition of multiplication, so we can find the area of a rectangle by multiplying the lengths of two sides sharing the same corner.
Thus we have a formula for finding the area of a square or a rectangle,
Area = Length x Width.
We can extend this formula to other shapes by noticing that if we cut the rectangle in half along a diagonal, it forms two triangles
So instead of one rectangle with area 12 square units, we have two equal triangles, each with area twelve divided by two, or 6 square units.
So the area of a triangle is half the base times the height. I've changed the term for width, since, with that slanted side, the "width" of a triangle becomes a somewhat slippery notion. We also need to distinguish the area of a triangle from the area of a rectangle when we write the formula.
But this is a special kind of triangle. It is a right triangle, that is to say, the corner where the base and height connect is a right triangle. What about other triangles, triangles where there are no right angles?
What about something like this?
Here is a triangle with a base of length B, and a height of length H.
We can show the formula works for this kind of triangle by dividing it into two right triangles
Same triangle, only now it has a "construction line" dividing it into two right triangles.
The height of both triangles is the same, but they have new bases, b1 and b2. The only thing we know about b1 and b2 is they add up to the original base length B.
So the area of this triangle is
We can use the distributive property to change this to
Since we know b1 and b2 add up to the original base length, we replace b1 + b2 with base, and we are back to the original formula.
Finally, there is one other case to consider. The triangle we just saw divided neatly into two right triangles. What if we have a triangle where that isn't possible? Something like this?
We can still make two right triangles out of this.
Only in this case, the large blue and white right triangle is larger than our original triangle, so we will have to subtract the area of the smaller white right triangle. I'll leave working this out to someone in the comments, but it, too, winds up being our original formula.
Have fun in the comments