 One of the main things that we study in geometry are areas. So this begins with the idea of a unit square, and in general what we're going to do to find the area of a figure is we're going to compare how the area of that figure compares to the unit square, which is a square that is one unit along each side. So for example here we might say that this is our unit square, and how many of these will fit into our figure. So if I take a unit square and I join another one, I have another figure, and this is if you can count, you can do this one, two, this has an area of two unit squares, and if I include another one, I now have one, two, three unit squares all together. That's all that area really is, is counting up unit squares. So we can count. So now we have this figure over here, and we want to find its area, and if you can count, you can do geometry. One, two, three, four, five, six, and so since there's six squares, then it's going to have an area of six square units. Now there's only two difficulties we're going to run into. The first is we might not want to always have to count each and every single unit squares, and so it'd be nice if we could find some sort of area formula that we could use to give us the area of the figure. Before we get there, let's go ahead and think about how we might approach the problem without an area formula. Here I have a rectangle that consists of a number of rows of unit squares, and if I wanted to find the area of this, I could count one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. The area of this is twelve unit squares, twelve square units. On the other hand, I might try to approach this problem a little bit more efficiently. So now let's take a closer look at this rectangle, and what I see is that it consists of one, two, three rows, each of which contain one, two, three, four of these unit squares. So my rectangle consists of three rows, each containing four squares, so if I want to find the area, I can just add four, plus four, plus four, and, well, that's a repeated addition. So repeated addition is a multiplication. This is three, fours added together, and I can do that computation one way or another. That's going to be my area of twelve once again. However, the important thing here is that it suggests that I can find the area of any rectangle whatsoever as the product of its length, the number of unit squares that forms a row, and the width, the number of unit squares that forms a column. So in this case, I have three rows consisting of four unit squares apiece, or if I want to do the multiplication in the other order, I could also say that there's four columns, each consisting of three unit squares apiece. And in either case, the total number of unit squares is going to be the product four times three. Now, remember that area is a conserved quantity and that a rigid transformation does not alter the area of a figure. I can move things, I can translate things, I can rotate things, I can even do reflections, and the figure still has the same area. So here's a good starting point. Assume that there's some rigid transformation, some transformation that turned this blue triangle into the green triangle. And so let's identify what that rigid transformation is, and suppose I know something about the blue triangle. What can I tell you about the green triangle? Well, let's see, I have this blue triangle here, and it looks like I've just moved it over some distance. And the type of transformation that is a move over some distance, that is a translation. So it is a horizontal translation of some length. I don't have the units printed here, so it's hard to tell exactly how far it is, so we'll leave that unspecified. Also, area is preserved under any of these transformations, so the area of the green triangle is exactly the same as the area of the blue triangle. Both triangles have an area of three square units. Now, the idea that transformations preserve areas is going to be essential for finding the area of any figure other than a rectangle. So, for example, let's consider a parallelogram. What about the area of a parallelogram? And so there's a number of ways I can do that. What if I kind of slice off the parallelogram, and then take this slice, which is a triangle, and I'm going to slide it over until it gets over here. And this slice will fit right over here on this side. Now, area hasn't changed. Again, the area of the piece after is the same as the area that I started with, which was the same as the area of the parallelogram. So what I've done here, this translation of the piece, has not changed the area of the parallelogram. But the new figure that I have is a rectangle, and I now know how to find the area of a rectangle. It's the product of the width and the height, which tells me that if I want to find the area of the parallelogram, it's also going to be the product of the width and the height. So the area of a parallelogram is the same as the area of a rectangle with the same width and height. Well, how about triangles? Again, area is conserved. Area does not change when we apply a rigid transformation. And this time let's do a couple of rotations. Now, there's a simple rotation that we could do, but for a variety of reasons, we're going to take a somewhat more circuitous route. I'm going to apply a couple of rotations. And this time I'll rotate about this point here, this vertex of the triangle, partly because that's a more natural way of doing the rotation. So I'll start a rotation a little bit, a little bit more, a little bit more, rotate it until it's in that position. Again, rotation preserves area. So the area of the original triangle is the same as the area of the green triangle. And now I'm going to let that triangle sort of slide into position. And again, the area hasn't changed. The area of the green triangle is exactly the same as the area of the original triangle. And so the area of the parallelogram is two of the triangles. And the triangle is half the area of the parallelogram. But I know how to find the area of the parallelogram. It's equal to the area of a rectangle at the same height and width. So if I can find the area of the parallelogram, I can find the area of the green triangle. I can find the area of the original triangle. And that tells me that in general, the area of the triangle is half that of the parallelogram, which means that the area of the triangle is half that of the rectangle with the same height and width. And this is a process we can do in general. In general, any figure whose area we can find, we can find it in one of two ways. If it's not something we can compute directly like the area of a rectangle, we can generally find the area by using scissors and either removing pieces of computable area. So for example, if I want to find the area of this triangle, I could start with a rectangle. I could start with a larger rectangle, looking something like this. So there's my width and my height, and then I can remove pieces that I don't want, which would be these two pieces out here. Or I might be able to use scotch tape, so I might join pieces to form the figure that I want. And again, as long as I can compute the area of the pieces, I can work with it. So for example, take a figure that looks something like that. If I wanted to tape pieces together to form this figure, I might start with two triangles and a rectangle. And again, as long as I could find the areas of the rectangle and the two triangles, the three pieces together would have an area equal to that of the trapezoid. And in general, either approach will work. Either approach could be used for the same figure. So let's take a look at that trapezoid again. And if I want to put this together from scotch tape, one thing I could actually do is I could take two triangles and find the area of the two triangles, and the area of the whole figure is going to be the sum. The other possibility is I can form this figure by cutting it down from a larger figure. So here's another way I can form that trapezoid. I can start with a big rectangle and get rid of two pieces that I don't want.