 Because multiplication and division are related, any multiplication method that we have will also double as a division method. So we might consider the area model we have for multiplication, and we can reverse it and produce a model for division using areas. So let's consider that again, remember what the area model for products and quotients looks like. If I have two sides for a rectangle, the area is the product, and because of the relationship between multiplication and division, what that means is if I have an area and one side, the other side is going to represent the quotients. And so this leads to an area model, which I can use in the following way. So let's consider the division 221 divided by 17, and I'll use a tape diagram to represent this area model. Tape diagram are really interchangeable terms. So let's take a look at that. So remember, area is the product is the dividend, so this dividend 221 is the area of my rectangle, and the divisor is going to be one side. So I have an area 221, one side equal to 17, and my question is how big is the other side? Okay, so how can I approach this? Well, I might start by taking some chunk of the rectangle whose area I can find easily. And the only requirement I have is that I want to make sure that this rectangle has an area of 221. So maybe I'll take a chunk of size 10, and so now I have this 10 by 17 sub-rectangle, and it has a nice area. That's easy to calculate, 170, and I know that that is smaller than what I need, so that's a good starting point. So let's take another chunk. How about, well, I can take another chunk of size 10, because that's going to give me a total area of 340. So how about width 2, for example? So I have another rectangle here that's 2 by 17, and I can find the area 34. And now I have area 170 plus 34, that's 204, which means that I have area 17 left over. And so I can take one more chunk, area 17, and there's my rectangle with area 221 with width 17 and other width 13. And so my quotient is going to be 10 plus 2 plus 1, and that's going to be 13. Well, let's look at a harder problem. So 1,537 divided by 29, and again what I have is the area, 1,537, one side of the rectangle, 29, and so I can draw that figure. And so I'm going to draw it looking something like this, and so here I've kept track of how big this rectangle is, how long one side is. Now if you go back to how we did the area model for a product, it's worth remembering that we could, and often did, break up the one side into smaller, more manageable pieces. So we might break up this 29 into two pieces, one of length 20 and the other of length 9. So, take a chunk of the rectangle. How big? I don't know. How about 20? Why 20? We'll try it, see if it works. So now I have two rectangles, 20 by 20, I could find the area of that, 20 by 9, I can find the area of that as well. And those areas 400, 180. So now we might try and determine how much area do we have left. So I can do the subtraction. Starting with this area, 1,537, if I remove the area 400, I have 1,137 left. Remove the area 180. Again, you could do the subtraction, or again that's, if we count past, that's 200 down return 20. So that's down to 937, back up to 957. So there's the area that I have left, and notice that this area left is still greater than the 400 plus 180, still greater than 580, so I can remove another chunk of size 20. And again, I'll figure out how much area I have left after those two chunks have been taken out. So again, 957 left, take out the 400, leaves 557 left. Take out the 180, get rid of 200, and return 20, that drops us down to 357, back 2377. And not enough. These are bigger than what's left over, so our remaining chunk has to be smaller. So how big can it be? Well, if I am taking with 29, I know I can take out at least 10. And again, doing the same computations, this 10 by 20 rectangle, that's 200, that's 90, and what's left over, 87. And how much can I take out now? Well, there's 87 left, so I definitely can't take a chunk of size 10 anymore. So I'll take a smaller bit, turns out that 3 works out to be the ideal size. This is 3 by 20, that's 60, 3 by 9 is 27, and at the end of it, there's nothing left. Actually, I wouldn't believe me. Let's go ahead and make sure that we've actually accounted for all of the area. So let's see, that's 400, 800, 1000. Let's see, what else can I group here? So how about these 200, so that's 1200, 8 in 8 is 160, so that's 1360, 1460 minus 10, 1450, plus 60, 1510, and 27, 1537. So all of our area, add up to 1537, which is all we have. So that says all of the area is accounted for, and the length is the quotient. So that's going to be 20 plus 20 plus 10 plus 3, and I can just add those together to get my quotient 53.