POW#12-
Problem Statement: In the picture above that's say that all of those dots are evenly spaced, you want to find out what is the area of the polygon? The area can be found by calculating the boundary points and the interior points using a formula. Your task is to find out what that formula is.
Process: The first thing we did was create at least 20 examples of different lattice polygons. Which at the start I did get confused because I was not there the day Jocelyn explained it so I had no idea what to do. Then in Erin's x-block Sarah explain to me what we’re suppose to do and gave me an example to help me understand a little more. Then she also re-explained what A,B,I. Also showed me the formula of how to find the area of just a triangle and a square.We all (meaning my whole class) made at least 20 examples, so that we could all share are different sets of data with each other and also so we would not have to create as many examples. Then we put all our data on a spreadsheet. So that we could look at all the data together and take off which numbers had repeated a lot. Then using the data we had on the sheet we created graphs and found the line of best fit. So far with the data we have found, we have found the line of best fit and in using the line of best fit we have found a formula that will work if all of your interiors points were zero. The next thing we did was for homework we all had to make a new formula for different number of interior points. I did a formal for one interior points. To find the formula I just repeated the same steps as what we did for zero just instead with the data that had one interior point. The next day we shared are formulas and with the person sitting next to you you checked the formula that Jocelyn told you to do. and see if you could find a counterfeit. If you found a counterfeit then we would take that formula off the list cause we knew it would not work. After about 10-15 minutes of do this we found there were only three equations that we had not found a counterfeit for so then again Jocelyn multiple groups the same equation to see again if we could find a counterfeit. After about another 10 minutes of trying to find a counterfeit and we could not find any. We looked at the equations we had to see if we could find an equation that would fit and that was A=½ B+(I-1)
Solution: The formal that we found was A=½ B+(I-1). I do think that this equation will work with any polygon because we did try mean other equations and got many different opinions and this equation is the only one that has worked. If you have 5 boundary points and 0 interior points and you plug that into the equation it we look like A=½ 5+(0-1) our area is 1.5
Evaluation:I thought this problem was educationally worth wild because it show me how I could make a equation with a lot of data. It was pretty difficult because I didn't know where to start so it really did help that we did it as a class.
POW#2-
Problem Statement: In this problem you have five bales of hay. But for some reason you weight them in pairs instead of individually. You weight them in the combinations of 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, and son on. When you wrote down the combinations you put the in numerical order. Without keeping track of which weight matched which pair of bales. The weights were 80,82,83,84,85,86,87,88,90,91. So now what you have to do is find out which weight went with which bales.
Process: In this problem the first thing that I did was write all the numbers out and at first I thought that they a lined already so I thought 1 and 2 went with 80 and so on. But what I found out later was that wasn't necessary true. The first thing that I tried to for the numbers was just divide them by two, but that would work for five of the bales pairs but not the other five. Then I tried subtracting one from one and add three to the other but that also did not work. Then when I finally got it the number for one was 39 than I add 2 an got 41 for 2 did the same thing for 3 so I got 43. Then for 4 I couldn't add two it would be one to many so I just added one an got 44 and for 5 I add three so I got 47. Before I got these numbers I read over the problem because I was confuse know why it was not working for the ones but not the twos and threes. When I went back to read the problem I notice that the numbers did not have to be in that order an that I when I started to really understand the question.
Solution: 1 = 39, 2 = 41, 3 = 43, 4 = 44, 5 = 47
1 and 2 = 80
1 and 3 = 82
1 and 4 = 83
2 and 3 = 84
2 and 4 = 85
1 and 5 = 8
3 and 4 = 87
2 and 5 = 88
3 and 5 = 90
4 and 5 = 91
Evaluation: In this problem I got confuse and had to go back and the problem which made it frustrating. Bu tin the end I was pretty confident with my answer and work. This was a good problem that made you think.