For the last week, our entire agenda was on areas and how we could find areas for different polygons on a Geo Board. We focused on the formulas of a triangle (b*h*1/2), and a rectangle (l*w). We also learned how to find all 3 heights for different kinds of triangles (obtuse, right, equilateral etc.) Now onto the problem, the problem starts with a person named Freddie Short who has a formula to find the area of any polygon that has no pegs on its interior. We then are asked to find his formula through finding patterns when we use an in/out table. After we find a pattern or an actual formula, we are asked to find a different formula that works for a polygon with exactly one peg in its center (I found a trapezoid). After we found the pattern for that constraint, we had to pick a number larger than 1 and use the number as the interior pegs. Once that was completed, it was easy to find a correlation between the formulas because they both had to follow the same pattern. We then had to come up with some other formulas with the same constraints in order to further back up our formula claim. For another person named Sally Shorter, she said that she has a formula for any polygon with exactly four pegs on the boundary. All we have to do is tell her the amount of pegs on the interior and she can tell us the answer. We also had to go through the same process as we did with Freddie Short (find the formula, pick a number other than the one on the first question, and further prove our claim.) For the final question, we had to find a super formula that Frashy Shortest came up with through the use of the in/out tables, that would find any polygon's area on the Geo Board. Now onto my process and formulas, I started with Freddy. I had to write down the number of pegs and the area of the polygon until I found some patterns Pegs Area 6
2 8 3 10 4 12 5 14 6 As you can tell as the pegs goes up twice, the area goes up by one if you kept on going. 1a. A= P/2-1
I then began the next problem where it stated that I had to find polygons with exactly one peg on the inside.
One PegArea 8 4 6 3 4 2If you can tell all the polygons with one peg on the interior, have an area 1/2 of the pegs on the boundary. 1b. A=P/2
Now that we have both those formulas, we had to begin experimenting with other values to see if our formulas worked.
3 Pegs interior Area 12 8 8 61c. A=P/2+2
4 Pegs Interior Area 14 10 12 9
1d. A= P/2+3
As you can tell, there is a pattern between the formulas because of correlation of the original formula. Now onto Sally Shorter I had to begin by writing out an original in/out table, which was peculiar because I could only find different shapes that had the same lengths due to the constraints.
Pegs Area 4,0 on inside 1This was the only length I could find due to the constraints and that had me end up with the formula: 2a. A=P+1
For the other length that I used (6 pegs on boundary), I created yet another in/out table.
Pegs Area 6, 0 on interior 2 6, 3 on interior 5I was able to create the formula of 2b. A=P+2 based on the table above and other tests I conducted which also related to the previous formula.
Now onto the next formula that I had to find in order to further prove my previous formulas. I used 7 pegs on the boundary this time.
Pegs Area 7, 0 on interior 2.57, 3 on interior 5.5Based on this, and my previous knowledge of the other formulas, I easily came to the conclusion of 2c.A=P+2.5
As you can see the correlation between all of the formulas have a 0.5 increase every time you add a peg. That's why when you have 4 pegs you have to add 1 to the peg for the area, and when you have 6 pegs you must add 2 pegs for the area etc.
Now for the super formula that is needed for any polygon took a lot of experimenting and use of the in/out table
Pegs Area 3 in 7 out 5.5 0 in 7 out 2.50 in 6 out 2 0 in 4 out 11 in 9 out 4.5 2 in 11 out 6.5
I noticed tat these patterns had a lot to do with 0.5 or halves so I kept that in the back of my head when trying to figure out the formula. I tried to half the interior and boundary and I tried multiple ways of adding until I added the Boundary*1/2 + the Interior. I was really close to getting the answer except I was one above each time. I tried to subtract 1 from each table set and the subtraction worked. I ended up with a final formula of
3. A=B*1/2+I-1
After doing this problem which used a lot of mental drainage, I found out that I used a lot of habits of a mathematician. I definitely used look for patterns because the problem required me to pick out different patterns that were in my in/out table in order to find a formula more efficiently. I also used staying organized because creating multiple tables and formulas could have got very confusing for me but I made sure I clearly mapped out where I put my tables down in order for me to spot them quicker. I also made sure to physically write down any patterns I saw so I could easily remember them when I did this write up.
2 8 3 10 4 12 5 14 6 As you can tell as the pegs goes up twice, the area goes up by one if you kept on going. 1a. A= P/2-1
I then began the next problem where it stated that I had to find polygons with exactly one peg on the inside.
One PegArea 8 4 6 3 4 2If you can tell all the polygons with one peg on the interior, have an area 1/2 of the pegs on the boundary. 1b. A=P/2
Now that we have both those formulas, we had to begin experimenting with other values to see if our formulas worked.
3 Pegs interior Area 12 8 8 61c. A=P/2+2
4 Pegs Interior Area 14 10 12 9
1d. A= P/2+3
As you can tell, there is a pattern between the formulas because of correlation of the original formula. Now onto Sally Shorter I had to begin by writing out an original in/out table, which was peculiar because I could only find different shapes that had the same lengths due to the constraints.
Pegs Area 4,0 on inside 1This was the only length I could find due to the constraints and that had me end up with the formula: 2a. A=P+1
For the other length that I used (6 pegs on boundary), I created yet another in/out table.
Pegs Area 6, 0 on interior 2 6, 3 on interior 5I was able to create the formula of 2b. A=P+2 based on the table above and other tests I conducted which also related to the previous formula.
Now onto the next formula that I had to find in order to further prove my previous formulas. I used 7 pegs on the boundary this time.
Pegs Area 7, 0 on interior 2.57, 3 on interior 5.5Based on this, and my previous knowledge of the other formulas, I easily came to the conclusion of 2c.A=P+2.5
As you can see the correlation between all of the formulas have a 0.5 increase every time you add a peg. That's why when you have 4 pegs you have to add 1 to the peg for the area, and when you have 6 pegs you must add 2 pegs for the area etc.
Now for the super formula that is needed for any polygon took a lot of experimenting and use of the in/out table
Pegs Area 3 in 7 out 5.5 0 in 7 out 2.50 in 6 out 2 0 in 4 out 11 in 9 out 4.5 2 in 11 out 6.5
I noticed tat these patterns had a lot to do with 0.5 or halves so I kept that in the back of my head when trying to figure out the formula. I tried to half the interior and boundary and I tried multiple ways of adding until I added the Boundary*1/2 + the Interior. I was really close to getting the answer except I was one above each time. I tried to subtract 1 from each table set and the subtraction worked. I ended up with a final formula of
3. A=B*1/2+I-1
After doing this problem which used a lot of mental drainage, I found out that I used a lot of habits of a mathematician. I definitely used look for patterns because the problem required me to pick out different patterns that were in my in/out table in order to find a formula more efficiently. I also used staying organized because creating multiple tables and formulas could have got very confusing for me but I made sure I clearly mapped out where I put my tables down in order for me to spot them quicker. I also made sure to physically write down any patterns I saw so I could easily remember them when I did this write up.