- Level: International Baccalaureate
- Subject: Maths
- Word count: 1176
Mathematics SL Parellels and Parallelograms. This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.
Extracts from this document...
Introduction
Jia-Der Ju Wang
Parallels and Parallelograms
Mathematics Coursework
This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.
Methodology
- We started out the investigation with a pair of horizontal parallel lines and a pair of parallel transversals. One parallelogram (A1) is formed (shown in Figure 1)
- A third parallel transversal is added to the diagram as shown in Figure 2. Three parallelograms are formed: A1, A2, and A1∪A2
- When a fourth transversal is added to Figure 2 (Figure 3), six parallelograms are formed. A1, A2, A3, A1∪A2, A2∪A3, A1∪A3
- Figure 4 has 5 transversals cutting the pair of horizontal parallels, forming ten parallelograms. A1, A2, A3, A4, A1∪A2, A2∪A3, A3∪A4, A1∪A3, A2∪A4, A1∪A4
- A sixth transversal was added to Figure 5, forming 15 parallelograms shown in Figure 6. A1, A2, A3, A4, A5, A1∪A2, A2∪A3, A3∪A4, A4∪A5, A1∪A3, A2∪A4, A3∪A5, A1∪A4, A2∪A5, A1∪A5
- When a seventh transversal is added, twenty-one parallelograms are formed (Figure 7). A1, A2, A3
Middle
Solve the two new equations as simultaneous equations
Answer:
Replace the values in the following formula
But when I tried out the formula I found out that the answers didn’t match:
U2 = 3 (3 +1) = 6
2
U3 = 4 (4 +1) = 10
2
The results we had have moved one term so instead of adding 1 to “n”, we need to subtract 1 to “n” so the values can match. Now our final formula is:
Un = n (n -1)
2
Once again I tested the above:
U2 = 2 (2 -1) = 1
2
U3 = 3 (3 -1) = 3
2
- In order to find the general formula for the parallelograms formed by m horizontal parallel lines intersected by n parallel transversals; I decided to further the investigation by considering the number of parallelograms formed by three horizontal parallel lines intersected by a pair of parallel transversals (Figure 8). Three parallelograms were formed: A1, A2, and A1∪A2
- When a third transversal was added to the above figure, nine parallelograms were formed (Figure 9). A1, A2, A3, A4, A1∪A2, A3∪A4, A1∪A3, A2∪A4, A1∪A4
- One more transversal was added to Figure 9, to form 18 parallelograms (figure 10). A1, A2, A3, A4, A5, A6, A1∪A2, A3∪A4
Conclusion
- When I compared all the formulas I got, I found the relationship which led me to find the general formula.
2 horizontal lines:1n (n -1)
2
3 horizontal lines: 3(n (n -1))
2
4 horizontal lines:6(n (n -1))
2
5 horizontal lines:10(n (n -1))
2
The number and sequence repeat the formula (n (n -1)) and multiplies by the first term. 2
- From this realization I was able to find that the final formula for calculating the number of parallelograms formed when m horizontal parallel lines are intersected by n parallel transversals:
m(m -1) x n(n -1)
2 2
- To test the validity of the formula I tested it against previously counted parallelograms (Figure 10), the intersection of 4 transversals with 3 horizontal parallel lines should form 18 parallelograms:
Using the formula:
m(m -1) x n(n -1)
2 2
3(3 -1) x 4(4 -1)
2 2
3(2) x 4(3)
2 2
6 x 12
2 2
3 x 6
= 18 parallelograms
Limitations
It is very difficult to test the validity of the formula when there are lots of parallel transversals intersecting lots of horizontal parallels, because even though we will get a number, to prove there are so many parallelograms is confusing and difficult; therefore we can only assume that the answer might be right.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
Found what you're looking for?
- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month