2 x 6
But can I use this rule to estimate the difference of a 2 x 2 square? I predict that the trend will continue, as a square is just a special form of rectangle, and the difference shall be 10.
2 x 2
From my results so far I can draw a table:
What Do I Notice?
The area increases by 2 each time. This is because the length is always being multiplied by the height of 2.
The difference increases in increments of 10. Also, the highest less the lowest corners when the longest side is on top increases by a factor of 1 while the Highest less the lowest corners when the shortest side is on top increases by a factor of 10.
Possible Formulas
L=Length H=Height D=Difference
L-1 x 10
Or
The length subtracted by one multiplied by ten. But from past experience I doubt this rule would work with other sized rectangles without 2 as a side length
L-1 x 5H
Or
The length subtracted by one multiplied by five multiplied by the height.
The lowest corner subtracted from the highest corner subtracted by one, but only when the rectangle is aligned so that the shortest sides are at the top and base.
Do These Rules Work With Other Rectangles?
3 x X
I will now look at rectangles where one of the sides has a length of 3 squares and see if the same patterns apply and investigate whether the above rules are true for rectangles of any shape.
I believe I shall require only one example of each and will proceed on this principle.
3 x 3
3 x 4
3 x 5
What Do I Notice?
The “L-1 x 5H” does not apply. But L-1 x 20 does.
I feel confidant that L-1 is correct but I still need to find how H can be used to calculate what to multiply L-1 by.
For 2 x X L-1 x 10
For 3 x X L-1 x 20
Could the universal rule be: D = L-1 x 10 x (H-1)?
Subtract one from the length, multiply by 10 and then multiply by the height subtracted by one?
Testing the Rule
4 x X
I will now look at 4 x X and see if my new found rule truly encompasses all rectangles.
4 x 4
4 x 5
The difference is advancing by 30. 4 x 6 will have a difference of 150.
4 x 6
(table calculated with a calculator)
Could the x 10 in my rule be related to width of my number grid, which is ten squares wide?
To test this idea I will try my rule on a different sized grid.
The Answer is forty, my rule works for any rectangle, on any grid.
L-1 x (Grid Width) x (H-1)
Or
Subtract one from the length, multiply by the width of the grid and then multiply by the height subtracted by one.
Using Algebra to Look Closer:
Calculating the rectangle’s values in relation to “X”
If the number in the top left hand corner is X, then the following squares will be....
(G = Grid width)
For a 3 x 2 rectangle when the first square is X the difference is:
The algebraic expression for the difference for this size of rectangle is: 2G, the width of the grid multiplied by two. This is correct as from my previous working on a rectangle of these proportions I know that on a 10 x 10 grid the difference is 20 (10 x 2 !)
For a 4 x 2 rectangle when the first square is X the difference is:
The algebraic expression for the difference for this size of rectangle is 3G, the width of the grid multiplied by three. This is correct as from my previous working I know that on a 10 x 10 grid the difference is 30 (10 x 3 !)
How can I produce an equation from the side lengths of this rectangle to give me 3?
Possible equations:
Or
I will look at the ?G rule in other rectangles to see if any of my above rules also apply and are true for rectangles across the board.
The algebraic expression for the difference for this size of rectangle is 8G, the width of the grid multiplied by six. This is correct as from my previous working I know that on a 10 x 10 grid the difference is 80 (10 x 8 !)
None of my rules apply, but I can now build a table with my results:
This must be because the L has increased by 1. This tells me that the L must be connected to the difference. But How?
How can I get 2 from 2 & 3, and 3 from 2 & 4 using the same formula?
If I subtract 1 from each of 2 & 3 I get: H=1 L=2
1 x 2 = 2.
When I subtract one from each 2 & 4 I get: H=1 L=3
1 x 3 = 3
The answers to both of these short sums give me the figure to multiply G with to get the difference.
I predict that this will work for 3 & 4:
3 - 1 = 2
4 - 1 = 3 2 x 3 = 6 So H-1 x L-1 x G = Difference
This methods work and it is also another way of writing my previous universal rule. “H-1 x L-1 x G” is the same as “L-1 x (Grid Width) x (H-1)”.
Conclusion
I have found the rule for any size of rectangle. I think that I can write down the expression to show this, that does not require the actual figures for the length:
(X + L - 1) x (X + (G(H - 1))) = XxX + X (G + X(H - 1)) + L-1 X + L-1 (G (H - 1))_
X x (X + (G(H - 1))) + L - 1) = XxX + X(G + X(H - 1)) + X L-1 _________________
L-1 (G (H - 1))
To Test This Expression
L-1 (G (H-1))
Let us substitute the value of X for 1 in a 2x2 rectangle on a 10x10 grid and check that it calculates the square values correctly.
So far so good
And now I will use the above expression to find the difference (which should be 10)
L-1 (G (H-1))
2-1 (10 (2-1))
1 x (10 x 1)
1x 10 = 10
It seems fine so far.
I will now check it with a 3 x 4 rectangle on a 10x10 grid. I know the difference should be 60 from previous workings.
L-1 (G (H-1))
4-1 x (10 (3-1))
3 x (10 (2))
3 x 20 = 60
The expression is true!
L-1 (G (H-1)) is the universal rule with the brackets in different positions:
L-1 (G (H-1)) = L-1 x (Grid Width) x (H-1) = Difference
Evaluation
I believe I have completed my aims and succeeded in attaining the goals I set myself. I have found the rule to work out the difference for any rectangle of any size on any grid. I have done this through the use of algebraic methods.
I have taken this course work as far as I can in the time that I have been allotted. I am happy with the work that I have done and can not think of anything I could have carried out better or improved upon.
