Opposite Corners of a Square on a Number Grid

---

Data Example 1 – 2*2 Square:

So, as I have shown, the difference between the diagonally opposite corner results of a 2*2 square, are always 10.

---

Data Example 2 – 3*3 Square:

        

So, as I have shown, the difference between the diagonally opposite corner results of a 3*3 square, are always 40.

---

Data Example 3 – 4*4 Square:

So, as I have shown, the difference between the diagonally opposite corner results of a 4*4 square, are always 90.

---

Data Example 4 – 5*5 Square:

        

So, as I have shown, the difference between the diagonally opposite corner results of a 5*5 square, are always 160.

---

Data Example 5 – 6*6 Square:

        

So, as I have shown, the difference between the diagonally opposite corner results of a 6*6 square, are always 250.

---

Results – Pattern:

Because the second difference is 20, the nth term is 202= 10n²

So we know that the final algebraic expression will be associated with 10n². Now we test the results to see how the algebraic expression relates to the results. We do this so we can form our final algebraic expression.

        

Testing the Expression:

We know that the difference between the results of the diagonally opposite corners of a 6*6 square is always 250. With the expression we have so far, we shall see what we need to do to it to get our final expression.

Now, we are testing the expression on a 6*6 square, so n, the term number, will be 6.

  10n²

=10 * n²

=10 * 6²

=10 * 36

=360

That answer is too high for our final expression to be 10n², but if we divide the expression by 10 and divide all the results by 10, we are left with n² and these results:

 

 

---

Forming an Expression:

The number before the term number is squared and then multiplied by 10 to reach the result. So this means the new expression would be written like this:

10(n-1) ²

Now we shall test this for a 6*6 square, so the term number will be 6:

10(n-1) ²

=10 * (n-1) ²

=10 * (6-1) ²

=10 * 5²

=10 * 25

=250

We already know that the difference in the results of the opposite corners of a 6*6 square is 250, o this algebraic expression works!

Analysis:

This investigation could have been limitless. I had a maximum square size of 10*10, but I would only be able to give 1 example of this, as my number grid would be too small. But now, I wouldn’t need a bigger number grid to work out the differences between the results of the diagonally opposite corners, because I have my algebraic expression now. I have tested it and proved it to be accurate and I am happy with my results.

Conclusion:

From this investigation, I have learned that there is a relationship between the size of the box on the number grid, and difference in the diagonally opposite corners. My prediction was correct, because I said that there would be a pattern and algebraic expression which would work on all box sizes, and there is:

10(n-1)²

I also predicted that there would be a lot to think about and explore, which there is. I also enjoyed discovering the results.

If I was to do the investigation again, I would branch out even more, and test if the expression worked on other quadrilaterals, like rectangles, or even 3D shapes such as cubes.