2X2 square = 12X16=192, 11X17= 187 192-187=5
2X2 square = 4X8=32, 3X9=27 32-27=5
I noticed that no matter where in the grid the 2X2 square was the diagonal difference was always 5. This is equal to the number of rows. Once I noticed this I realised that the 3X3 grids diagonal difference was equal to the number of rows X4.
Also the number of rows in the 4X4 square are timed by 9 And in a 5X5 square the rows are timed by 16.
I then noticed that if you take the number of rows in the square and take away one and square the answer you can times this by the number of rows in the grid you have the diagonal difference. This equation is written
(rows in area used - one)squared X rows in grid = diagonal difference
or
(x-1)2 * r = d
To prove this equation I will use it to work out more diagonal differences:
(2-1)2 X 4 =4
(3-1)2 X 4 =16
(4-1)2 X 4 =36
Now I will work them out by cross multiplying.
2X5 - 1X6 = 4
3X9 - 1X11= 16
4X13 – 1X 16= 36
Now I will test my equation on a grid using algebra.
((4+a)X(13+a))-((1+a)X(16+a))= (52+17a)-(16+17a)=36
(4-1)2 X4=36
((3+a)X(9+a))-((1+a)X(11+a))=(27+12a)-(11+12a)=16
(3-1)2 X4=16
((2+a)X(5+a))-((1+a)X(6+a))=4
(2-1)2 X4=4
Extension Work: Diaganal Differences in Rectangles
Yellow =a
Green =b
Blue =c
Purple =d
D= diagonal differences
In “a” D is (2X6)-(1X7)=4
In “a+b” D is (3X6)-(1X8)=12
I noticed that D of “a+b” was 3 times “a”.
In “c” D is (13X21)-(11X23)=20
In “c+d” D is (15X21)-(11X25)=315-275=40
I noticed that D of “c+d” was 2 times “c”.
I will now try this on some different squares.
Yellow=a
Green=b
Blue=c
Purple=d
14X
