Pebbles

A group of children from Manorfield Primary School, Stoney Stanton sent in lots of ideas:

T.K and A.H said:

The pattern: As you double it each time, the shape changes from a square to a rectangle repetitively as each time you add a square to make a rectangle (square $+$ square = rectangle) and add a rectangle to make a square (rectangle $+$ rectangle = square).

A.H and E.R said:

We have noticed that each shape changes from a square to a rectangle because if you add a square to a square you get a rectangle e.g.:
$4\times4 + 4\times4 = 8\times4$ - a rectangle

S.B. and N.L. produced the following table of results:

Number	Shape	Number of pebbles on side	Area of shape	Perimeter of shape
1	Square	2x2	1cm$^2$	4cm
2	Rectangle	2x3	2cm$^2$	6cm
3	Square	3x3	4cm$^2$	8cm
4	Rectangle	3x5	8cm$^2$	12cm
5	Square	5x5	16cm$^2$	16cm
6	Rectangle	5x9	32cm$^2$	24cm
7	Square	9x9	64cm$^2$	32cm
8	Rectangle	9x17	128cm$^2$	48cm

PATTERNS AND FORMULAE

It was said in whole class discussion that the pattern for the size of the shapes was:

Squares: each side is the same length as the longest side of previous rectangle

Rectangle: one side is the same length as the side of the previous square, the other side is a "new length"

A.H and E.R also said:

The formula for the area = $2$ to the power of $(n-1)$.

The pattern for the new side of the rectangles is $+1$, $+2$, $+4$, $+8$, $+16$ ... (it doubles)

The pattern for the perimeter is $+2+2$, $+4+4$, $+8+8$, $+16+16$ ... (it doubles)

PREDICTIONS

S.W. and L.B filled in their table for the first six shapes, and then predicted:

For shape number seven, the area will be double the area of shape six and the perimeter will be $8$ more than the perimeter of shape six. And for shape eight, the area will be double the area of shape seven and the perimeter will be $16$ more than shape seven.

Well done all of you - you obviously worked hard on this problem.