# Activity: Drawing Squares

For this activity all you need is a grid of dots, a pencil and your brain.

### Let us discover how many squares you can make on different grids:

**Note:** "1 by 1" means how many **sides** (not how many dots).

###

So, let's try drawing in some squares and count how many:

### 1 by 1

Well, that's easy, there's just one: |

### 2 by 2

That seems to be easy too. There are four of them, aren't there? | ||

But wait, that's not the complete answer. There's also this bigger one: |

That makes five squares altogether - four 1 by 1 squares and one 2 by 2 square

## Your Turn !

### 3 by 3

Over to you now. Here's the grid: |

**Hint:** For the 3 by 3 case, you will expect to get 1 by 1 squares, 2 by 2 squares and 3 by 3 squares. How many of each?

**Now you can start to fill in a table:**

How Many 1 by 1 squares | How Many 2 by 2 squares | How Many 3 by 3 squares | How Many 4 by 4 squares | How Many 5 by 5 squares | Total | |

1 by 1 Grid: | 1 | 1 | ||||

2 by 2 Grid: | 4 | 1 | 5 | |||

3 by 3 Grid: | ||||||

4 by 4 Grid: | ||||||

5 by 5 Grid: |

## Did you notice anything about the numbers in the table?

They are all square numbers:

- 1
^{2}=**1**, - 2
^{2}=**4**, - 3
^{2}=**9**, - etc ...

and the totals are found by adding together square numbers.

### Formula to The Rescue ... !

There is actually a formula for adding the first **n** square numbers:

S_{n} = n(n+1)(2n+1) / 6

### Example: The number of squares in the **5 by 5** case

Try substituting **n = 5** into the formula:

_{n}= n(n+1)(2n+1) / 6

_{5}= 5 × (5+1) × (2×5+1) / 6

_{5}= 5 × 6 × 11 / 6

_{5}= 55

## So, we seem to have solved the question. Yipee!

### But wait ... there's more!

I said you would need to use your brains. Let's go back to the 2 by 2 case:

### 2 by 2

There is another square too, this one:

**Why is it a square? **It has four equal sides and four right angles, so that's a square.

So, that makes six squares altogether.

Four 1 by 1 squares, one 2 by 2 square and one**x by x** square.

What is the value of **x**? We can use Pythagoras' Theorem to find it:

^{2}= 1

^{2}+ 1

^{2}= 1 + 1 = 2

So, we have **four 1 by 1 squares**, **one 2 by 2 square** and **one √2 by √2 square**.

## Your Turn !

### 3 by 3

- Are there any more squares?

**YES**! Can you find them?

### 4 by 4 and 5 by 5

Also try the 4 by 4 grid, and the 5 by 5 grid

As you proceed, you will find squares like these:

## What are the lengths of the sides of these squares?

You can use Pythagoras' Theorem to work that out yourself

### In each case, how many do you get of each one?

Here is a table to help you:

How Many 1 by 1 | How Many 2 by 2 | How Many 3 by 3 | How Many 4 by 4 | How Many 5 by 5 | How Many √2 by √2 | How Many √5 by √5 | How Many √8 by √8 | How Many √10 by √10 | How Many √13 by √13 | How Many √17 by √17 | Total | |

1 by 1 Grid: | 1 | 1 | ||||||||||

2 by 2 Grid: | 4 | 1 | 1 | 6 | ||||||||

3 by 3 Grid: | ||||||||||||

4 by 4 Grid: | ||||||||||||

5 by 5 Grid: |

## Advanced

Can you find a formula to calculate the number of squares that have lengths that are square roots?

Can you then find a formula for the total number of squares in each case?

Are there any more squares lurking in there that we've missed?

## Conclusion

What started off as seeming to be a simple exercise turned out to be quite complex. You really **do** need to use your brains to think this one through, but it's a challenging and rewarding exercise.