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The Truth About Triangles and Squares

Achievement Objectives:

Achievement Objective: NA4-9: Use graphs, tables, and rules to describe linear relationships found in number and spatial patterns.
AO elaboration and other teaching resources


This unit explores some of the algebra behind triangular and square numbers.

Specific Learning Outcomes: 
  • find a rule to describe any member of a number sequence and express it in words
Description of mathematics: 

Triangular and square numbers belong to a family called the figurative numbers that arise from the shapes numbers of objects form. Figurative numbers are believed to have originated from the Pythagorean society of ancient Greece(c. 550-350 BC).

Triangular numbers most commonly arise in probabilistic situations, most notably in establishing the number of outcomes for given events. Square numbers are the simplest example of a quadratic function. Quadratic functions are used frequently to model real world situations where the rate of growth or decline is not constant.

Links to Numeracy

This unit can be used to develop student knowledge of basic facts, in particular square numbers and their corresponding roots.

As students work with square numbers and their roots create a list of known square numbers. Challenge the children to find other numbers that can be added to the chart and have students check the additions others have made.

What is the largest square number you can think of?
What is the smallest square number you can think of?
How can you check whether a number is a square number?

Required Resource Materials: 
Copymaster 1
Copymaster 2
Key Vocabulary: 

 figurative numbers, sequential patterns, spatial patterns, square numbers, triangular numbers, roots, consecutive, conjecture, proofs, generalisation



Session 1

Build this triangular pattern with counters, and ask the students to tell you how many counters were used.

diagram of counters.

Discuss the methods they used like adding the layers 1 + 2 + 3 + 4 + 5 = 15.
Comment that fifteen is a triangular number since it forms a triangular pattern.
Challenge the students to find how many counters would be needed to make the tenth triangular number, that is a triangular pattern with ten counters on the bottom row.

Allow the students access to counters if need be but encourage them to use mental arithmetic to find a solution. Discuss ways in which they solved the problem.
Suggest that one way to find the number of counters in the five-layer triangle may have been to transform it like this:


So 1 + 2 + 3 + 4 + 5    = (1 + 4) + (2 + 3) + 5
                            = 3 × 5
                            = 15

Encourage the students to use the same strategy for the tenth pattern:


(1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6)           = 5 × 11

                                                                                    = 55

Challenge the students to make up a list of the triangular numbers for as far as they can, beginning with the known numbers; 1, 3, 6, 10, 15, …, 55, …

Discuss what patterns they can see that allows them to work out the next number without making the figure with counters. Note that the pattern goes:


Challenge the students to find the 100th triangular number, i.e. the sum of the counting numbers to 100:

1 + 2 + 3 +… + 98 + 99 + 100            = (1 + 100) + (2 + 99) + (3 + 98) + …

                                                            =  50 × 101

                                                            = 5050

Make this square figure using counters.


Ask the students to work out how many counters, in total, were needed to make the figure.

Tell them that 4 × 4 = 16 is a square number as sixteen counters can be arranged into a square figure.

Challenge them to find other square numbers. Compile a collective list as more square numbers are found. Note that the differences between consecutive square numbers are all odd.


Ask the students to look at the square figures of counters and explain the odd differences.


The number of new counters added on to an existing square is always a multiple of two (black counters) plus one (white counter). So the difference is always an odd number.

Challenge the students to find the 100th square number which is 100 × 100 = 10 000.

Session Two

Revisit the set of triangular numbers. Ask the students to find the sum of consecutive triangular numbers and organise their results in a systematic way. For example:

1 + 3 = 4                      Triangular numbers       1          3          6          10        15        …
3 + 6 = 9                      Consecutive sums               4          9          16          25            …
6 + 10 = 16
10 + 15 = 25

Ask the students to describe any pattern they see and record it as a conjecture. For example:
“If you add any two consecutive triangular numbers you always get a square number.”

Suggest to the students that they need to come up with a way to prove that the conjecture is true. Their proofs are likely to be geometric. For example:


This may lead to the generalisation:

diagrram of generalisation.

Require the students to be more specific in terms of expressing their generalisation.
“Can you predict the size of the square number if you are given the size of triangular numbers? For example, if you know that the first triangular number is the sum of,
1 + 2 + 3 + 4 + 5 + 6, can you say how big the square number will be?”
Use more specific examples like that to encourage students to describe the link between triangular numbers and square numbers. Ask them to express their ideas in words first, such as:
“The first triangular number plus the second triangular number equals the square number that is the equivalent to the second triangular number”

Students might recognise that inventing notation like T


for the sixth triangular number, and S


for the seventh square number (7


) can be used to describe the generalisation:

T6 + T7 = S7 , leading to Tn + Tn+1 = Sn+1 = (n + 1)2  

Session Three

Use Copymaster 1 to provide the students with a sequence of problems related to square numbers. Allow the students to solve the problems independently or in small co-operative groups. Assemble the class to discuss the common properties of the problems.

Problems one and two both contain geometric patterns in which an odd number of triangle or squares is added to the previous pattern. Consideration of the layers in the triangular pattern shows that one, then three, then five, then seven triangles are added. This is the property of the sequence of square numbers found in lessons one and two. Students should realise that the tenth square number is 100 so the window would have ten layers of triangles.

Similarly, rearranging the staircase patterns shows that they can be reformed into squares. For example, here is a reformation of the 4-high staircase.


Therefore the number of small squares found in a 15-high staircase would be the fifteenth square number (15 x 15 = 225).
The garden problem (Problem Three) also involves square numbers, as students should detect from the 9m2 and 49m2 square fields. The problem requires connections to the square root, that is the side length that produces a square of given area. So the length of a side in the 49m2  plot is 7 metres since (√49 = 7). Comparing the side lengths of the given squares 7 metres and 3 metres suggests that the smallest plot must have a side length of 2 metres (area 4m2) so the top left square has a side length of 7 – 2 = 5 metres (area 25 m2). The other plot areas can be worked out from there as 16m2 and 36 m2
In Problem Four students should recognise similar number sequences in the “Dance till You Drop” competition. If someone dances for twenty-four hours the last hour will be worth $47 and they will receive a total of 242 (twenty-four squared) dollars. That is $576.
For problem five students will need to be aware that there are different sized squares involved and this is helpful in systematically finding them all. Consider the 4 x 4 checkerboard.
Counters can be used to mark the centre of each square as shown below.


Sixteen 1 × 1 squares                    Nine 2 × 2 squares                  Four 3 × 3 squares  

…and the large square itself.
So the 3 × 3 checkerboard has 1 + 4 + 9 + 16 = 30 squares (The sum of the first five square numbers) By similar reasoning the 8 × 8 checkerboard will have 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204 different squares.
Exactly the same reasoning can be applied to problem six which shows successive squares being joined together. Students may recognise that the number of floors below each square join is 1 then 3 then 6 then 10, and so forth. These are the triangular numbers so a fifty-five floor building must contain 10 different square sections since 55 is the tenth triangular number. So the total number of windows will be found by 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 windows.

Session Four

Give the students copies of Copymaster Two that has problems related to triangular numbers. Allow them to solve the problems individually or in co-operative groups.
After a while gather the class together to share their solution strategies.
In solving problem one the students might note that the welds form matching triangular patterns allowing the triangular numbers to be used an efficient way to get the answer. For example, the welds for four joined squares is twice the third triangular number ( 2 x 6 = 12). For ten square twice the ninth triangular number, forty-five gives the number of welds (90).

joined squares.

Problem two involves a pattern in which the next distance in the spiral is the previous distance plus one metre. Students should recognise that this pattern of differences is found in the triangular numbers. The spiral will have a total distance of the tenth triangular number, fifty-five.
This pattern is less recognisable in problems three to five that involve combinations. In the icecream problem each new flavour creates an extra number of icecreams equal to the previous number of flavours plus one. For example, combining Uncool with the other flavours X, Y, and Z, gives four more possible icecreams. They are XU, YU, ZU, and UU. The next flavour Vex will combine with the others to create five more double scoop icecreams XV, YV, ZV, UV, and VV. So the total number of possible icecreams will be the sum of 1 + 2 + 3 + 4 + 5 + 6 = 21 (the sixth triangular number).
Similar reasoning can be used to solve the other two problems. Each new town creates a number of roads equal to the number of existing towns. Consider the addition of the first new town, Dag, results in three new roads, AD, BD, and CD.

roads and towns.

So the number of roads with seven town in total will be the sum of 1 + 2 + 3 + 4 + 5 + 6 = 21 (the sixth triangular number).
Problem Five involves a different context but the same mathematics where each new person added must play games with each of the current players. The number of games that must be played is the sum of 1 + 2 + 3 + …+7 + 8 + 9. This is forty-five, the ninth triangular number.

Session Five

Provide the students with the following sequential patterns. Invite them to find out how many counters will be needed to form the tenth member of each pattern.

Swimming Togs

swimming togs.

size one            size two                        size three                                  size four

Town Houses

town houses.

Plan 1               Plan 2                           Plan 3                           Plan 4

Note that both patterns involve figures that can be partitioned into triangular and square arrays. The swimming togs pattern is made of two triangular arrays joined together and therefore the number of counters is twice the matching triangular number:























two times second triangular number.


The townhouse pattern consists of square (body of the house) and triangular (roof line) patterns being joined.













































As a final problem pose this challenge:

triangular pattern.This triangular pattern can be changed into a square by moving only a few counters.What is the smallest number of counters that need to be moved?

Students may realise that the figure has 36 counters that could be used to form a 6 x 6 square pattern.


 triangular pattern.


Superimposing the size of a 6 x 6 counter square over the triangular array reveals that only six counters need to be moved.

TruthAboutTriangles&SquaresCM1.doc29.5 KB
TruthAboutTriangles&SquaresCM2.doc72 KB