Lesson video

Hello everyone, and welcome to another maths lesson.

It is great to see you joining me.

Mr. Gratton here for a lesson where we will be exploring relationships between squares that are joined together at their vertices to create different types of triangles, including right-angled ones.

We won't be looking explicitly at Pythagoras' theorem until much later on in the lesson, but pause here if you want a sneak peek at what we'll be looking at today.

First up, let's look at a creative task about physically creating triangles from combinations of squares.

Let's have a look.

Lucas here can create a triangle by joining up some of the vertices of three squares, like this.

The triangle is in that empty space in the middle of those three squares.

We can use a protractor to measure these angles.

This will help us to see what type of triangle it is.

In this situation, Lucas is correct, it is an equilateral triangle.

Laura, however, says that this is obvious.

Of course, it is an equilateral triangle since Lucas used three congruent squares to create that triangle.

If each square is congruent, the area of each square is equal.

So, is it possible to make different types of triangles with squares that are not congruent with each other? Well, for this check, let's find out.

Lucas tries to create a triangle from two congruent squares, each with an area of 20 centimetres squared, and one smaller square with an area of four centimetres squared.

Pause here to consider what type of triangle has Lucas made.

And the answer is an isosceles triangle.

Similarly, for this next check, what type of triangle is made from these three squares joined at their vertices? And pick which explanation or explanations, explain and justify how you know what type of triangle this is.

Pause now to do this.

This is, once again, an isosceles triangle.

This is because there are exactly two squares that are congruent to each other, each with an area of five centimetres squared.

There is one square that is different in size.

This time, larger in this area compared to the other two squares.

And this time, I now have the three angles in this triangle.

One angle is 115 degrees, then 30 degrees, and 35 degrees.

Can you figure out what type of triangle this is, and what do you know about these three squares from the information about the angles given above? Pause now to do both of these questions.

This triangle is a scalene triangle.

We know this for certain because a scalene triangle has three angles that are all different in size.

Furthermore, a scalene triangle has three sides of equal length.

And this means that there are three squares here, each with a different area.

Onto the first practise task.

For this task, it is very helpful to print out these squares that you can see on screen drawn to scale.

This is so that any triangles that we make from these squares can be as accurate as possible.

You will need to cut out each square separately and as accurately as possible.

When you've done that, experiment for a little bit, take any three squares and join them together at their vertices to see if you can create a triangle in that empty space in the middle of those three squares.

When you've played around a little bit, let's have a look at question one.

Create a triangle by joining the vertices of the squares with areas one, three and five units squared.

Then, take a protractor and measure each of the angles in that triangle.

And for question number two, explain why you cannot create an equilateral triangle with the set of squares that I've given you.

Pause now to do questions one and two.

And for question three, using the same set of squares, create two triangles.

This time using squares with areas two, five and six units squared and a separate triangle with squares four, 10 and 12 units squared.

When you've done that, measure the angles in both of those triangles, and see if you can spot any patterns between the two triangles.

For question number four, create a triangle that is similar but not congruent to a triangle made from the squares with areas 21, 27 and 39 units squared.

You may need to use your observations from question three to do this.

Pause now to do questions three and four.

For question five, let's look at isosceles triangles only.

Create these three isosceles triangles, and take a protractor and measure what you think is the largest angle in each of those three triangles.

Which triangle has an acute angle as its largest angle? Which one has a right angle as its largest angle? And which one has an obtuse angle as its largest angle? Pause now to do question five.

And for question six, take as long as you can to experiment with all of these squares.

Let's complete the table.

Pick any three squares and join them at their vertices to create a triangle.

Judging by eye, is the largest angle in that triangle acute, obtuse, or right? If you are not sure, you can use a protractor, but for ones that are obvious, you can just say "the largest angle is", for example, "acute".

Keep a note of the combinations of three squares in that table and fill out as many rows as you can.

Pause now to do this and see if you can spot a pattern between the areas of the squares that you've chosen and the size of that largest angle.

Great effort so far in experimenting with all of those squares.

Here are the answers.

For question one, the interior angles of that triangle are approximately 25, 48, and 107 degrees.

If you are around two degrees above or below those values, then that's okay.

It can be pretty tricky to be super precise with your protractor when the squares can shift around a little bit.

For question two, it is impossible to create an equilateral triangle because I've only given you at most two copies of each sized square, and you need three congruent squares in order to create an equilateral triangle.

Oh, for question three, I've spotted that both of those triangles have angles of approximately 35, 64, and 81 degrees.

But, why is this? This is because the triangles are similar to each other as their angles are the same.

Did you notice that the squares that make up triangle Y are twice the area of the squares from triangle X? I wonder if this would also work if I tripled the areas of the squares in triangle X rather than doubled them.

Wait a second.

That's exactly what I can do with question number four.

Notice how I have not given you any squares with areas of 21, 27, or 39 units squared.

I can divide the areas of each square by three, so seven, nine, and 13 are a third of 21, 27, and 39 units squared.

If I were to join those three squares together at their vertices, I would create a triangle that was similar to one that I could create with squares of 21, 27 and 39 units squared.

For question five, triangle A had a largest angle of 100 degrees, B, 90 degrees, and C, 80 degrees.

The triangle with the largest angle that was acute was triangle C.

The triangle with a right angle as its largest angle was triangle B.

And so triangle A as its largest angle had an obtuse angle.

And for question number six, there are many, many possible triangles that you could have created from different combinations of squares.

Here are some examples of ones that I have done.

Did you come up with a theory for the relationship between the areas of the squares and the size of the largest angle of the triangle? Let's find out if your theory is correct.

So, now that we've seen that combining different squares can result in different triangles, the next question is, are there specific properties of those triangles that depend on the size of the squares we choose, and can we reliably control the properties of the triangles we make by changing the size of those squares? Well, let's find out.

Lucas agrees, we can create a range of different triangles from differently sized squares.

But Alex wants to dig deeper into this.

Is there a pattern in the squares that we choose and the largest angle within that triangle? Here are some of Laura's observations from the previous task.

Pause here to think or discuss.

Can you spot a pattern in the areas of the squares that make up each type of angle? Alex notices that the area of the largest square is smaller than the sum of the areas of the two smaller squares.

That is to say, the largest square is not significantly larger than the other two squares.

In all cases that this occurs, the largest angle inside that triangle is acute.

So, for example, seven plus 10 is 17, and the largest square with area of 13 is less than 17.

And again, four plus seven is 11, and the largest square with area of nine is less than 11 as well.

Is there a similar pattern for obtuse and right angles that requires you to look at the sum of the two smaller squares and compare it in size to the largest square? Pause here to adapt Alex's observations to these other angles and think about or discuss your observations.

Dynamic geometry software can be a great way of showing many triangles very quickly, and perhaps spot a pattern more easily.

Here's a link to one.

See if this helps you find or confirm your observations.

As Alex observed, if the area of the largest square is less than the sum of the two smaller areas, meaning, the largest square isn't significantly bigger than the other two squares, then the largest angle in that triangle is always acute.

For example, seven plus four is 11, and eight is less than 11.

However, the opposite is also true.

If the area of the largest square is greater than the sum of the two smaller areas, meaning, that the largest square is significantly larger than the other two squares, so big in fact, that it's bigger than the other two combined, than the largest angle in that triangle is always obtuse.

For example, seven plus four is 11, but the largest square is so much bigger at 18.

So, we've seen that the largest angle in a triangle can be either acute or obtuse.

But can it be right-angled? It can.

If the sum of the two smaller areas is exactly equal to the largest area, the largest angle in that triangle is always right-angled.

Like in this example, the two smallest squares have an area of seven and four centimetres squared, and the largest square has an area of exactly 11 centimetres squared, which is seven plus four.

For this check, let's use our observations by adding together the areas of the two smallest squares and comparing that sum to the area of the largest square, find out which of these triangles has a largest angle that is acute, obtuse, and right-angled.

Pause now to do this for all three diagrams. And the largest angle in triangle A is right-angled, as 12 plus eight is 20, which is exactly the area of that largest square.

For B, 10 plus six is 16.

The largest square is larger than the combined area of the two smallest squares.

Because the largest square is just that big, the largest angle in the triangle is obtuse.

Obtuse angles are larger than right angles and acute angles.

For C, eight plus seven is 15.

The largest square is still smaller than the combined area of the two other smaller squares.

Because the largest square is still fairly small in size when compared to the other two smaller squares, the largest angle is also quite small being an acute angle instead.

For this check, the largest angle in this triangle is confirmed to be an obtuse angle.

Which of these are possible values of X units squared, the area of that largest square? Pause now to answer this question.

And note, there may be more than one correct answer.

For my answer, I need a square with an area greater than 14 plus 11.

Since 14 plus 11 is 25, any answer greater than 25 will do, and so 30 and 36 units squared are suitable areas.

This triangle is right-angled.

Pause now to find the value of Y.

And the answer is 16 plus 12, which is 28.

Alex sensibly asks, is the location of the largest angle random, or is there a relationship between the largest angle and the squares? Laura's observation is that the largest angle never seems to touch the largest square.

Maybe the largest angle is always opposite the largest square.

Alex seems to agree, as it works for these three triangles.

But this pattern that they have observed still seems to hold true no matter how big or small that largest square in the diagram is, assuming that it still is the largest square and that the three squares still make a triangle.

This is helpful.

The largest square is always opposite that largest angle.

And so using the observation, in this check, which angle in this triangle is the largest? Pause now to have a look.

And the largest angle is angle B because it is opposite the largest square with an area of 35 units squared, rather than 30 or 28 units squared.

Okay, for each of these triangles, draw on an angle marker at the largest interior angle of each triangle and write down whether it is acute, right, or obtuse.

Pause now to do this question.

For these two questions, write down the areas of the two unlabeled squares.

For question two, there is a range of possible answers.

However, question three has only one valid correct answer.

Pause now to do questions two and three.

Each group shows the areas of three squares that creates a triangle.

Place each group into the correct part of the table that describes the size of its largest angle.

Pause now to do question four.

Question 1A, is acute, as the largest square is smaller than the sum of the two smaller squares.

For B, we have a right angle, as the largest square is equal in size to the sum of the two smaller squares.

And for C, we've got an obtuse angle, as the largest square is larger than the sum of the two smaller squares.

For question two, that unlabeled square is the largest square as its opposite the largest angle.

Therefore, its area must be larger than 14 units squared.

But because the largest angle is still acute, it must be less than the sum of 11 and 14, and so its answer must be less than 25.

So any answer greater than 14, but less than 25, is completely valid.

However, for question number three, the one correct answer is nine units squared.

This is because 24, take away 15, is nine.

And for four, pause here to compare your table with the completed one on screen.

We have joined different combinations of three squares, sometimes where some of the squares are congruent and sometimes where none of them are congruent.

Depending on the combinations chosen, we can create equilateral, isosceles and scalene triangles.

We've also observed that the largest angle in the triangle can either be acute, right, or obtuse, and its size depends on how large the area of the largest square is when compared to the sum of the areas of the two smallest squares.

And importantly, if the areas of the two smallest squares are equal in size to the area of the largest square, a right-angled triangle is created, and that is Pythagoras' theorem.

Thank you so much for making the decision to join me today.

I hope you can use this understanding for the rest of the Pythagoras' theorem topic.

I've been Mr. Gratton, and until our next maths lesson together, take care and enjoy the rest of your day.

Goodbye.