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 4 centimetre 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 5 centimetres squared.

There is one square that is different in size, this time larger in it's 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.

On to 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 the 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 1, 3 and 5 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 2, 5 and 6 units squared and a separate triangle with squares 4, 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 area 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 2 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 3, so 7, 9 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 120 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, 7 plus 10 is 17, and the largest square with area of 13 is less than 17.

And again, 4 plus 7 is 11, and the largest square with area of 9 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, 7 plus 4 is 11 and 8 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, then the largest angle in that triangle is always obtuse.

For example, 7 plus 4 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 7 and 4 centimetres squared, and the largest square has an area of exactly 11 centimetres squared, which is 7 plus 4.

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 8 is 20, which is exactly the area of that largest square.

For B, 10 plus 6 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, 8 plus 7 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 it's 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 it's answer must be less than 25.

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

However, question number three, the one correct answer is 9 units squared.

This is because 24 take away 15 is 9.

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

So far we focused on all different types of triangles, but let's refine our focus onto just right angle triangles and see if we can consistently generate them from different squares.

Let's go.

Laura says if she picks any two squares and those are are areas of those two squares, it is possible to instantly know the size of this third square needed to create a right angle triangle.

This is because the areas of the two smaller squares must add up to make the area of the larger square, as we can see here, 16 plus 24 equals 40.

The area of the largest square must have an area of exactly 40 square units.

But Alex raises a very good point.

What if you do not know the area of the two smaller squares? Well, it's possible to calculate the area by measuring one side length of a square, in this case, 5 units in length, then squaring that length as the area of a square is its base multiplied by its height, which is of equal length to each other.

So the area of this square is 5 times 5 equals 25 square units.

Therefore, the area of this third square needed to create a right angle triangle is 30 plus 25 equals 55 square units.

So, for this check, what is the area of square B in units squared? Pause now to have a think.

I take the length, which is 4 units, and square it.

4 squared is 16 units squared.

For the next question, what is the area of square C, now that I know the area of square B? Pause now to do this next question.

And as usual, I take the areas of the two smaller squares, 18 and 16 in this case, and add them together.

Therefore, 18 plus 16 is 34, and so the area of square C is 34 squared units.

And for this next check, find the area of each of these 3 squares.

Pause now to do this.

The area of square A is 4 squared, which is 16.

The area of square B is 3 squared, which is 9.

The area of square C is those two areas added together, 16 plus 9 is 25 units squared.

Next up, what is the value of X, the side length of square C? Pause now to do this.

The side length of square C is going to be 25 square rooted, which is 5.

Therefore, the side length of square C is 5 units.

And for this last check, once again, find the areas of squares A, B, and C.

Pause now to do this.

For A, 3 squared is 9, for C, 7 squared is 49, and for B I've got 49 take away 9, which is 40 units squared.

For B, I had to subtract rather than add because I know for certain that square B is not the biggest square in this diagram because it is not opposite the right angle, which is the largest angle in that triangle.

And on to the practise.

For question one, find the area of each unlabeled square.

Pause now to do this.

And similarly for question two, pause here to find the area of every square in each diagram.

And lastly, for question three, pause here to find the area of each square and find the missing length of each square that is labelled with a letter.

On to the answers for question one.

A had a missing area of 22 units squared, B had a missing area of 30 units squared, and C had a missing area of 15 units squared.

And for question two, pause here to see if your answers match those that are on screen.

And once again, for question three, pause here to match your answers to those that are on screen.

We've looked so much at triangles and squares, but where does Pythagoras' theorem come into all of this? Let's see if it's anything new.

Sam seems to think not.

They believe that we've already done Pythagoras' theorem.

They say that the area of square A plus the area of square B equals the area of square C, which seems familiar when we think about the work that we've done in the previous cycles in this lesson.

And Alex agrees and identifies that we've all already done this, but only when dealing with right angle triangles.

This is because we're looking at an equality, the area of square A plus the area of square B equals the area of square C.

We've seen triangles with, for example, obtuse angles where the area of the largest square is massive compared to the combined area of the two other smaller squares, not simply equal to them.

Lucas raises a very good point.

What if we don't know the area of the squares? Oh, but haven't we asked this question before? What was the solution last time? Alex remembers, find the length of a side of the square, then square it to find its area.

And so there are many equivalent, but different ways of looking at Pythagoras' theorem, area of square A plus area of square B equals the area of square C.

Or if we do not know the areas of the squares, we can consider the length of side A, then square it, plus the length of side B then square that equals the length of side C also squared.

And this way of describing Pythagoras' theorem is a lot closer to one that you might have heard of before.

This rule can be further condensed down into a form like this.

Rather than saying the length of a side squared, simply say, A squared plus B squared equals C squared, where A squared, B squared and C squared are just the areas of each square found by squaring the length of each of those three squares.

But it is important to note that C squared must be the area of the largest square, since the largest square is equal in area to the sum or addition of the other two squares.

However, Lucas is a bit sceptical.

We've seen it's true for some right angle triangles, but the following proof can be adapted to any right angled triangle, no matter the size of the squares that make it.

And here's the proof.

We can take the side of the largest square that is shared by the triangle and translate it down onto one of the smallest squares.

This is usually the middle sized square, not the largest or smallest one.

We can then repeat this exact same process, but for one of the perpendicular sides of the largest square, like so, again, translating it down onto the middle sized square.

What we've done now is split that middle square into four congruent quadrilaterals.

We can then take each of these four quadrilaterals and translate them into the largest square, fitting them within each corner of that square.

Doing this will leave a square size gap in the middle.

And, what do you notice? Oh, the area of that gap is exactly the area of the smallest square.

So what we've shown is a method of manipulating the two smallest squares from our right angle triangle, in such a way that they fit perfectly into the largest square from that same triangle.

This visual proof is not something that you need to learn or replicate, but it is important in visually showing why Pythagoras' theorem actually works.

And onto the check.

Select the statement that completes this sentence.

For Pythagoras' theorem to be used, three squares must join together to make what? Pause now to choose the correct option.

We use Pythagoras' theorem when we are dealing with a right angled triangle.

And for this next check, which of these are suitable ways of describing Pythagoras' theorem for three squares which join at their vertices to create a right angle triangle? Pause now to look through all four options.

And actually, all four of them are correct.

Each of these four options are things that we've looked at throughout this lesson.

And on to the practise questions.

On which of these three diagrams is Pythagoras' theorem true and explain your reasoning and what does that tell you about those triangles? Pause now to do this question.

Which of these four options are valid representations of Pythagoras' theorem for these three squares which enclose that right angle triangle? Pause now to do this question.

And lastly, pause here to answer question number four.

And here are the answers.

Only on diagram B is Pythagoras' theorem true? This is because Pythagoras' theorem suggests that the sum of the areas of the two smaller squares equals the area of the largest square, and 14 plus 14 is 28.

This means that only B creates a right angle triangle.

And for question three, A, B, and D are all valid representations of Pythagoras' theorem for this diagram.

C is not valid because it suggests that square F is the largest square.

We can see here that square H is the largest instead.

And pause here to have a look at the answers for question number four.

And thank you all so much for your effort today.

We have explored so many triangles from a range of different squares, in a lesson where 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 smaller 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 math lesson together, take care and enjoy the rest of your day.

Goodbye.