Lesson video

Welcome everyone, it is great to see you here for another maths lesson with me, Mr. Gratton.

For this lesson, we'll be building upon and using Pythagoras' theorem to find the length of the hypotenuse of a right-angled triangle.

Pause here if you want to check the definitions of high hypotenuse as well as right-angled triangle and Pythagoras' theorem, the other keywords and ideas that we'll be using today.

In order to use Pythagoras' theorem effectively, let's have a quick look at the squares created from the sides of a right-angled triangle.

Aisha remembers that you can join three squares at their vertices to create a triangle.

But what conditions were met that made this triangle a right-angled one? The condition was that the areas of the two smallest squares had to equal the area of the largest square.

So the smallest squares with areas of 10 and 15 centimetres squared is equal to the area of the largest square.

10 plus 15 is 25 centimetres squared.

Andeep correctly points out that if you square root the area of a square, you can also find out the length of that same square.

In the case of the larger square with an area of 25 centimetres squared, its side length is the square root of 25, which is five centimetres.

This means five centimetres is also the length of the longest side of that triangle as each side of a square is equal in length.

Every side of the larger square here is five centimetres long.

Andeep is using the correct keyword here.

The hypotenuse is the longest side of a triangle, but only if it is a right-angled triangle.

And the hypotenuse is easy to spot since it is always opposite the right angle itself.

So using that information for this check, which square shares a side length with the hypotenuse of this triangle? Pause now to check and answer this question.

And the answer is square C.

This is because square c is opposite the right angle and not adjacent to or touching the right angle.

And similar again, which of these three squares shares a side with the hypotenuse of that triangle? Pause now to have a think.

Even though this diagram is in a different orientation to the last one, the rule is still the same.

Square Q is the correct answer because it is opposite and not adjacent to the right angle itself.

For the squares made from the sides of a right-angled triangle, which of these statements is correct? Pause now to look through all four options.

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

And both the correct answers are all to do with the largest side or largest area.

The square with the largest area is always opposite the right angle and the hypotenuse of the triangle is always the longest side of that triangle.

And for this next check, what is the area of square C? Pauses now to have a look at all three squares and think about the relationship between all three of them.

The area of square c is 81 centimetres squared because it is the sum of the areas of the two smaller squares.

58 plus 23 is 81 centimetres squared.

And for this same diagram, what is the length of the hypotenuse of that triangle? Pause now to have a look at which area you need to use in your calculation of the length of hypotenuse.

The hypotenuse shares a side with the largest square and so my focus is on the 81-centimeter-squared square.

To find the length of the hypotenuse, I square root that area, and so the square root of 81 is nine.

The hypotenuse of this right-angle triangle is nine centimetres.

Okay, let's remove the squares from the sides of this triangle for a moment.

Andeep wonders if it is still possible to find the length of the hypotenuse given only the side lengths of the two shorter sides of the triangle.

Sam is correct, if there are no squares, well, you can just draw them on yourself.

In this instance, we can draw on this square.

The area of the square is six squared or 36 centimetres squared.

The same can be done for this eight-centimeter side.

The square attached to it has an area of eight times eight, which is 64 centimetres squared.

We know that the square from the hypotenuse has an area equal to the sum of the two smaller squares.

So this square has an area of 36 plus 64, which equals 100 centimetres squared.

The length of the hypotenuse is the same as the length of any side of that largest square.

And so to find the length of the hypotenuse, we must square root the area of that largest square, and the square root of 100 is 10.

So the hypotenuse is 10 centimetres long.

Let's go through a series of checks for this diagram.

Izzy wants to draw a square with a side shared by the nine-centimeter side of this triangle.

What is the area of this new square? Pause now to have a think.

And the answer is I take the length of that square, nine, and square it.

Nine squared is 81 centimetres squared.

Next up, Izzy then draws a square with a side shared by the hypotenuse of the triangle.

What is the area of this even bigger square? Pause now to have a think.

The area of the square that shares a side with the hypotenuse of the triangle is always the sum of the areas of the two smallest squares on that same triangle.

So 40 plus 81, in this case, is 121 centimetres squared.

And now we know the area of that largest square.

What is the length of the hypotenuse of that triangle? Pause now to answer this last check question.

I take the area of the square, 121 centimetres squared, and square root it.

The square root of 121 is 11.

So the hypotenuse is 11 centimetres long.

Okay, onto the practise task.

For question one, find the area of each square in each of these three diagrams. Pause now to do this.

And for question number two, by, first of all, finding the area of a square, find the length of the hypotenuse of each of these triangles.

Also try and figure out which side of that triangle is the hypotenuse.

Pause now to do question two.

And finally, question three.

First of all, sketch on any missing squares on each of those triangles.

These squares do not need to be accurately drawn.

Then find the area of each square and hence find the length of the hypotenuse of each of these three triangles.

Pause now to do question number three.

And on to the answers.

For 1a, the area of the largest square is 72 centimetres squared.

For b, the area of the largest square is 51 centimetres squared, which was the sum of 15 and 36, 36 being six squared, the area of that middle-sized square.

And for c, we've got the smallest square with an area of one centimetres squared.

Then three squared is nine, so the middle-sized square is nine centimetres squared, and nine plus one is 10.

So the largest square has an area of 10 centimetres squared.

For question two, pause here to see if your answers match all of those that are onscreen.

And for question three, this square has an area of 324 centimetres squared.

And so the largest square has an area of 400.

The square root of 400 is 20, which is the hypotenuse of that triangle.

For part b, these two squares have areas of 400 and 441 centimetres squared respectively.

Therefore the area of the largest square is 841 and the square root of 841 is 29 centimetres, which is the length of the hypotenuse.

And for part c, we've got 576 and 49 centimetres squared as the areas of the two squares.

And the largest square has an area of 625 centimetres squared, which is 25 times 25.

Therefore the hypotenuse itself is 25 centimetres long.

So we can use squares from the sides of the triangle to find the length of the hypotenuse.

But we can make our method more efficient by using Pythagoras' theorem in a different way, let's see how.

Sam explains that it is time-consuming and takes up lots of space to always draw on the squares for each of the three sides of every triangle.

Laura responds though that you can calculate the area of the square without actually drawing on the square itself.

As you can see here, I can take the 11 centimetres and say 11 squared is 121.

And so the area of the square attached to that side of the triangle has an area of 121 centimetres squared.

Similarly, this square has an area of 81 centimetres squared because nine times nine is 81.

I know this without even drawing the square itself.

Furthermore, I know that the largest square, the one that shares a side with the hypotenuse itself, always has an area that is the sum of the squares on the other two sides of the triangle.

Therefore, this square has an area of 121 plus 81, which is 202 centimetres squared.

When we know the area of the square from the hypotenuse, we must square root that area to find the length of the hypotenuse.

But Sam identifies that the square root of 202 doesn't give an integer answer.

In fact, it's a decimal that continues on and on and on.

Sam wonders if this is okay.

Of course it is, any length doesn't need to be a whole number or integer.

It is possible and sometimes very likely that the lengths will be decimals instead.

Just make sure to round that number to a suitable number of decimal places, such as 14.

21 centimetres for the length of the hypotenuse.

Okay, for this check, what is the sum of the areas of these two squares? Pause now to have a think.

25 plus 49 is 74.

So the sum of the areas of these two squares is 74 centimetres squared.

Now that you know this value, what is the hypotenuse of this triangle rounded to two decimal places? Pause now to use all of this information to find the length of the hypotenuse.

That 74 centimetres squared also represents the area of the largest triangle that shares a side with the hypotenuse.

Therefore, I can square root 74 to give 8.

6 centimetres, which is the length of the hypotenuse.

There are many forms of Pythagoras' theorem, but they all achieve the exact same goal: describing the relationship between the squares made from the three sides of a triangle.

Or equivalently, rather than physical squares made from the sides of a triangle, rather a relationship between the square of the lengths of the three sides of the triangle.

So let's compare these two representations of Pythagoras' theorem.

The first one being the calculation of the areas of the squares attached to each of the three sides of a triangle.

This can be done either with or without the physical drawing of the squares themselves.

The second method is the algebraic method.

By substituting length of the triangle into the Pythagoras' theorem formula, a squared plus b squared equals c squared where a and b are the lengths of the two shortest sides of the triangle and c is the length of the hypotenuse of the triangle.

These two representations are equivalent and you'll spot the same values being used in the same way in both methods.

Let's see how.

Squaring the lengths of the two shorter sides of the triangle is the same as algebraically saying a squared and b squared.

So 18 squared and 30 squared in both representations gives 324 and 900.

Adding the results together is the same as adding a squared plus b squared together for a total of 1,224 centimetres squared in both representations.

This is both the square of the hypotenuse of the triangle and the value that equals c squared.

Since c always represents the length of the hypotenuse, c squared represents the area of the square from the hypotenuse, a big connection between these two methods.

To find the length of the hypotenuse, which is also the side length of the square attached to the hypotenuse, square root 1,224, giving approximately 34.

99 centimetres after rounding.

This is the same as square rooting both sides of the Pythagoras' theorem equation, which gives you the exact same answer of c equals 34.

99.

The hypotenuse c has a length of 34.

99 centimetres.

Okay, for this check, let's go through the algebraic method step by step.

Step one, fill in the blanks to set up a Pythagoras' theorem equation for this triangle.

Pause now to do this, what numbers go in which boxes? And the answer is 10 squared plus 15 squared equals c squared where c is the hypotenuse of that triangle.

Note that you could have put the 15 and the 10 the other way around 'cause they are both the shorter two sides of the right-angle triangle.

Okay, next up, what goes into these two blanks? Pause now to have a think.

What goes into those two blanks are 10 squared, which is 100, and 15 squared, which is 225.

Pause now to figure out what goes into this blank.

The answer is 100 plus 225, which is 325.

Last up, I've now got c rather than c squared.

What goes into that final blank? Pause now to have a think.

Because I've square rooted c squared, I also have to square root 325 and the square root of 325 is approximately 18.

03.

If you've got a number that is close to 18.

03 because of different rounding, that is also equally correct.

And therefore the hypotenuse of this right-angle triangle is about 18.

03 centimetres.

Sam looks at our methodology.

Whilst it is often helpful or expected to show your full algebraic method and also totally okay to still draw the squares from the triangle itself if it helps you, there is a more efficient way of using a calculator to find the hypotenuse rather than using a calculator to do each individual calculation one by one.

For this example, a and b, the two sides whose squares are added together, always represent the two shorter sides of a right-angle triangle.

They are 23 and 35 on this triangle.

Pythagoras' theorem shows that we must first square each side, then add the results together.

On a calculator, this looks like 23.

Then I press the squared button, that looks like this, plus 35, again, squared.

This reads 23 squared 35 squared and those results added together.

I then press the execute function to calculate this giving a result of 1,754.

This is the result that you get from squaring both the shorter sides and then adding the squares together.

This gives us the area of the square from the hypotenuse or c squared where c is the length of the hypotenuse itself.

To find the value of c rather than c squared, we press the square root button.

Then 1,754 and then execute the calculation.

This gives us 41.

88 when rounded.

Therefore 41.

88 centimetres is the length of that hypotenuse.

Whilst what Sophia says is correct, you do have to square the two short lengths of the triangle, what she has typed into the calculator is actually incorrect.

Can you figure out why? Pause now to have a quick look at what she's typed.

Laura correctly notices that only the 43 has been squared.

You also need to explicitly square the 41 as well.

Like so, notice how Sophia has used the squared button after both the 41 and the 43.

Sophia now asks if 3,530 centimetres is the true length of the hypotenuse.

This is not the case.

3,530 is the square of the hypotenuse, or algebraically, c squared.

You must, first of all, square root 3,530 to find the length of the hypotenuse.

And so she correctly types the square root of 3,530 into a calculator, giving her the correct answer.

Sophia's observation is also very clever.

59.

4 as a hypotenuse seems sensible as it is bigger than, but not ridiculously greater than, the other two side lengths of 41 and 43 centimetres.

3,530 would be far, far too big for it to be any sort of triangle alongside the other two sides of 41 and 43 centimetres as well.

Okay, for this check, whose calculation to find the hypotenuse of this right-angle triangle is correct? Pause now to make a decision between Lucas and Jun.

And the correct answer is Lucas, not Jun.

But what might Jun have done wrong in their method? Pause now to either think or discuss what he might have done wrong.

Jun has done the order of operations incorrectly.

Jun added together the two shorter sides before squaring where you should have squared the two shorter sides first and then added together the two resultant squares.

And next up, whose final calculation to find the hypotenuse of this triangle is correct? Pause now to make a decision between Izzy and Alex.

And Izzy's answer is correct.

Whilst Alex has chosen the correct number, his typed the number in before the square root.

Sometimes you may have to change the format of your answer so that the hypotenuse is a length given as a decimal.

So for this question, I correctly calculate the square of the two shorter sides added together at 648 centimetres squared.

Then I square root this to get the hypotenuse, but this is my result.

This is called a surd form.

We don't want to deal with that at the moment.

We want to change its format to decimal form.

To do this, find the format button, press down to highlight decimal, then press OK to activate that decimal form.

The decimal form of this hypotenuse is 25.

5 centimetres.

This process may be a little bit different for different types of calculator.

For some calculators, you might see a button with S arrow D on it.

Pressing this once also converts from surd form into decimal form.

For this last check, find the hypotenuse of this right-angle triangle, giving your answer in decimal form accurate to one decimal place.

Pause now to do this.

On my calculator, I would type in 20 squared plus 24 squared, then execute to get 976.

I then square root 976, oh, it will give me four, then square root 61.

This is the surd form, I don't want the surd form, so I convert it into decimal form to get a hypotenuse of 31.

2 centimetres.

Okay, onto the practise questions.

For question one and two, fill in all of the blanks.

And for question two, use what you filled in to show that the hypotenuse of this triangle is 17.

2 centimetres.

You may need to use a calculator to do this.

Pause now to do questions one and two.

And for question number three, each triangle matches to a value that represents the square of the hypotenuse, or c squared, and the length of the hypotenuse itself.

Pause now to match each triangle to two values underneath it.

And for question four, each of these triangles have two shorter sides that sum to 64 centimetres.

Starting with the shortest, put the hypotenuses of each of these four triangles in order of their lengths.

Pause now to do question four.

And on to the answers.

For question one, there are two valid answers.

As long as 61 squared is equal to, it does not matter whether you put the 60 squared or the 11 squared first.

And for question two, I take the two side lengths, 10 and 14, and put them into the blanks.

So 10 squared plus 14 squared, typing that into a calculator, I get 296.

But the question says show that the hypotenuse is 17.

2.

This is because I then have to square root 296 and convert it to decimal form to get 17.

20465, et cetera, which rounds to 17.

2.

And for question 3a, the square of the hypotenuse is 149 centimetres squared.

Square rooting that gives you the length of the hypotenuse of 12.

2.

For b, the square of the hypotenuse is 146 and so the hypotenuse itself is 12.

1.

And for c, the square of the hypotenuse is 173 and therefore the hypotenuse itself is 13.

2.

And for question four, oh, w has the shortest hypotenuse at 45.

25.

x is next up, a little bit longer at 45.

61.

Then y is next up at 47.

85.

And z is the longest hypotenuse at 54.

92.

The correct order was w, then x, then y, then z.

Oh, the order that I gave it to you in, why is that? What about those triangles made the order the way it is? It's because the bigger the difference between the two shorter side lengths, the larger the hypotenuse became.

So far, we've looked at all sorts of right-angle triangles.

But there's a subset of right-angle triangles that have a particular property, called the Pythagorean triples.

Let's have a look at what they are.

Whilst many if not most of the right-angle triangles that you will come across have at least one side length that is a decimal, there are some triangles with three sides that are all integers, or whole numbers.

These are called the Pythagorean triples, a trio of numbers, when used as lengths of a triangle, always make a right-angled triangle.

The most common of these is the 3, 4, 5 triangle.

Let's see why.

We can use the representation of Pythagoras' theorem, a squared plus b squared equals c squared.

I know that a and b are the lengths of the two shorter sides and I know that three and four are shorter than five.

And so three squared plus four squared equals c squared or the hypotenuse squared.

So nine plus 16 is three squared and four squared.

So I know that c squared equals 25.

25 then square rooted is five, giving me the third side of my triangle.

So side lengths three and four and the hypotenuse five centimetres are all integers.

And because they're all integers, we say that they are a Pythagorean triple.

To see if a right-angle triangle produces a Pythagorean triple, ask yourself, are both shorter sides integers? If so, then calculate the hypotenuse and see if it is also integer.

If it is, the triangle produces a Pythagorean triple of the three integer sides.

For example, I've got a triangle of nine and 40.

They are both the two shorter sides of the triangle and c is the hypotenuse.

By setting up an equation, nine squared plus 40 squared equals c squared, I can square both the nine and the 40, add them together, then square root to get an answer of exactly 41.

This triangle produces the Pythagorean triple, nine, 40, 41.

For this check, a right-angled triangle has two shorter side lengths of 39 and 80 centimetres.

Fill in the blanks of this method to find the hypotenuse of this right-angle triangle using Pythagoras' theorem.

I will pause after each stage of working.

Pause now for the first stage.

The two shorter side lengths, 39 and 80, go in the two blanks, they can occur in either order.

Pause now to think what goes into these two blanks.

The square of 39 is 1,521 and the square of 80 is 6,400.

Pause now to think what goes in this blank.

And the sum of 1,521 and 6,400 is 7,921.

Pause now to think what goes in this final blank.

And the square root of 7,921 is exactly 89.

Pause now to answer this question, does this triangle produce a Pythagorean triple? And the answer is yes.

It produces the Pythagorean triple 39, 80, and 89 because all three side lengths are whole numbers.

Okay, onto the final practise task.

For question one, use Pythagoras's theorem to figure out which of these triangles produce a Pythagorean triple.

Pause now to do this for all three triangles.

And for question two, triangles, A, B, and C all produce a Pythagorean triple.

Pause now to figure out what the hypotenuse is, making sure that they are integer in length and spot a pattern or relationship between each of the three triangles.

And then use this relationship to write down the two shorter side lengths for a triangle with a hypotenuse of 169 centimetres.

Oh, and you're not allowed to use a calculator for that part c.

Pause now to do question two.

Okay, here are the answers to question one.

The triangles g and j both produce Pythagorean triples because all of the side lengths are integers at 7, 24, 25, and 20, 21 and 29.

However, triangle h does not.

And for question two, the hypotenuse of triangle A is 13, triangle B is 39, and triangle C is 65.

The relationship is all three triangles are similar to each other.

For example, all three sides of triangle B are triple the length of triangle A.

I can use this relationship to figure out the two shorter side lengths of the triangle with a hypotenuse of 169 centimetres.

I know that 13 times 13 is 169, and so I can multiply all three side lengths of triangle A to produce a triangle whose hypotenuse is 169 centimetres.

Therefore the two other shorter sides are 65 and 156 centimetres long.

Amazing work everyone, great job with all of your effort in combining algebraic methods alongside the use of squares and triangles in a lesson where we've looked at several strategies at using Pythagoras' theorem, represented in different ways in order to calculate the hypotenuse of a right-angle triangle when given the length of its two shorter sides.

The first representation involved adding the areas of the two squares from the two shorter sides of a triangle to find the area of the square from its hypotenuse.

We then square rooted this area from the hypotenuse to find the length of the hypotenuse itself.

We've also looked at algebraic approaches using the common a squared plus b squared equals c squared representation to help us, and become more efficient at Pythagoras' theorem when using a calculator.

We've also briefly looked at Pythagorean triples, special right-angle triangles whose side lengths are all integer.

Once again, I appreciate you choosing to join me here for today's lesson.

And until our next time together, take care and have a wonderful rest of your day.