Lesson video

Hello and welcome.

Thank you for making the choice to join me, Mr Gratton, for today's lesson.

Together we'll be looking at the use of Pythagoras' theorem when finding one of the shorter sides of a right-angled triangle and how we can use this information in a range of contexts.

Pause here to have a quick look at some of the keywords and ideas that we'll be using today.

First, let's check the relationship between the sides of a right-angled triangle and the squares that you can form from those sides.

Andeep remembers that it is possible to draw squares from the three sides of a right-angled triangle and finding the area of each square is possible by squaring each length.

So for example, six squared is 36, 8 squared is 64, but that largest square, the one formed from the hypotenuse of the triangle has an area equal to the sum of the two smaller squares.

So 36 plus 64 equals 100, and so the area of that square is 100 centimetres squared.

Andeep's final piece of advice is that you can square root the area of a square to calculate the length of its side.

And so, for the largest square, square rooting 100 gives 10.

The length of the hypotenuse is therefore 10 centimetres.

Sam then queries whether it is possible to calculate the length of side labelled a using the same method.

Pause here to think about or discuss how the side labelled h centimetres on the left is different from the side labelled a centimetres on the right.

The side labelled h centimetres is the hypotenuse.

We know this because it is opposite the right angle, and so the area of the square formed from the hypotenuse is the sum or addition of the two other squares.

81 plus 144 is 225 centimetres squared, and as always, the square root of the area of the square gives you the length of one side of that square.

And so, the square root of 225 is 15.

15 centimetres is the length of that hypotenuse.

However, here on this diagram, we know the length of the hypotenuse of that triangle is 13 centimetres.

The area of the square formed from it is 13 squared or 169 centimetres squared.

The side labelled a centimetres is not the hypotenuse.

It is one of the two shorter sides.

Because we know the area of the square from the hypotenuse is the sum of the other two areas, we have 169, the area of the hypotenuse, is equal to the area of that bottom square, 144, plus the area of square A, which we do not currently know.

How can we find the area of square A? We don't know the length of its side and the square isn't the hypotenuse, so we can't add together the areas of the other two squares, but what we can do is solve the equation that we just wrote.

How can we do this? Well, we can subtract 144 from both sides to make the area of square A equal to 25.

Rather than adding when finding the area of the hypotenuse, the area of one of the smaller squares is found from subtracting or finding the difference between the area of the other two squares, the area of the largest square from the hypotenuse and the area of the other smaller square.

And as always, the length of the side of a square is found by square rooting the area of that square.

So the square root of 25 is five, five centimetres being the length of one side of that square and also the shorter side of that triangle.

Right.

Let's see if you can apply this information in this check.

Which of these calculations is correct, when trying to find the area of that smaller square B? Pause now to look through all of these options.

The correct answer is 1,681 take away 1,600.

But why? Which of these explanations is correct? Pause to choose the correct one.

We have a subtraction, because square B is the square from one of the shorter sides of the triangle, not from the hypotenuse.

And now we know the calculation to find the area of square B, how can we find the length of one of its sides, b centimetres? Pause to follow through the two correct calculations.

We know from the previous check that the correct area is 1,681 takeaway 1,600, which is 81.

The area of square B is 81 centimetres squared.

To find the length of square B, I square root that area, so square root of 81 is nine, and so the length b is nine centimetres long.

It is the length of the shortest side of that triangle.

For this check, in the order a, then b, then c, match the value to its missing side length or area.

Pause now to look through all three parts of this question.

For a, 16 squared is 256.

For b, since it is a smaller triangle, since the square root is not opposite the right angle, I subtract the other two areas.

So 256 that we got from part a takeaway 60 equals 196.

For c, I then square root that same area, and so the square root of 196 is 14.

Brilliant.

Onto the practise.

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

Pause now to do this.

And for question two, by drawing on any missing squares first and finding the areas of any square that doesn't have its area already labelled, find the length of each side that is labelled with a letter.

Pause now to do all of this for question two.

And here are the answers for question one.

Pause now to compare your answers to the ones that are on screen.

And similar again for question two.

Pause here to match your answers with these ones.

So now we're familiar with the relationship between the areas of squares made from the sides of a right-angled triangle, let's compare this method to Pythagoras' theorem.

Since c squared, the square of the hypotenuse, is the subject of the Pythagoras' theorem formula, we know we can use Pythagoras' theorem to find the length of the hypotenuse.

But can we use the theorem to also find the length of a shorter side of a triangle rather than the hypotenuse? Laura's attitude is great.

Let's give it a go by first substituting in the known lengths of this triangle into Pythagoras' theorem formula and see if we can figure out anything from there.

But before we do, we need to be careful not to fall into the same trap that Sam has.

Sam says the length x should be substituted into the variable c on the right hand side of the Pythagoras' theorem formula.

This is definitely incorrect.

C squared, the subject of the formula, represents the square of the hypotenuse, since c always represents the hypotenuse itself.

We can see that, in this triangle, 23 centimetres is the length of the hypotenuse, not x centimetres.

In this conventional representation of Pythagoras' theorem, c always represents the hypotenuse, whether it is a known length or not, whether it is represented by a number or a letter.

And so, now that we know that 23 centimetres is the length of the hypotenuse, I can start the equation by squaring the two shorter sides, which are nine squared and x squared, and this equals the square of the hypotenuse, 23 squared in this case.

So after making those substitutions into the Pythagoras' theorem formula, I evaluate any squares of numerical length.

So nine squared is 81 and 23 squared is 529.

Since x is an unknown length for now, x squared remains x squared.

We can then rearrange this equation to solve for x.

Take the 81 and subtract it from both sides.

Now we are back to this familiar step from the previous cycle.

The square of a shorter side is found by subtracting or finding the difference between the square of the hypotenuse and the square of the other shorter side.

Now we can perform the subtraction 529 takeaway 81 equals 448, and the square root of 448 is 21.

17.

So 21.

17 centimetres is the length of that shorter side of the triangle.

Okay, let's check.

These four students substitute the lengths of this triangle into the Pythagoras' theorem formula.

Pause here to figure out who has done so correctly and more than one person may have done so.

Both Sofia and Alex have done so correctly.

This is because they have both identified that 18 centimetres is the length of the hypotenuse, and so, in their equation, they've got 18 squared on one side of the equation, with r squared and 10 squared on the other side of the equation.

For this next check, which of these are the correct calculations for and value of p squared? Pause now to look through all of these options.

Because p is a shorter side of that triangle, I have a subtraction between the squares of the other two sides, so 24 squared take away 15 squared.

24 squared is 576, whilst 15 squared is 225.

That is why b and d are correct.

I then perform the subtraction.

576 take away 225 gives you 351.

However, 351 is not the length of p.

It is still the square of it.

To find the length of p itself, you must then square root 351.

Here, Lucas claims that the length of side labelled k is around 36.

67.

Which of these statements help explain why Lucas' statement is incorrect? Pause now to look through all three options.

Both b and c are the correct answers.

This is because, in this triangle, 33 centimetres is the length of the hypotenuse, and so the length of any of the other sides must be shorter than 33.

Now, k, he says, is 36.

67, which is longer than the hypotenuse and so is incorrect.

This might be because Lucas did not subtract the square of the other shorter side from the square of the hypotenuse itself, rather he added them by mistake.

This following conversation will highlight why you must be flexible with the conventional representation of Pythagoras' theorem, a squared plus b squared equals c squared, and understand what the theorem is trying to achieve, rather than as a number and letter matching or substituting exercise.

Andeep is certain that 16 squared plus 14 squared equals c squared is the correct equation for Pythagoras' theorem for this triangle.

His logic is, because c represents the length of one of the sides of that triangle, that one near the top, surely this matches the c where the c squared is in the Pythagoras' theorem formula.

Confusingly, Laura is actually correct and not Andeep.

The c in the triangle and the c in the theorem are different.

The c in the Pythagoras' theorem formula always represents a hypotenuse, regardless of what is labelled on any triangle.

No matter whether there is a length labelled c on the triangle or not, the priority is to always substitute the length of the hypotenuse into the location of c in the conventional representation of the Pythagoras' theorem formula.

Andeep is correct.

It is messy and easy to make a mistake when there is both a c in a triangle and using the c in the a squared plus b squared equals c squared representation of Pythagoras' theorem.

And so, whilst it's not as neat and tidy as the a squared plus b squared equals c squared representation, this way of representing the formula is correct.

Shorter side squared plus the other shorter side squared equals the hypotenuse squared.

This is an alternative, but equally correct way of representing Pythagoras' theorem that does not lead to any misconceptions or mistakes when a triangle is labelled in a way that could trick you.

Using this representation, c is a shorter side.

This is correct.

14 is the other shorter side and 16 has been correctly identified as the hypotenuse.

We can use a calculator to find the length of a shorter side in a more efficient way, rather than using a calculator to do each individual calculation one by one.

This involves a little bit of algebraic rearrangement first.

So let's see how.

For this triangle, 68 is the hypotenuse, whilst t and 39 are the other two shorter sides.

Rather than using a calculator to find the values of 39 squared and 68 squared separately, we can simply subtract 39 squared itself from both sides to get the expression 68 squared minus 39 squared in an equation that equals t squared.

And so, 68 squared take away 39 squared can be typed into the calculator in its entirety.

Pressing the execute button then gives 3,103, which is the value of t squared.

To find t, I then square root 3,103, and then execute to get 55.

7.

55.

7 centimetres is the length of the shorter side labelled t.

For this check, a triangle has shorter sides with length 28 and w centimetres.

Its hypotenuse is 60 centimetres.

Pause here to figure out whose calculated display shows a correct step in the calculation of w.

There may be more than one correct answer.

Jun shows the correct first calculation.

60 squared is the square of the hypotenuse take away 28 squared, the square of one of the other shorter sides.

Sofia, on the other hand, shows the correct final calculation, the square rooting of the result that Jun himself got.

Sofia can convert her answer into decimal form by changing the format of her answer using the format button.

So w in this instance equals 53.

07 centimetres when rounded.

Okay, onto the practise.

For question one, by first of all identifying the hypotenuse of this triangle, fill in all the blanks to complete the method to calculate the length of side e.

Pause now to do this question.

For question two, which of these equations are correct for this triangle when trying to find the length of side c? For questions three and four, show full method using Pythagoras' theorem to find the length of w and to show that the perimeter of the triangle in question four is 84 centimetres.

Pause to do these three questions.

And for question five, pause here to match each triangle to a correct calculation and the correct answer, the length of each missing side.

And here are the answers.

Pause now to check your answers for question one and know that, for the first and second lines of working, it is okay if you wrote e and five and e squared and 25 the other way around.

For question two, b, c, and d were all correct.

And for question number three, w equals 58.

52 centimetres and the perimeter is 84 centimetres, because the length of that shorter side is 35 centimetres.

For triangle a of question five, g was the correct calculation and i is the correct answer.

For triangle b, d and h were correct.

And for triangle c, e was the correct calculation.

F was the square of the correct answer, which was j.

So far, we've used Pythagoras' theorem to find the length of a side, but in doing so, we can also find other properties of a triangle, such as its area.

Let's see how.

Here, Andeep makes a correct observation.

The area of this right-angled triangle can be found, because its base and height are known.

The area of a triangle is the base times its perpendicular height divided by two.

The 15 and eight are perpendicular to each other, so the formula can be followed simply, and then divided by two to get 60 centimetres squared.

However, Andeep is concerned that it isn't possible to find the area of this triangle, since 35 and 12 are not perpendicular lengths.

Andeep is correct that 35 and 12 are not the values used when finding the area of this triangle, but Laura is correct.

It is possible to find its area.

You just need to use Pythagoras' theorem first in order to find the second perpendicular side length.

If the length of a side needs to be found, but isn't labelled, you can label it with a letter yourself.

It does not matter which letter you choose, as long as it doesn't get confused with any other length which is also labelled with the same letter.

For this question, let's call that base side length b centimetres.

We can use Pythagoras' theorem to get 12 squared plus b squared equals 35 squared, which can be rearranged to get b squared equals 1,081, and square rooted to get b equals 32.

87856.

In the middle of a calculation, you must be as precise as possible and make sure to write down any values on a calculator to several decimal places to avoid any rounding errors that you may see later on in your calculation.

And since we now know both the base of 32.

87856 centimetres and the perpendicular height of 12 centimetres, we can find the area of the triangle.

So we can substitute the base, times 12, and don't forget the divide by two.

The final answer can then be rounded to a more reasonable number of decimal places, at 197.

3 centimetres, if rounded to one decimal place.

For this check, which two sides do you need to know the length of in order to find the area of a triangle without needing to use Pythagoras' theorem? Pause now to have a think.

And the answers are the sides labelled a and c, because they are both perpendicular to each other, as shown by the right angle between them.

Currently, we cannot calculate the area of this triangle.

Which side do we need to find the length of first? Pause now to consider.

We know the base and the hypotenuse.

We do not know the perpendicular height.

Now we know that we need to find that perpendicular height, let's label it, so that we can use Pythagoras' theorem on it more easily.

Pause here to find the value of the perpendicular height p.

And the answer is 36.

This is done by 39 squared take away 15 squared, and the answer to that subtraction is then square rooted to get 36.

Now we know the appropriate side lengths, pause here to find the area of that triangle.

The base of 15 times by the perpendicular height of 36, then divided by two is 270 centimetres squared.

For this single practise task, pause here to find the area of each triangle.

You will only need to use Pythagoras' theorem for some of these triangles.

To find the area of a, you could just do nine times 20, the base times the perpendicular height, divided by two to get 90 centimetres squared.

For b, you did have to use Pythagoras' theorem to find the other side length of about 17.

86057 centimetres.

The area of this triangle is 80.

4 centimetres when rounded.

For triangle c, both the base and the perpendicular height are 18.

So 18 times 18 divided by two is 162 centimetres squared.

And for d, you also had to use Pythagoras' theorem to find that shorter length of about 13.

41640 centimetres.

The area of this triangle is therefore 80.

5 centimetres squared when rounded.

Whilst Pythagoras' theorem is a theorem specifically related to right-angled triangles, let's have a quick look at how we can work with other types of triangles in a way where Pythagoras' theorem can also be used.

Sam is correct.

Most of the time, Pythagoras' theorem can only be used when two of the three sides of a right-angled triangle are known.

But in this specific example, we have an isosceles right-angled triangle with only one side length known, the hypotenuse.

Let's see if it is still possible to find the length of the other two sides.

Since both of those two hash marks show that both of the shorter sides are of equal length, we can label both the base and the perpendicular height of this triangle as x centimetres.

So our Pythagoras' theorem equation will look like this, x squared, with this x representing, for example, the base, plus x squared again, with that x representing the perpendicular height, equaling 200 squared, 200 being the hypotenuse.

We can collect the like terms to get 2x squared and evaluate 200 squared to get 40,000.

We can divide both sides by two to get one lot of x squared equals 20,000, and therefore we can square root 20,000 to get 141.

42.

Both shorter sides of this isosceles triangle, both the base and the perpendicular height, have a length of 141.

42 centimetres.

For this check, there will be several steps.

After each step, choose the correct option in each pair.

Pause here to choose which of a and b are correct for this first step.

A is correct because those hash marks show that j is both that left-hand and right-hand shorter side of that triangle.

Pause again for this second step.

D is correct.

I can collect the like terms of j squared and j squared.

I now have two lots of j squared.

The 56 squared has not changed.

Pause here for step three.

Option f is correct.

I've divided both sides by two.

So the times by two and divide by two on the left-hand side of the equation have cancelled out, and I now have a divide by two after the 56 squared.

And pause here to have a look at step four.

And the answer is g.

This is because 56 squared is 3,136, and then I half that, because of the divide by two, to get 1,568.

And for the last time, pause here to find the final answer, the length of j.

And the correct answer is i.

This is because we round to 39.

60.

39.

59 is an incorrectly rounded answer.

Is Pythagoras' theorem useful for any non right-angled triangle? Well, actually yes.

For example, this right-angled triangle can be reflected with the line of reflection at one of its two shorter sides to create an isosceles triangle.

If given this isosceles triangle, we can split it into two congruent right-angled triangles by drawing a line segment perpendicular to its base, which creates a right angle that also intersects the opposite vertex of the right-angled triangle.

Because these two are congruent right-angled triangles, any calculation we find for one triangle can be applied to the other.

So let's ignore one of them.

The horizontal side length of this triangle is half the base length of the original isosceles triangle.

From there, we can treat this half triangle, the right-angled one, like we've done with any other right-angled triangle.

We can calculate the length of the missing shorter side, now labelled w, but you can pick any letter, of this right-angled triangle using Pythagoras' theorem.

21 squared plus w squared equals the hypotenuse, 30 squared.

We can then solve this to get w equals the square root of 459, which is 21.

42.

So that perpendicular height of this right-angled triangle is 21.

42 centimetres.

However, we're not dealing with a right-angled triangle.

We're dealing with this isosceles triangle.

Well, the height of this isosceles triangle is also 21.

42 centimetres.

For this final check about this isosceles triangle, match the letters to their value.

Pause now to do this.

The base length of that right-angled triangle is half the base length of the isosceles triangle.

Half of 24 is 12.

I can then use Pythagoras' theorem to find that the height of that isosceles triangle, the height of that right-angled triangle as well, is 29.

7 centimetres.

And for this final practise task, pause here to answer all four questions.

Here are the answers.

For question one, both of these triangles are isosceles, and so I have two shorter side lengths that are the same, x and x for that first triangle, and y and y for that second triangle.

Using that within Pythagoras' theorem, I get x equals 42.

4 centimetres and y equals 14.

8 centimetres.

For question two, by splitting this isosceles triangle into two right-angled triangles with a base of 22 centimetres each, I can calculate that the height is 82.

1 centimetres.

For this isosceles right-angled triangle, I can use Pythagoras' theorem to find that the length of one of the shorter sides is 76.

37 centimetres.

Therefore, I've got two sides of length 76.

37 and one length of 108 centimetres, giving a total perimeter of 260.

735 centimetres when rounded.

And finally, for question four, I can split this isosceles triangle up into two congruent right-angled triangles with a base length of 10 centimetres.

By using Pythagoras' theorem, I can calculate that the height of the isosceles triangle is 12.

4899 centimetres and 12.

4899 times by the base length of 20 centimetres, then divided by two gives you a total area of 124.

9 centimetres squared.

And well done.

That is some amazing work.

Thank you so much for your effort in this varied lesson, where you've had to adapt Pythagoras' theorem to a range of different triangle properties, such as finding the area of a smaller square from a right-angled triangle by looking at the difference in size of the areas of the other smaller square and the largest square, the one from the hypotenuse of a triangle.

We've also used Pythagoras' theorem in a more algebraic form used to find the shorter side from first rearranging an equation.

We've also found the areas of triangles using Pythagoras' theorem, which can be used to find the length of the base or perpendicular height of a right-angled triangle.

And lastly, we've looked at Pythagoras' theorem in different ways across different types of isosceles triangles.

Once again, thank you for joining me today in today's learning.

I wish you all the best for the rest of the Pythagoras' theorem topic and beyond.

I've been Mr Grattan and until next time, take care and goodbye.