Lesson video

Hello everyone and welcome to another maths lesson with me, Mr. Gratton.

It is great to see you joining me here for this lesson where we will be using Pythagoras's theorem across a range of right-angle triangles where we have to first figure out whether the hypotenuse or a shorter side has a missing length.

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

The most important thing to look at is whether the missing side is the hypotenuse or not.

Let's look at strategies for identifying those sides.

Both Sam and Andeep are in agreement.

It is possible to use Pythagoras' theorem for this triangle because it is clearly a right angle triangle.

However, here, Sam says you can use Pythagoras' theorem because it is a right angle triangle whilst Andeep is not so certain because a right angle marker isn't clearly labelled.

Whose statement is correct? Pause now to think about or discuss this.

Sam is correct.

Pythagoras's theorem can be used to find the length of a missing side of any right angle triangle.

It does not matter whether a right angle marker is shown or not, as long as you can clearly tell or figure out that the triangle is definitely right-angled.

If a right angle marker isn't clearly shown, then it is helpful to find the right angle and identify which side of the triangle is the hypotenuse.

Because we know for certain, the angles in a triangle sum to 180 degrees, we can find the size of this angle here.

180 take away 28 degrees take away 62 degrees equals that's 90 degree right angle.

So this angle is a right angle and therefore, this is a right-angled triangle.

Also, the side opposite the right angle is always the hypotenuse.

And now that we know for certain that this is a right-angled triangle, we can use Pythagoras' theorem to calculate the length of the missing side.

This time, the length of the hypotenuse.

The hypotenuse squared equals 15 squared plus eight squared, where 15 and eight are the two lengths of the shorter sides.

We can then type 15 squared plus eight squared into a calculator to get to 289 and the square root of 289 is 17.

So the hypotenuse is 17 centimetres.

For this check, which of these four triangles are right-angled triangles? Pause here to inspect and calculate all interior interior angles of each triangle.

Both A and C are right-angled triangles.

Here's a tip that you might have noticed.

If you add together the two non-right-angled angles, if that sum equals 90 degrees, then you know for certain that there is 90 degrees left over for the one remaining angle.

That's 90 degree right angle.

For example, for triangle A.

Nine plus 81 is 90 and so there is also 90 left over for the right angle.

However, for triangle B.

48 plus 47 is 95, meaning there is less than 90 degrees left over, meaning that final angle is definitely not right-angled.

And so, for these two triangles that we now know are definitely right-angled, which side is the hypotenuse? Pause to inspect both of these triangles.

Both C and E are the hypotenuses.

The hypotenuse is always the side opposite the right angle.

This means it is the one and only side of the triangle that isn't a leg of the 90 degree angle.

That is to say it is the only side that isn't perpendicular to another side of the triangle.

Sam wonders why we can't use Pythagoras' theorem for triangles like this that are almost right-angled but not quite where in this case the angle is 91 degrees, not 90 degrees.

Laura correctly remarks that Pythagoras' theorem explicitly needs a hypotenuse and there isn't one if it is not a right-angled triangle.

For triangles that are nearly right-angled, incorrectly using Pythagoras' theorem may get you an answer that is close to the correct one, but it will never be precisely correct.

We can see this visually here for this almost right angle triangle.

Let's draw on the three squares from the length of the three sides.

The areas of the two smaller squares are 1,296 and 1,156 centimetre squared.

Furthermore, the area of the largest square is 50 squared, which is 2,500.

However, the sum of the areas of the two smallest squares is 2,452, but we just said that the area of the largest square is 2,500, but since 2,452 obviously isn't 2,500, this is not a right-angled triangle and so Pythagoras' theorem just simply does not work for it.

As a consequence, there is no hypotenuse on this triangle.

And so, we know that Pythagoras' theorem only works if we have a right-angled triangle.

In which of these triangles is Pythagoras' theorem true? Pause now to observe all six triangles.

And Pythagoras' theorem works for these triangles.

A is clearly a right angle triangle because it has a right angle marker.

B is a right-angled triangle because it has a hypotenuse.

For C, the area of the two smallest squares equals the area of the largest square.

And for F, well, this is a bit more tricky.

45 degrees is one of the two base angles, but base angles in an isosceles triangle are both equal and so 45 degrees is also on the other side as the other angle in this triangle.

Since 45 plus 45 is 90, there is also 90 degrees remaining for the right angle in that triangle.

And okay onto the practise questions.

For question one, complete each statement and for question two, write a sentence or two explaining why Pythagoras' theorem cannot be used to find the value K in this triangle.

Pause now to do both of these questions.

For question number three, tick each of these triangles that are right-angled.

Cross each triangle that is definitely not right-angled and put a question mark by each one where you simply cannot tell.

For each right angle triangle, circle or highlight its hypotenuse.

Pause now to do this.

And for question four, which of these triangles can you definitely use Pythagoras' theorem.

Where possible then calculate the length of the hypotenuse and label that length.

Pause here to look at all of the details of each of the three triangles and the squares that are created from the sides of each triangle.

The answers for question one.

Pythagoras' theorem can only be used to calculate the length of a missing side of a triangle if one of its sides is a hypotenuse, and only right angle triangles even have a hypotenuse.

The hypotenuse of a right-angled triangle is always opposite the right angle.

For question two, the missing angle is 89 degrees, not a 90 degree right angle and you cannot use Pythagoras' theorem on triangles that are not right-angled.

For question three, only A and D are definitely right angle triangles and I've highlighted the hypotenuses for you.

B has very little information in it and so we cannot tell if it is right-angled or not.

C is also not very labelled, however, we know that one angle is 92 degrees leaving 88 degrees for the other two angles.

88 degrees isn't enough to make a right angle and so we definitely know that is not a right-angled triangle.

For question four A, you definitely cannot use Pythagoras's theorem because the areas of the two smaller squares does not add up to the area of the largest square.

However, B and C, you can use Pythagoras's theorem because they are both clearly labelled as right-angled triangles.

Let's look at how congruence or incongruence of two right-angled triangles will influence an equation we can write from Pythagoras' theorem.

Andeep claims that these two triangles are congruent because they both have sides of 10 centimetres and 14 centimetres.

Is his statement correct? Pause here to think about or discuss what he said.

Laura thinks not.

This is because on triangle B, 14 centimetres is the known length of the hypotenuse.

Whilst on triangle A, the hypotonus is unknown and the 14 centimetres is instead the length of one of the shorter sides.

Whilst Andeep agrees, he does not understand how this affects their congruence.

The hypotenuse is always the longest side of a right-angled triangle.

Therefore, this hypotenuse, X must be greater than 14 centimetres, greater than both the other two shorter sides.

Furthermore, this shorter side Y must be less than 14 centimetres since this side must be shorter than the hypotenuse of 14 centimetres.

Since X must be greater than 14 and Y must be less than 14, they are definitely different sizes.

X definitely does not equal Y.

Therefore, these two right angle triangles are not congruent as they do not satisfy the SSS rule for congruence.

Triangles can only be congruent if all three of their sides are of equal length.

Let's see if Laura's query is correct.

Will the fact that these two triangles aren't congruent even though they have two sides of the same length affect how equations formed from Pythagoras' theorem will look when compared to each other.

Let's focus on the hypotenuse first.

Since X is the hypotenuse, it's square is equal to the square of the other two sides.

Given the equation, 10 squared plus 14 squared equals X squared.

Whilst for this triangle, the hypotenuse is known at 14, giving us the equation Y squared plus 10 squared equals 14 squared.

The equation formed from Pythagoras' theorem is different for each triangle.

As for triangle A, X squared is equal to the other two squares.

Whilst the triangle B, it is 14 squared that is equal to the other two squares.

And for this check, pause here to match the triangle to the equation formed to find the length of its missing side.

For triangle A, equation E is correct.

We do not know the length of the hypotenuse of triangle A and so that unknown letter, X in this case, representing the hypotenuse is correct.

Furthermore, 18 and 11 are the two shorter sides and so the sum of the squares of the two shorter sides is 11 squared plus 18 squared.

Conversely, for triangle B, 18 is the hypotenuse and so 18 squared being on its own on one side of the equation makes sense with the two shorter sides, 11 and the unknown length, which we can call X being 11 squared plus X squared.

And Andeep takes what we've just learned and tried to apply it to these two triangles.

He claims these two triangles are not congruent since their two equations formed from Pythagoras' theorem look different.

However, Sam observed that the bottom triangle is just a rotation of the top one and congruency is preserved when an object is rotated.

In more detail, both triangles are right-angled, have a hypotenuse that is exactly the same length and have one shorter side that is also exactly the same length.

These two triangles are definitely still congruent because they satisfy the RHS rule for congruence.

The difference between this example and our previous one is again down to the hypotenuse.

In both of these triangles, the hypotenuse is the same length at 60 centimetres even if the orientation of the hypotenuse has changed with the triangle.

Both equations to find the length of the missing side of each triangle are the same even though they look like they've been written differently.

Notice how the only real difference here is that 41 squared and X squared have been written the other way around.

Addition is commutable, meaning the order of these two squares when they are added together can be changed without changing its value or what the equation itself represents.

Let's compare this to a more simple example.

For example, is five plus seven any different to seven plus five? Of course not.

They both equal 12.

This same commutativity exists for the sum of two squares.

As usual, the focus is also on the hypotenuse.

The order of the squares of the two shorter sides can be rearranged as long as the hypotenuse of 60 squared remains the same at 60 squared on the opposite side of the equation to the sum of the two smallest squares.

Sam's query is fair.

There are other ways of writing the same equation from Pythagoras' theorem.

All of these are equivalent.

This is because they all say at some point, 60 squared equals and then 41 squared and X squared added together on the opposite side of the equation to the 60 squared.

And you can even rearrange terms as you start to solve the equation.

The equation still correctly relates to the same original triangle.

For this check in two parts.

Pause here to find the length of the hypotenuse of this triangle.

And the answer is 100 centimetres.

That side with the length 100 centimetres is opposite the right angle.

And so, now that we know that the hypotenuse is 100 centimetres, which of these equations are correct for this triangle? Pause now to look through all four equations.

A and D are correct.

In each of these two equations, the P squared and the 52 squared have commuted or swapped around.

This is okay because they are both the two shorter sides whose squares are added together.

And the sum of these two smaller squares still equals the square of the hypotenuse regardless of whether this 100 squared is a term on its own on the left hand side of the equation or the right hand side of the equation.

And for this final check, these four equations aim to find the length of the missing side of four triangles, three of which are congruent with each other.

Pause now to consider which three equations belong to the three congruent triangles.

And the three congruent triangles are A, B and D.

This is because 24 squared is a term on its own side of the equation regardless of whether it is the left or the right side representing a hypotenuse of length 24.

15 squared and the A, B or D squared can commute.

The triangle formed from B is not congruent to the others as the hypotenuse in this triangle has length C rather than the known length of 24 like the other three.

And for this one question practise task, match each equation to the correct triangle.

Pause now to do this.

For triangle A, A and D are the correct equations.

For triangle B, C, E, and F are the correct equations.

However, B and G do not apply to any triangle at all.

Now that we've looked rigorously at congruent and different triangles, let's see if we can find the lengths of missing sides of all different types of right angle triangles.

Let's go.

Andeep now correctly notices that these two triangles are different.

This is because the top triangle has a hypotenuse of 24 centimetres.

Whilst we do not know the hypotenuse of the other, Lucas agrees as the 24 centimetres is opposite the right angle for the top triangle whilst it is adjacent to the right angle on the bottom one.

But why does this matter? Because as we discovered in the previous cycle, the equations formed from Pythagoras' theorem will change for non-congruent triangles even if they have some sides of equal length.

Therefore, as we solve these equations, the solutions may be different.

And so, define the length of the shorter side of this top triangle, I used Pythagoras' theorem to start with X squared plus nine squared equals 24 squared and solve that to get 22.

5 centimetres as the length of the shorter side.

Whilst on this bottom triangle, I use Pythagoras' theorem to find the length of the hypotenuse this time with 24 squared plus nine squared equals Y squared and I can solve this as well to get a hypotenuse with a length of 25.

63 centimetres.

Notice how the two solutions are different to each other even though the two original lengths were both 24 centimetres and nine centimetres.

For this first part of this check, which side A, B, or C is the hypotenuse of this triangle? Pause now to consider where the hypotenuse is in relation to the right angle.

C is the hypotenuse as it is opposite the unlabeled right angle.

Next up, now that we know the hypotenuse, pause here to consider which of these equations is correct for this triangle.

Actually, there are two correct answers, both D and F are correct since they both show R to be the length of the hypotenuse and 30 and 31 to be the length of the two of the sides.

And lastly, pause here to calculate the length of the hypotenuse of this triangle rounded to one decimal place.

And by typing in on my calculator, 31 squared plus 30 squared, I get R squared equals 1,861 and then by square routing that number, I get 43.

1 rounded to one decimal place.

And so, the length of the hypotenuse of that triangle is 43.

1 centimetres.

Pythagoras' theorem can be used multiple times on composite shapes made from many right angle triangles such as this shape made from one right angle triangle and another stacked upon the hypotenuse of the first.

Our goal is to use Pythagoras' theorem to calculate a side that is shared by both of these triangles.

Lucas is correct.

There is quite a lot to look at here.

One triangle is most certainly better to focus on than the other.

Let's see which one and why.

And so, if we try to start with triangle A, we only have one known side length, the 13 centimetres.

This is very unhelpful.

We cannot do much at all with only one side length.

We cannot use Pythagoras' theorem to calculate the length of that unknown side labelled W centimetres as we usually need two known side lengths to use Pythagoras' theorem.

However, we do know the two side lengths of triangle B, the hypotenuse of 26 centimetres and one of the shorter sides at 15 centimetres.

Therefore, we can use Pythagoras' theorem to calculate the length of the third side of triangle B.

Let's have a look at these calculations.

For the moment, we can ignore triangle A completely and only look at triangle B.

If we want to find the length of an unlabeled side, we can label it with whatever letter we want as long as it is an unused letter elsewhere in this equation.

So I've chosen the letter X for the shorter side, but you can choose any other letter except for W, which is a length on triangle A.

Using Pythagoras' theorem we can form the equation, X squared plus 15 squared equals the hypotenuse squared of 26 squared, which can be used to get a value of X equals 21.

23676.

This is the length of that shorter side currently labelled X centimetres.

As we are only partway through a longer question, it is sensible to keep the precision on any long decimal rather than rounding it as we may need to use this more precise value later on in our calculations.

That side length 21.

23676 is a shared side that is a shorter side of triangle B and the hypotenuse of triangle A.

So let's go back and check triangle A.

Ah, we now know a second side length and so we can now use Pythagoras' theorem to calculate that third side labelled W centimetres.

We could not do this before because we did not know the length of that side, the hypotenuse now labelled 21.

23676 centimetres.

The equation we can form is W squared plus 13 squared equals the hypotenuse squared of 21.

23676 squared.

Notice how this value is the hypotenuse this time rather than the shorter side of triangle B.

We can then solve to get the W equals 16.

79.

And so, the length of that shorter side W is 16.

79 centimetres.

Whilst this method works, it is long-winded and really messy with all of those long decimals.

It certainly was tricky for me to say all of them correctly.

And there is a way to avoid them.

Here's an equation partway through our previous calculation.

Rather than converting the square root of 451 into a decimal, we can keep it in third form.

This is because the square root of 451 squared.

Oh, it's just 451, which is a lot less messy than writing down all of those decimal places and it's a lot more useful later on in our method.

Oh, and furthermore, it's actually more accurate as well as our calculator still had to round to get that long messy decimal that we dealt with before.

As squaring and square rooting are inverse operations, the square root of 451 then squared is just 451.

The two operations cancel each other out.

And from this line of working, our method is functionally the same as before, just a lot neater and it shows as we get exactly the same answer as before with a length of 16.

79 centimetres.

For this final check, which of these are the correct lengths of the shared side labelled case centimetres? Pause now to consider which triangle is most important to focus on and therefore which lengths must be used in the first set of calculations.

And the answers are B, D and E.

This is because I focus on the top triangle because I know two of the three sides, the hypotenuse at 39 centimetres and one of the shorter sides at 20 centimetres.

Onto the final practise task.

For question one, match the triangle to the length of its missing side.

Pause now to match all six.

And for question two, focusing first on triangle X, which of these seven other triangles are congruent to triangle X? Note, you may have to use Pythagoras' theorem to find whether some of them are or are not congruent to triangle X.

Pause now to explore all seven triangles.

And for question three, I have three sets of composite shapes.

Calculate the length of each unknown side that is labelled with a letter rounding your answer to two decimal places.

Pause here to give this really challenging set of questions a go.

And here are the answers.

Triangle A matches with I, B with K, C with G, D with H, E with L and J with F.

Well done if you got all six.

And for question two, only triangles A, E, and F are congruent to triangle X.

And for question three, A was 32.

39 centimetres, B was 17.

66 centimetres, but C was exactly five centimetres, no rounding necessary.

That is of course, if you used the third form rather than the decimal form of your calculations partway through that question.

Thank you all so much for your flexibility in dealing with a diverse and increasingly tricky set of questions in a lesson where we have identified the hypotenuse from angles in a right angle triangle that aren't the 90 degree right angle itself.

We've also shown that addition is commutative, meaning that you can write down the sum of the squares of the two shorter sides of a right angle triangle in either order in an equation constructed from Pythagoras' theorem.

We've also used Pythagoras' theorem multiple times in a row in composite shapes with right-angled triangles as their component parts.

Once again, I appreciate your presence here for this lesson.

I hope you are all well and until our next maths lesson together, have a great day.