Lesson video

Hello, I'm Mrs. Lashley, and I'm gonna be working with you as we go through the lesson today.

I really hope you're ready to try your best as we go through the lesson.

Our lesson today is working with congruent triangles and by the end of it, we're hoping we can prove that two triangles are congruent by using the criteria RHS.

Two key words that you would've met in your earlier learning is about a right-angled triangle and about shapes being congruent.

It might be that you wish to pause the videos just so that you can read those definitions before we make a start in the lesson.

This lesson, we'll be using the key word of hypotenuse, which is a side of a right-angled triangle, which is opposite the right angle.

We'll look at this more when we get to it.

This lesson's got two learning cycles.

The first learning cycle is called finding the right angle, and the second learning cycle is justifying congruence by RHS.

So we're gonna make a start on that first learning cycle where we're gonna try and find the right angle.

So you may be aware that when two sides and an angle, which is not between them are known, this can lead to two different triangles.

And for this reason, it is not sufficient to prove congruence between two triangles.

If we've got two sides and the angle between them, then that is a proof of congruence SAS, but having two sides and an angle that's not between them is not enough information.

And you can see on this diagram the reason why.

If you have a 50 degree angle, an edge length of six centimetres, and the edge that is opposite 50 degree as five centimetres, there's in fact actually two positions that the third vertex of the triangle could be.

And therefore this is not enough information to be able to be sure that two triangles are congruent.

Alex wants to know, does this always lead to two triangles? So in this circumstance where we've got the 50 degree, the six centimetres and the five centimetres, then we do get two different triangles.

But do we always? So we're gonna investigate Alex's question by keeping the angle on one edge the same and changing the second edge.

So we're gonna keep that 50 degree angle.

We're gonna keep AB as six centimetres, and the opposite edge, which will be B C is going to change.

Vertex C has to be somewhere along that ray line in order for the angle to stay fixed at 50 degrees.

So if I use a pair of compasses, which is the most efficient and effective way of knowing a fixed distance.

So my radius is going to be five centimetres.

When I draw my arc, it intersects the ray here, but it also intersects the ray here and that leads to two different triangles with the same properties.

So we're gonna start by increasing the length of the side opposite the known angle.

So we just did it with five centimetres, so we're gonna increase it.

We've increased it to 5.

5 centimetres.

It intersects once.

It intersects twice.

So this would still lead to two very different triangles despite the fact they would both have a 50 degree angle, a six centimetre edge, and the opposite edge to the angle as 5.

5 centimetres.

Thinking about that then a check for you.

If an arc of radius 5.

3 centimetres was to be drawn from B, how many times would it intersect with the ray line? So pause the video, you might wanna rewind and rewatch and then when you think you know the answer, press play and we'll check.

Well it will intersect two times and that's because when we had it at five centimetres, it intersected at two times.

When we had it at 5.

5 centimetres it intersected twice.

So 5.

3 is between the two of those, so it still would intersect two times.

So so far we've had two triangles being created.

So if we increase the radius again to six centimetres, does this change anything? I want you to think about that just for a moment before any animation takes place.

This would be an isosceles triangle.

Did you notice that that the one of the edge that we was keeping fixed was six centimetres.

We've now increased our radius to six centimetres, which is another edge of our triangle.

So it's isosceles.

So the arc intersects with the ray twice or does it? Well the second point is actually the vertex A, and that wouldn't create a triangle because that just makes C equal to A, just makes them the same line.

So there is actually only one point of intersection here.

So only one triangle is created.

If we increase the radius to over six centimetres, what's gonna happen now? So just think about that again.

So when it was 5.

5, it intersect twice, when it was six centimetres, it was an isosceles and therefore it only intersected once because the second intersection was back onto vertex A.

So if we now increase it to over six centimetres, what's gonna happen? Well this only intersects with the ray at one point as well because the second intersection is no longer on the ray.

If you made the ray align then the angle wouldn't be 50 degrees anymore.

So we're only intersecting the ray at one point.

And once again, only one triangle is created.

What happens if we make it less than five? So we started at five centimetres and that was where it intersected twice, we increased it to 5.

5, it intersected twice.

You'll check at 5.

3, which was between the two intersected twice.

Six centimetres was an isosceles intersected once over six centimetres only intersected once.

So what about now if we start going the other direction? So we're gonna start on four centimetres.

Well this doesn't intersect at all.

No triangle has been created.

Alex says this will be the case for all lengths less than four centimetres.

Just think about that for a moment.

If you reduce that to three centimetres, your circle is getting smaller, it's gonna get further from the ray line.

So Alex is right, that any radius of four centimetres or less is definitely not going to intersect the ray.

So in this construction, so have a look at it, how many triangles will there be? So will there be no triangles? Will there be one triangle or will there be two triangles? Because we've seen that that is possible.

We can have two triangles, we can have one triangle and we can have none.

But in this setup, how many would there be? Pause the video and then when you're ready to check press play.

I'm hoping you went for B.

And the reason we know it will be B is because both of those edges are five centimetres.

So it's an isosceles triangle that's being constructed.

Alex has said, we know that five centimetres intersects twice.

At four centimetres it doesn't intersect at all.

So there must be a length where it only touches the ray line at one point and that length is gonna be the minimum distance.

Anything less than it, it will not touch the ray line anything more than it, it will intersect twice.

So there is a point that it's got to go from no intersections to one intersection and it is the minimum length.

So your task is using a pair of compasses, find and mark, the location of vertex C, which makes the minimum length of BC.

So we are looking for, using a pair of compass, the distance of your radius where it will only intersect the ray at one point and not because it's an isosceles, not because it is longer than an isosceles edge, but because it is the minimum length.

And then question two, use your protractor to measure the angle BCA on each of the diagrams. What do you notice? So press pause whilst you're doing that and when you're ready to check, press play.

So on question A, B, and C, on question one, you needed to find the point C, the vertex C, and they should have been in those sort of locations.

And when you use your protractor to measure, you hopefully notice that they were all 90 degrees or they were the right angle.

So we have found the right angle.

The right angle or right angle triangle is constructed when the minimum distance is marked.

So we've now got to the second learning cycle of the lesson, which is just justifying congruence by RHS.

So if we have two triangles that we can prove they are congruent using the criteria of RHS.

So here are two right angle triangles.

We know they're right-angled because there is the symbol that indicates that there is a right angle.

We are not assuming because they look like it, we know it because of the notation.

From this given information we can show that they are congruent and we can say they are congruent because AB is equal to DE, they're both 21 centimetres, angle ABC and angle DEF are equal, they're both 90 degrees and angle, and side BC is equal to side EF, they're both 20 centimetres.

So we can say that they are congruent by SAS.

We've got two sides and the angle between them.

But if we were given this information instead, can we still be sure that they are congruent? So there's two right-angled triangles.

We've got two edges that are equal, BC and FE are 20 centimetres and AC and DF are both 29 centimetres.

So normally having two sides and the angle which isn't between them is not sufficient.

So this is SSA.

And that's normally not sufficient to prove congruence because you could get two different triangles.

However, when there is a right angle, only one triangle is formed.

And we saw that in your task from the earlier part of the lesson.

It only intersected once at the minimum distance, which created a right angle.

So consider placing a ladder against a wall on a level ground.

So the level ground and the wall means that they are perpendicular to each other.

The ladder is a fixed length, it's not a telescopic ladder, it's a, it's a fixed length.

If the foot of the ladder, the bottom part is put six feet away from the wall, how many different places will the top of the ladder touch on the wall? So just think about that for a moment as a scenario.

You've got a wall and a level ground that are perpendicular to each other.

You've got fixed ladder length and you are placing the bottom, the foot of it six feet away from the wall.

How many different places can it touch? I'm hoping you said only one place.

So the only one right-angled triangle can be formed with those two sides with fixed lengths by keeping it six foot away and the ladder not being able to change size, there is only one place it can touch the wall and this is SSA, but only one is formed, it's a special case.

So in a right-angled triangle, the 90 degrees is the largest interior angle.

The side opposite the right angle is also the longest edge and it's called the hypotenuse.

So we are gonna be using this word in this lesson and in your future learning of mathematics you'll use it in many other places as well.

So it's really important that you remember the hypotenuse is the edge opposite the right angle in a right-angled triangle.

If it's a non right-angled triangle, then it isn't called a hypotenuse.

So here you've got three examples of where the hypotenuse is located.

So it's the longest of the three edges, it's opposite the largest angle.

The largest angle is the 90 degree angle, which makes it a right-angled triangle.

So let's just check on that one.

Which edge is the hypotenuse in this right-angled triangle? Press pause and then when you're ready to check, press play.

It's C.

So C is opposite the right angle.

If it's opposite the right angle, then it's a hypotenuse.

Another check.

The longest edge of a triangle is called the hypotenuse.

True or false? Pause whilst you consider that one.

And then when you're ready to check and then get the justifications, press play.

So this is false.

Why is it false? Is it because the hypotenuse is the longest edge in a right-angled triangle or is it because the longest edge is always opposite the largest angle? Press pause whilst you consider the justifications and then when you're ready to check it, press play.

Okay, I'm hoping you went for A, so the hypotenuse is a special edge on a right-angled triangle.

So going back to our right-angled triangles and proving of congruence, we've seen that there is a special case when you know two sides in an angle and it's holds only for right-angled triangles.

So if we know it's a right-angled triangle and you know the hypotenuse and you know one of the other two sides, then we can prove two triangles are congruent by RHS.

So despite the fact that the angle is not between the edges, it's a special case.

And so it's the RHS condition.

R for right angle, H for hypotenuse and S for side.

So triangle ABC and triangle MNO are congruent.

What is the length of BC? So just think about that for a moment.

We are told that they are congruent, which means that they would fit exactly on top of each other.

If we could cut them out, we might need need to rotate them, we might need to reflect them, we might need to translate them, but they are exactly the same size and shape.

Okay, well BC would be 12 centimetres.

The 35 centimetres, which is marked on both of the triangles, is the hypotenuse is opposite the right angle.

On triangle MNO, NO is marked as 12 centimetres.

So BC would be the corresponding edge of 12 centimetres.

Are these two triangles congruent? So have a look at them.

Are they congruent? At first glance, none of the criteria for congruent supplies.

So let's just go through what they are.

The first one is SSS.

So that's when you know the three side lengths, which we don't know all three of them.

We know two of them but not three.

So we can't use SSS.

The next one is SAS.

So when you have two sides and the angle between them, you can prove two triangles are congruent.

Here on triangle ABC, we do have 21 centimetre edge, then the 44 degree angle and then the 29 centimetre edge.

But on triangle IJK, we do not know the angle between the edge of 29 centimetres and the edge of 21 centimetres.

So we can't use SAS.

The next proof of congruence is a ASA or AAS.

And on triangle, ABC, we could use ASA 'cause we've got 46 degrees, then the side 29 centimetres and then the 44 degrees.

Whereas on IJK, we do not have those two angles either side of the 29 centimetre edge.

And RHS, this new one, right angle, while on the right hand triangle on triangle IJK, it is a right-angle triangle.

On ABC, it's not marked to be a right-angled triangle.

So at the moment RHS does not seem to apply.

However, we can calculate the third angle when you know two of the angles and it comes out to be 90 degrees.

So now that we do have a right angle, we can think about RHS again.

So we've got a right angle in both triangles.

Is the hypotenuse the same length? Well the hypotenuse is the edge opposite the 90 degree angle on ABC, it's 29 centimetres, on IJK, it's 29 centimetres.

So the corresponding hypotenuse are the same.

And then lastly, RHS, we need another side.

And here we've got BC is 21 centimetres and IJ is also 21 centimetres.

So we can actually say that these two are congruent by RHS.

So I've removed the information we haven't needed for RHS from the triangle ABC, however we did need it to calculate the 90 degrees.

Using proofs of congruence, sometimes we need to use properties of shape, we won't know the lengths or the size of the angles, but we can still prove that two triangles are congruent and here is an example of that.

So how do we know that triangle ABC, and triangle DAC are congruent in this rectangle? Well the important part here is that it's a rectangle.

Angle ABC and angle ADC are 90 degrees because it's a rectangle.

AC is a shared edge and is the hypotenuse of each triangle.

It's opposite the right angle in both of those triangles.

AB equals CD as they are opposite sides of a rectangle.

And hence, we can say that triangle ABC and triangle ADC are congruent by RHS, we have the right angle, we have the hypotenuse and we have a side.

The corresponding sides are equal, the corresponding hypotenuse are equal.

Here's a check for you.

You need to fill in some blanks.

So ABCD is a kite.

Prove the triangle DAC and triangle ABC are congruent by RHS.

Pause the video and then when you're ready to check what you have put for the blanks, press play.

So angle ADC, and angle ABC are 90 degrees.

They are labelled on the diagram using the symbol for a right angle.

AC is the hypotenuse of both triangles.

It is the edge that is opposite the right angle.

DC is equal to BC as they are adjacent sides on a kite.

Hence, the triangle DAC and triangle ABC are congruent by RHS.

We're up to the task part of your lesson.

Task B, so question one, two parts.

You need to use RHS to prove the following pairs of triangles are congruent.

Pause the video whilst you're working through parts A and part B and then when you are ready for question two, just press play.

Question two, triangle TUV is isosceles, where TU equals TV.

TW is perpendicular to UV.

Prove the triangles TUW and TVW are congruent by RHS.

Pause the video whilst you're writing down your proof and then when you're ready for question three, press play.

And here we're at question three.

So using known facts about rhombi and their diagonals, prove that triangle AED is congruent to triangle BEC by RHS.

If you've suddenly panicked because you thought I don't remember any facts about rhombi or their diagonals, then I suggest you ask your teacher if you're in a classroom to give you some facts about rhombi and diagonals, or point you in the direction of a textbook or a revision guide that might help you.

If you're at PC, you might wanna pause the video and then go and Google, look it up, find out and research the facts before you try this question.

So press pause, you might be doing some research first and then trying the question.

And then when you're ready to go through the answers to questions one, two, and three, press play.

Question one, part A, you needed to prove the following pairs of triangles were congruent.

So angle ABC and KLM were both 90 degrees, they were labelled in the diagrams. AC and KM are each triangles hypotenuse and are equal as 35 centimetres is equal to 0.

35 metres.

CB equals KL as 12 centimetres is 120 millimetres, so they are congruent by RHS.

So here we had the right angle, we had the hypotenuse and we had the side, but they were mixed units, so we needed to check that they were equivalent to each other.

Part B, first of all, you needed to calculate the angle MON was a right angle, which it was.

So then you could say that angle MON was equal to angle EDF, that EF and MN were equal and the hypotenuse of each triangle.

And lastly that DE and MO were equal so that they were congruent by RHS.

Question two, you needed to prove that the triangles that had been made by TW were congruent by RHS.

So angle TWU equals angle TWV as we were told that they are perpendicular.

So perpendicular means that they meet at 90 degrees.

TW is the altitude of that triangle.

TU equals TV as triangle TUV is isosceles, you were told it was isosceles and they also told you TU equal TV.

That is the hypotenuse for both of those triangles.

TW is a common side, both triangle TUW and TWV and therefore triangles TUW and TWV are congruent by RHS.

Okay, so question three, using facts about rhombi and their diagonals prove that AED and BEC, the two triangles were congruent.

So angle AED and angle BEC are 90 degrees as the diagonals of a rhombus are perpendicular.

So that is a fact about the diagonals in any rhombus.

AD equals BC as they are edges of a rhombus and all the edges of a rhombus are the same length.

AE equals EC as the diagonals of the rhombus bisect each other.

So again, that's a property of the diagonals of the rhombus.

And so triangle AED and BEC are congruent by RHS.

AD and BC were the hypotenuse in each of those triangles and we know that they are equal in their lengths.

So to summarise today's lesson, which was looking at congruent triangles by using the criteria of RHS.

So for a right-angled triangle only you can prove the congruence by using the hypotonus and one other side.

This is known as RHS.

R for right angle.

It only applies if it's a right-angled triangle.

H is the hypotenuse, which was our new keyword, which is the edge opposite that right angle, and finally side.

Really well done today and I look forward to working with you again in the future.