# Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 3 • Question 8

The shape is made of four congruent trapeziums and two triangles. 𝐴𝐶 and 𝐻𝐽 are straight lines. For each statement, tick the correct box. a) The triangles are isosceles. [A] must be true [B] could be true [C] must be false b) The triangles are congruent. [A] must be true [B] could be true [C] must be false

04:01

### Video Transcript

The shape is made of four congruent trapeziums and two triangles. 𝐴𝐶 and 𝐻𝐽 are straight lines. For each statement, tick the correct box. a) The triangles are isosceles: must be true, could be true, must be false. b) The triangles are congruent: must be true, could be true, must be false.

Before we answer this question, let’s analyse the information we’ve been given. The first word we’re interested in is the word “congruent.” Two congruent shapes are shapes that are completely identical. They must have the same-size sides and the same-sized angles.

We’re told that the congruent shapes in our diagram are trapeziums. These are four-sided shapes which have two parallel sides only. The other two sides are never parallel. But sometimes we do have something called a right trapezium. And this is a trapezium for which two of its angles are 90 degrees.

For part a), we’re trying to establish whether the triangles are isosceles. A triangle is isosceles if it has two equal sides and two equal angles. These are often called the base angles. Let’s consider what would need to be true for triangles 𝐵𝐸𝐹 and 𝐸𝐹𝐼 to be isosceles.

One way that this could be true is if the side 𝐵𝐸 was equal to the side 𝐵𝐹. And this is entirely feasible. Our trapeziums are congruent. And if we imagine they’re not in the same orientation, that is, one of the trapeziums has been reflected in a line 𝐵𝐼, we can see that the lines 𝐵𝐸 and 𝐵𝐹 are indeed of equal length. So this is one situation in which the triangle would be isosceles.

But what about if they were in the same orientation? Well, this time, 𝐴𝐷 and 𝐵𝐹 would be the same length and 𝐵𝐸 and 𝐶𝐺 would be the same length. This triangle certainly doesn’t look isosceles. It could be and this would be if the line 𝐸𝐹 was equal to one of the other two sides, so say 𝐵𝐸 was equal to 𝐸𝐹. This would be isosceles. There’s no way for us to know if this is true.

And one other way we could guarantee an isosceles triangle was if the trapeziums themselves were isosceles. That is to say, what we think of as almost the sloping sides were of equal length. Then 𝐴𝐷 would be equal to 𝐵𝐸, which is equal to 𝐵𝐹, which is equal to 𝐶𝐺. And we do indeed have an isosceles triangle. We have found a situation for where the triangles are isosceles and one for which they aren’t. So this could be true. The triangles could be isosceles.

Now let’s consider part b. The triangles are congruent. The same definition for congruency stands. They must have the same sides and the same angles. We don’t need to prove it for all sides and all angles though. We can use these four conditions for congruency.

SSS stands for side side side. If all the sides are the same, then the triangles must be congruent. SAS, that’s two sides and an angle. ASA is two angles and a side. And RHS means both triangles must have a right angle and their hypotenuse on one other side must also be of equal length.

Let’s look at a few scenarios. We’ll imagine that the trapeziums on the top row are a reflection of one another in a line between 𝐵 and 𝐼. We’ll then imagine that the trapeziums on the bottom row are a reflection of the ones on the top row in the line 𝐷𝐺. It’s quite easy to see that we have two sides of equal length in both triangles. 𝐵𝐸 is equal to 𝐸𝐼, and 𝐵𝐹 is equal to 𝐹𝐼.

In fact, there is another side of equal length. 𝐸𝐹 is what we call a shared line. It’s a line that joins both triangles. And by definition, it must be the same size. So by the condition for congruency SSS, we can show that 𝐵𝐸𝐹 and 𝐸𝐹𝐼 are congruent.

Now let’s imagine the trapeziums on the top row are oriented in the same direction. So 𝐴𝐷 is equal to 𝐵𝐹 and 𝐵𝐸 is equal to 𝐶𝐺. And let’s assume the triangles on the bottom row are also oriented in the same direction, but they’re not a reflection in the line 𝐷𝐺. This time, we can see that the lines 𝐵𝐸 and 𝐹𝐼 are of equal length. And we can also see that 𝐵𝐹 is equal to 𝐸𝐼. Once again, they share that third side.

So by the condition SSS, the triangles are congruent. And if we change the trapeziums on the bottom row so that they are reflections of the ones on the top, once again, we still have three sides that are the same. This must be true then. The triangles must be congruent.