Finding similar triangles using ratios

If necessary I can do a drawing, anyone helping please ask if you wish me to draw a similar triangle for this explantion.

We start with an equilatral triangle and label the vertices top A, bottom left B and bottom right C.

We draw a parallel line to BC.

We have;

A to Y 2 units, Y to B 1 unit.

I should think that A to Z will also be 2 units and Z to C 1 unit.

I am asked to find the lengths of the parallels lines using ratios, however the following idea has too many unknowns I think and am wondering if another method is available?

(AY) / (YB) = (YZ) / (BC) = (2) / (1) = (x) / (x)

The problem here is that I can't solve the problem with more than one uknown?

Any ideas please

Re: Finding similar triangles using ratios

I don't follow, what do you mean by "We draw a parallel line to BC." and what is Y & Z?

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**pickslides** I don't follow, what do you mean by "We draw a parallel line to BC." and what is Y & Z?

I'll upload a diagram with the details on it.

Hope this makes it easier to understand

Thanks

David

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** We start with an equilatral triangle and label the vertices top A, bottom left B and bottom right C.

We draw a parallel line to BC. We have;

A to Y 2 units, Y to B 1 unit.

**I should think that A to Z will also be 2 units and Z to C 1 unit.**

**Correct**

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**Plato** **Correct**

But I have not found the ratios of YZ and BC?

This is the part I am struggling with?

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** But I have not found the ratios of YZ and BC?

This is an equilateral triangle so $\displaystyle BC=3$.

Carry on.

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**Plato** This is an equilateral triangle so $\displaystyle BC=3$.

Carry on.

I worked it out at YZ = 1.5

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** I worked it out at YZ = 1.5

I don't know what that means.

$\displaystyle \frac{YZ}{BC}=\frac{2}{3}$

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**Plato** I don't know what that means.

$\displaystyle \frac{YZ}{BC}=\frac{2}{3}$

Sorry I think I got the fraction wrong way round?

2 / 3 x 3 = 2

The ratio 3:2 I think but there are suppose to be two answers also?

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** I worked it out at YZ = 1.5

look at your diagram and describe triangleAYZ

Re: Finding similar triangles using ratios

If the parallel BC = 3, and the triangle is equilateral, then each vertex should be 60 degrees, if this is so then;

tan 60 x 3 = AB = 5.2?

However the given information advises AB = 3

So I am confused?

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** Sorry I think I got the fraction wrong way round?

@Mr. Green

You are the one who struggles with even the most basic mathematical concepts. Therefore, if I were you I would pause and wounder why someone with over 13000 posts might be wrong about something so trivial.

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**Plato** @Mr. Green

You are the one who struggles with even the most basic mathematical concepts. Therefore, if I were you I would pause and wounder why someone with over 13000 posts might be wrong about something so trivial.

Trivial?

My last maths course teacher was a PHD.

One of my questions on that course was similar to this question here posted. The PHD gave a question about similar triangles and asked for the length of a side and the information given was a parallel side 5cm and an angle 65 degrees.

I worked out the other parrallel length which was 10cm, then I said tan 65 (10) = 21.4cm, which was the side length AB on that example.

The PHD said it was wrong and gave 0 out of 4 marks, then on the next example which was exactly the same the question asked for me to identify mistakes a student had made, I did and got 6 out of 6 marks, the PHD then wrote out the full solution and it was a mirror image of my last example?

How could that be wrong first and second time same thing its OK?

Solution;

We all can and do make mistakes, even with over 13000 posts.

Now before you get your back up, please remember that I am learning from first hand basics, and I am saying to you that I don't understand how you got BC = 3 when I am getting something different?

I am not saying you are wrong although it might seem that way, but as previously said I am somewhat confused because I don't understand?

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**David Green** I worked it out at YZ = 1.5

If it isn't OBVIOUS to you that YZ = 2, (and are trying to make it equal to 1.5)!,

then I agree with Plato: you need to get back to basics.

Re: Finding similar triangles using ratios

Quote:

Originally Posted by

**Wilmer** If it isn't OBVIOUS to you that YZ = 2, (and are trying to make it equal to 1.5)!,

then I agree with Plato: you need to get back to basics.

I agree that I don't understand, however you are the experts!

I am advised that the length of BC = 3 from previous, however I have struggled with the basics as you all point out, but I ask again is the information given correct?

Similarity ratio;

(AY) / (AB) = (2) / (1) = 2/1

If I now use the product rule by BC = 3, then YZ = 6????

The parallel line YZ in the triangle above cannot ever be longer than the parrallel line BC?

Like I said I don't fully understand and you are the experts, but to me YZ = 3 and BC = 6

If you now believe that I am wrong, then please show the working out to justify your solution.

Thank you

David