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Math Help - Finding similar triangles using ratios

  1. #1
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    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
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  2. #2
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    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?
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by pickslides View Post
    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
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    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
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  5. #5
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Plato View Post
    Correct
    But I have not found the ratios of YZ and BC?

    This is the part I am struggling with?
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    But I have not found the ratios of YZ and BC?
    This is an equilateral triangle so BC=3.
    Carry on.
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Plato View Post
    This is an equilateral triangle so BC=3.
    Carry on.
    I worked it out at YZ = 1.5
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    I worked it out at YZ = 1.5
    I don't know what that means.

    \frac{YZ}{BC}=\frac{2}{3}
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Plato View Post
    I don't know what that means.

    \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?
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  10. #10
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    I worked it out at YZ = 1.5
    look at your diagram and describe triangleAYZ
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  11. #11
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    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?
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  12. #12
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    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.
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  13. #13
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Plato View Post
    @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?
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  14. #14
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    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.
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  15. #15
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Wilmer View Post
    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
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