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Math Help - Finding similar triangles using ratios

  1. #16
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    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?
    It's an equilateral triangle so all sides must be the same length, it is one of the basic rules of geometry.

    Quote Originally Posted by David Green View Post
    Similarity ratio;

    (AY) / (AB) = (2) / (1) = 2/1
    AB = AY+YB = 2+1 = 3 so you get \dfrac{AY}{AB} = \dfrac{2}{3}


    Quote Originally Posted by David Green View Post
    If I now use the product rule by BC = 3, then YZ = 6????
    I don't get how you're using the product rule. Can you explain how you're using line BC to get line YZ? Surely there is another line you're using.

    edit: If I'm understanding correctly you've done BC \times \dfrac{AY}{AB} using your carried forward error in \dfrac{AY}{AB}.

    If you do YZ = BC \times \dfrac{AY}{AB} = 3 \times \dfrac{2}{3} = 2


    Quote Originally Posted by David Green View Post
    The parallel line YZ in the triangle above cannot ever be longer than the parrallel line BC?
    True, hence my confusion earlier

    Quote Originally Posted by David Green View Post
    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
    I shall use the cos rule to show YZ = 2.

    AZ = AY = 2 - we are told this. We also know angle A is 60 degrees as it's a property of equilateral triangles.

    c^2 = a^2+b^2 - 2ab\cos(C). In this example c = YZ - the side we want to know, a = AZ = 2, b = AY = 2 and C = A = 60 degrees (angle A)

    c^2 = 2^2 + 2^2 - 2(2)(2)\cos(60) = 4+4-4 = 4 and so c = YZ =2

    Thus YZ is 2 units across. It has been established that BC = 3 and so the ratio of YZ to BC is 2:3

    Similarly the ratio of BC to YZ is the reciprocal which is 3:2
    Last edited by e^(i*pi); November 14th 2011 at 04:08 AM. Reason: adding part in bold
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  2. #17
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by e^(i*pi) View Post
    It's an equilateral triangle so all sides must be the same length, it is one of the basic rules of geometry.



    AB = AY+YB = 2+1 = 3 so you get \dfrac{AY}{AB} = \dfrac{2}{3}




    I don't get how you're using the product rule. Can you explain how you're using line BC to get line YZ? Surely there is another line you're using.

    I was reading up on similar triangles and thought I found a good example, however after the event, "as always" I found that the similar traingles I was using were not equilateral, where after working through the maths I found that the sides in that example AC and AB were not equal, therefore a different method had been used, which was no good for my example.

    edit: If I'm understanding correctly you've done BC \times \dfrac{AY}{AB} using your carried forward error in \dfrac{AY}{AB}.

    If you do YZ = BC \times \dfrac{AY}{AB} = 3 \times \dfrac{2}{3} = 2



    True, hence my confusion earlier



    I shall use the cos rule to show YZ = 2.

    AZ = AY = 2 - we are told this. We also know angle A is 60 degrees as it's a property of equilateral triangles.

    c^2 = a^2+b^2 - 2ab\cos(C). In this example c = YZ - the side we want to know, a = AZ = 2, b = AY = 2 and C = A = 60 degrees (angle A)

    c^2 = 2^2 + 2^2 - 2(2)(2)\cos(60) = 4+4-4 = 4 and so c = YZ =2

    Thus YZ is 2 units across. It has been established that BC = 3 and so the ratio of YZ to BC is 2:3

    Similarly the ratio of BC to YZ is the reciprocal which is 3:2
    I must thank you for your time and explanation of the above. My workbook does have trigonometry in as you have used in this example, however that is in unit 12 and geometry is unit 8, where similarity is only discussed in unit 8 with one example, very limited example, and no reference to trigonometry as you have used, therefore based on my complete lack of experience using geometry I would never have solved this problem because the course books don't have sufficient good examples for us to learn from.

    Thank you

    David
    Last edited by David Green; November 14th 2011 at 04:31 AM. Reason: if possible I want to highlight some text
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  3. #18
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by David Green View Post
    (AY) / (AB) = (2) / (1) = 2/1
    C'mon David: LOOK at the diagram; can you not SEE that AB = 3;
    AB = AY + YB = 2 + 1 = 3 ; what can be more basic than that?

    And, since triangle ABC is EQUILATERAL, this means ALL sides are equal,
    so AB = 3 and AC = 3 and BC = 3.
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  4. #19
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    Re: Finding similar triangles using ratios

    Quote Originally Posted by Wilmer View Post
    C'mon David: LOOK at the diagram; can you not SEE that AB = 3;
    AB = AY + YB = 2 + 1 = 3 ; what can be more basic than that?

    And, since triangle ABC is EQUILATERAL, this means ALL sides are equal,
    so AB = 3 and AC = 3 and BC = 3.
    I saw that AY = 2 and that YB = 1 therefore 2 + 1 = 3.

    I knew that AZ = 2 and that ZC = 1 therefore 2 + 1 = 3.

    At some point I realised that the triangle was an equilateral, and that the angles were 60 degrees, but as I pointed out eariler, the course book does not give enough in the way of examples, and as I pointed out previously I had been told by my tutor that a similar triangle problem that I solved with an answer 21.4cm previously was incorrect and then in the next same example the tutor worked out the same problem with the same answer 21.4cm?

    So maybe I was looking at this problem all wrong, and yes I do need more practice to learn this topic.

    Thanks again

    David
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  5. #20
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    Re: Finding similar triangles using ratios

    Ok; good luck...
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