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Math Help - Similar triangles - calculating the length

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Similar triangles - calculating the length

    Similar triangles - calculating the length-another-maths.jpg
    AS=2cm, SB=4cm, BC=15cm

    *DQ is a typo. It should be SQ. Many apologies.

    Greetings everyone. PythagorasNeophyte here.

    I am doubtful of part 1 - I don't know what formula(s) must be used in finding the answer in part 1. But I am very categorical that there must be similar sides in this diagram. For example, SQ is similar to BC. That is true. Nevertheless, SQ cannot be found.

    Is there any way to find SQ?

    As for part 2, the answer would be triangle SQA and triangle RQA.
    Last edited by PythagorasNeophyte; September 23rd 2010 at 05:44 AM.
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  2. #2
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    Where is D in the diagram?

    If SQ is parallel to BC, then the triangles ASQ and ABC are similar. Then

    SQ/BC = AS/AB.

    Find SQ
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Where is D in the diagram?
    Just a typo, I'm sure: DQ is really SQ.

    You can go this way also:
    SQ / AS = BC / AB : SQ = AS * BC / AB ; P.N. did you look up "similar triangles"?
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  4. #4
    Junior Member PythagorasNeophyte's Avatar
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    Oh! I am very sorry that I had made a typo error in naming SQ as DQ. Please omit DQ. I'm very sorry.

    So:

    \dfrac{SQ}{BC} = \dfrac{AS}{AB}

    <br />
\dfrac{SQ}{15} = \dfrac{2}{4}

    SQ = 7.5cm

    Is this correct?
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  5. #5
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    NO. AB=6. Try again.
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  6. #6
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    No. ur answer is wrong,Pythagoras..
    AQ/QC=1/2
    So, BP/PC=1/2
    Let BP be x, x+ 2x=15 => BP=5.
    SQBP is a parallelogram.
    so BP=SQ=5cm.
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  7. #7
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    P.s.
    Similar triangles to PCQ are
    ASQ and ABC
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