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General geometry problem

  1. Mar 13, 2013 #1
    Hi Friends,
    I am getting problem in a geometry problem. Please help me to find the answer.
    The problem is as follows:

    AB, BC, CD, AD are the tangent of circle of radius 10 cm. and center O. If the length of BC = 38 cm and CD = 27 cm. Then find the length of AB. Here tangent AB and AD are perpendicular to each other.

    https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/488010_3019784550925_1195152407_n.jpg

    ATTEMPT :

    Here AP=AQ

    &

    OP = OQ = 10 cm

    ∵ ∠ PAQ = 90°

    ∴ quadrilateral AQOP will be a square

    ∴ AQ = AP = 10 cm

    Now, BQ = (x - 10)

    Now, BQ = BR = (x-10)

    So, CR = (11 + x) = CS

    Now, SD = (16 - x) = DP
    …….

    But after this I am unable to complete this. The answer which the book is showing is,

    x = 21 cm.

    Please friends help me in finding out the answer. Thank you all in advance.
     
  2. jcsd
  3. Mar 13, 2013 #2
    Strange, x comes out to be 21 cm if CS=27 cm.
     
  4. Mar 13, 2013 #3
    how?
     
  5. Mar 14, 2013 #4

    NascentOxygen

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    Can you explain how you got from this line:

    to this line:
     
  6. Mar 14, 2013 #5

    haruspex

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    Yes, you could deduce that if CS=27cm, but it is not physically possible.
    Thunderhadron, consider the angles BQ, DP and CR subtend at O. Call these α, β, γ. Can you see how to write γ in terms of α and β? What equations can you write for the tangents of these angles? (But I'm not sure this is the best way... Seems to lead to quartics.)
     
  7. Mar 14, 2013 #6

    haruspex

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    Managed to get it down to a cubic, and found that the answer to the question as posed is about 23.4. It's easy to show that it must be more than 21. At 21, you'd have CR=CD, making SD zero.
     
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