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Math Help - Triangle Problem!

  1. #1
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    Triangle Problem!

    Triangle Problem!-tr.jpg

    ABC right triangle
    AC=6
    AD=DB
    Radius = 5

    We must find: AB

    Please help. thanks
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  2. #2
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    Re: Triangle Problem!

    label intersection of BC and circle E connect DE and AE. See the solution?
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  3. #3
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    Re: Triangle Problem!

    please explain i dont know what you mean
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  4. #4
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    Re: Triangle Problem!

    Draw a line between A and the point where the circle crosses the line BC. What do you know about right angled triangles in circles?
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  5. #5
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    Re: Triangle Problem!

    how much did you get AB?
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  6. #6
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    Re: Triangle Problem!

    AB=AD+DB=2AD
    Meanwhile AD=AC=6 (the base of right angled triangle in a circle is on the diameter, congruent triangles. draw that line and you'll see what I mean.)
    so AB=2AC=2x6=12
    I cannot be bothered to write a full proof right now as this question is quite simple.'

    Oh! This is interesting, it's a 30 60 90 triangle. And the triangles inside the circle are also 30 60 90. Beautiful!
    Last edited by Azzie; July 12th 2012 at 03:28 PM.
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  7. #7
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    Re: Triangle Problem!

    thanks azzie but the answer says AB = 6*sqr(10)
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  8. #8
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    Re: Triangle Problem!

    Quote Originally Posted by Azzie View Post
    Meanwhile AD=AC=6 (the base of right angled triangle in a circle is on the diameter, congruent triangles. draw that line and you'll see what I mean.)
    The fact about the right angle and the diameter is correct (Thales' theorem), but it does not follow that triangles ADE and ACE are congruent and that AD = AC.

    Instead, ED is both a median and an altitude in triangle ABE, so triangle ABE is...
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  9. #9
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    Re: Triangle Problem!

    thanks alot. i get the answer
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  10. #10
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    Re: Triangle Problem!

    Oops! What a foolish assumption I made, sorry about that! I obviously need to go to bed :P thanks emakarov!
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  11. #11
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    Re: Triangle Problem!

    BEC perpendicular to AC given AC is adiameter (subtends arc of 180
    Triangle AEC 10^2 -6^2= EC^2 EC =8
    ED is perpendicular bisector of AB (AD=DB given) and angle ADE =90 (subtends arc of dia)
    AE=EB=10 BC = 18 18 ^2 +6^2 = AB^2 AB=18.97
    Last edited by bjhopper; July 12th 2012 at 07:00 PM.
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