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Find the Circumradius for a Triangle

  1. May 1, 2013 #1

    AGNuke

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    Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC.

    Given AH.BH.CH = 3 and (AH)2 + (BH)2 + (CH)2 = 7, answer the following
    Q1.
    [tex]\frac{\prod \cos A}{\sum \cos^{2}A}[/tex]Q2. What is the value of R?

    ANS 1. From properties of triangle, the distance of Orthocentre from a point A is given by AH = 2R.cosA. Using the values of cosines and from information in the question, I solved the first question to get the answer 3/14R.

    Now I have no clue on how to approach Q2. I can't seem to find any relation between the value of R and information given to me. BTW, from what I know, the answer mentioned is 3/2.
     
  2. jcsd
  3. May 1, 2013 #2

    Simon Bridge

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    What does the answer to Q1 tell you?
    Why is it important that the triangle is acute angled?
     
  4. May 1, 2013 #3

    AGNuke

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    Acute angled triangle means all the cosines are greater than zero?
     
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