Thursday, February 8, 2007

"GEOMETRY QUESTION OF THE DAY"

What is the measure of the circum radius of a triangle whose sides are 9, 40 and 41?

(1) 6 (2) 4
(3) 24.5 (4) 20.5
at

6 comments:

  1. Pinu13February 8, 2007 at 11:30 AM

    4

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  2. AnonymousFebruary 8, 2007 at 11:53 AM

    how???

    ReplyDelete
  3. Pinu13February 8, 2007 at 10:45 PM

    let a=9, b= 40, c= 41

    [ Area of traingle=sqrt(s(s-a)(s-b)(s-c)) where s=a+b+c/2, in this case s=41+40+9/2=45
    Hence area= sqrt 45(45-9)(45-40)(45-41)=sqrt 36*900=6*30=180 ]

    Circumradius of a traingle
    R=abc/4(area of the traingle)
    hence R = 9*40*41/4*180 = 41/2 = 20.5..hence R=20.5

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  4. Vishal NaiduFebruary 9, 2007 at 1:13 AM

    The hypotenuse will be the diameter.
    for every right angled triangle

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  5. Pinu13February 9, 2007 at 7:20 AM

    arrey haan yaar...i didnt realise it was a rt triangle and went thru the trouble.
    thnx

    ReplyDelete
  6. AnonymousFebruary 11, 2007 at 5:22 PM

    it is 20.5

    ReplyDelete