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
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6 comments:

Pinu13 said...

4

February 8, 2007 at 11:30 AM
Anonymous said...

how???

February 8, 2007 at 11:53 AM
Pinu13 said...

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

February 8, 2007 at 10:45 PM
Vishal Naidu said...

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

February 9, 2007 at 1:13 AM
Pinu13 said...

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

February 9, 2007 at 7:20 AM
Anonymous said...

it is 20.5

February 11, 2007 at 5:22 PM

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