In a triangle ABC, if s-a/4=s-b/3=s-c/2 and the radius of the circum-circle be 35/4root6, then
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Given : In a triangle ABC, if s-a/4=s-b/3=s-c/2 and the radius of the circum-circle be 35/4root6
To Find : Side of triangle
Solution:
s-a/4=s-b/3=s-c/2 = k
=> s - a = 4k
s - b = 3k
s - c = 2k
3s - (a + b + c) = 9k
=> 3s - 2s = 9k
=> s = 9k
9k - a = 4k => a = 5k
b = 6k
c = 7k
ar(ΔABC) = √s(s - a)(s - b)(s - c)
= √9k * 4k * 3k * 2k
= 6k²√6
R = Circumradius
R = abc/4 ar(ΔABC)
35/4√6 = 5k*6k.7k/ 4*6k²√6
=> k = 1
Hence sides are 5 , 6 and 7 units
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