IN A TRIANGLE ABC , sides a:b:c = 4:5:6 ,

then ratio of circumradius to inradius is???

Given

a:b:c = 4:5:6

so    a=4k,   b=5k,    c=6k

we know   cosA=b^2 + c^2 - a^2

                                                  2bc

on solving we get

      cosA=3/4      so            (sinA)^2=7/16------(1)                                                                                                                    <(sinA)^2=1-(cosA)^2>

circumradius, R=abc/4*area of triangle -----------(2)

inradius, r=area of triangle/semi perimeter of triangle -----(3)

area of triangle= bc*sinA/2 = (15k^2)sinA ---------(4)

semi perimeter, s=a+b+c/2 = 15k/2 -----------(5)

we are required to find  R/r

on solving (2) and (3) we get

R/r = s*a/bc(sinA)^2

using (1), (4), (5)

we get

R/r = 16/7

Approved

hi Rahul

PFA this will definetly help

Approved

correct but it took a long while indeed...........................................