Arun
Last Activity: 7 Years ago
Area = 24
perimeter = 24, hence semi perimeter (s) = 12
r1, r2, r3 are in H.P.
hence 1/r1, 1/r2, 1/r3 will be in A.P.
1/r2 – 1/r1 = 1/r3 – 1/r2
2/r2 = 1/r1 + 1/r3
2(s – b)/ = (s –a)/ + (s –c)/ on solving
2s – a –c = 2(s –b)
a+b+c –a –c = 2(s –b)
b = 2(s –b)
3 b = 2s
b = 24/3
b = 8
as we know,
a+b+c = 24
a+c = 16
c= 16 – a
when applying heron’s formula,
we get,
(a –6) (a –10) = 0
a= 6 or a= 10
if a =6 , c = 10
if a =10, c= 6
hence three sides are 6,8, 10