Question

30. In AABC, r2 +rž +r3 +r2 =​

Answer 1

Answer:

The radii r1,r2,r3 of escribed circles of trisngle ABC are in HP . if area =24, perimeter =24, find a,b,c

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)/ \Delta = (s –a)/ \Delta + (s –c)/ \Delta

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