The radius of the incircleof a …

Ans. 

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x. 

\(In\space \triangle ABC, \)

CF = CD = 6cm (Tangents on the circle from point C)
BE = BD = 8cm (Tangents on the circle from point B)
AE = AF = x (Tangents on the circle from point A)
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
\(=> s = {AB + BC + CA \over 2} = {x + 8 + 14 + 6 + x \over 2}= {28 + 2x \over 2} = x+ 14\)
\(=> Area \space of \triangle ABC = {\sqrt{s(s-a)(s-b)(s-c)} }\)

\(=> {\sqrt{(x+14)(x+14-14)(x+14 - x- 6)(x+14-x-8)} }\)

\(=> {\sqrt{(x+14)(x)(8)(6)} }\)  \(=> {\sqrt {16({3x^2+42x})}} = 4{\sqrt {({3x^2+42x})}} \)      .... ...... (1)

\(Area \space of \triangle OBC = {1\over 2}\times OD\times BC = {1\over 2}\times 4\times 14 = 28 \)

\(Area \space of \triangle OCA = {1\over 2}\times OF\times AC = {1\over 2}\times 4\times (x+6) = 2x+12\)

\(Area \space of \triangle OAB = {1\over 2}\times OE\times AB = {1\over 2}\times 4\times (x+8) = 2x+16\)

\(Also, Area\space of \triangle ABC = Area \space of \triangle OBC + Area\space of \triangle OCA + Area \space of \triangle OAB\)

\(=> 4{\sqrt {({3x^2+42x})}} = 28 + 2x + 12 + 2x + 16 \)

\(=> 4{\sqrt {({3x^2+42x})}} = 4(x + 14) \)  

\(=> {\sqrt {({3x^2+42x})}} = (x + 14) \) 

Squaring Both Sides, We get 

\(=> 3x^2+42x = x^2 + 196 + 28x \)

\(=> 2x^2+14x -196 = 0\)

\(=> x^2+7x -98 = 0\)

\(=> x^2+14x-7x -98 = 0\)

=> x(x+14)-7(x+14) = 0

=> (x+14)(x-7) = 0

Either x+ 14 = 0 or x-7 = 0

x= -14, 7 

Length cannot be -ve, So  x= 7 

So Sides of triangle are 
AB = 7 + 8 = 15 cm
CA = 6 + 7 = 13 cm
BC = 14 cm