A triangle ABC is drawn to …

Let the sides BC, CA, AB of {tex}\triangle{/tex}ABC touch the incircle at D, E, F respectively.
Join the centre O of the circle with A, B, C, D, E, F
Since, tangents to a circle from an external point are equal
{tex}\therefore{/tex} CE = CD = 6 cm
BF = BD = 8 cm
AE = AF = x cm (say)
OE = OF = OD = 4 cm [Radii of the circle]
AB = (x + 8) cm and AC  = (x + 6) cm and CB = 6 + 8 = 14 cm
Area of {tex}\triangle{/tex}OAB = {tex}\frac12{/tex}(8 + x) × 4  = (16 + 2x) cm² ........(i)
area of {tex}\triangle{/tex}OBC = {tex}\frac12{/tex}×14 × 4 = 28 cm² ............(ii) 
area of {tex}\triangle{/tex}OCA = {tex}\frac12{/tex}(6 + x) × 4  = (12 + 2x) cm² ...........(iii) 
{tex}\therefore{/tex} area of {tex}\triangle{/tex}ABC =  16 + 2x + 12 + 2x + 28 = (4x + 56) cm² ...........(iv)
Again, perimeter of  {tex}\triangle{/tex}ABC = AC + AB + BC
= 6 + x + (8 + x) + (6 + 8)
= 28 + 2x = 2(14 + x) cm
S = {tex}\frac{2(14+x)}2{/tex} = 14 + x
Area of {tex}\triangle{/tex}ABC = {tex}\begin{array}{l}\sqrt{s(s-a)(s-b)(s-c)\;}\;\\\end{array}{/tex} 
{tex}=\sqrt{(14+\;x)(14+\;x-14)(14+\;x-6-x)(14+\;x-8-x)\;}{/tex}
{tex}=\sqrt{(14\;+x)48x}{/tex}
{tex}\;\sqrt{672x+\;48x^2}{/tex}...........(v)
{tex}\therefore{/tex} (4x + 56) = {tex}\sqrt{672x+\;48x^2}{/tex}[By 4 and 5]
{tex}\Rightarrow{/tex} (4x + 56)² = 672x + 48x²
{tex}\Rightarrow{/tex} 16(x + 14)² = 16(42x +3x²)
{tex}\Rightarrow{/tex} (x + 14)² = (42x +3x²)
{tex}\Rightarrow{/tex} x² + 28x + 196 = 3x² + 42x
(x + 14) (x -7) = 0
x = 7 , x = -14
But x = -14 is not possible
{tex}\therefore{/tex} x = 7
AB = x + 8 = 7 + 8 = 15 cm
and AC = x + 6 = 7 + 6 = 13 cm