The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and
8 cm. Determine the other two sides of the triangle. (CBSE 2014)
(Hint: Equate the areas of the triangle found by using the formula √[s(s-a)(s-b)(s-c)] and also found by dividing it into three triangles.)

Given: The radius of the in circle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm.
To find: AC and BC.

Proof:

Area of a triangle
=
=
= …………(i)
Area of triangle = 2 area of Δ AOP + 2 area of Δ COR + 2 area of Δ PBO
                   = 2 × × 4 × 6 + 2 × × 4 × 8 + 2 × × 4 × x
                   = 24 + 32 + 4x
                   = 56 + 4x ………..(ii)
Equating (i) and (ii)
48x(x + 14) = (56 + 4x)²
48x(x + 14) = 16(x + 14)²  
         48x  = 16(x + 14)        or  x + 14 = 0
         48x  = 16x + 224
         32x  = 224
             x = 7                      or  x = -14(which we ignore)
AC = 6 + 7 = 13 cm
BC = 8 + 7 = 15 cm
The other two sides of the triangles are 13 cm and 15 cm.