235.

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.        

 
Fig, 10.14

Let O be the incentre of ΔABC such that      
OD = OE = OF = 4 cm.
Also, BD = 6 cm, CD = 8 cm.    
Since, length of tangents drawn from an external point are equal.  

 
Fig. 10.14 

So,    BD = BF = 6 cm
and,    CD = CE = 8 cm
Let the length of tangents drawn from first vertex be x.
⇒    AF = AE = x
[Tangents from external point A]
Now, sides of triangle arc
 
              AB  = x + 6 = c 
              BC  = 6 + 8 = 14 = a
and         AC =  x + 8 = b

We know that :   S =  
                   S = 
                    S = 

Therefore,
Area of ΔABC



Also, area of  ABC,
= Area of  BOC + Area of  AOC + AOB 




= 28 + 2 (x + 8) + 2 (x + 6)
= 28 + 2x + 16 + 2x + 12
= 4x + 56                                                ...(ii)

Comparing (i) and (ii), we get

        
Squaring both sides, we get
      48x (x + 14) = (4x + 56 )²    
  48x (x + 14) = (4 (x + 14)]² 
  48x (x + 14) = 16 (x + 14)]² 
   3x (x + 14) = (x + 14)² 
  3x (x + 14) - (x + 14)²  = 0
 (x + 14) [3x - (x + 14)] = 0
   3x - x - 14 = 0
   2x - 14 = 0
          2x = 14
            x = 7

Hencem    AB = x + 6
                     = 7 + 6 = 13 cm
and           AC = x + 8
                     = 7 + 8 = 15 cm.