  Question

# In figure,a triangle ABC is drawn to circumscribe a circle of radius 3 cm,such that the segments BD and DC are respectively of lengths 6 cm and 9 cm.If the area of ΔABC is 54 cm2,then find the lengths of sides AB and AC. Solution

## Given, in ΔABC, circle touch the triangle at point D, F and E respectively and let the lengths of the segment AF be x. So,  BF=BD=6 cm [Tangent drawn from same point are equal]      CE=CD=9 cm [Tangent drawn from same point are equal] and  AE=AF=x cm [Tangent drawn from same point are equal] Now,  Area of △OBC=12×BC×OD                                       =12×(6+9)×3                                       =452cm2     Area of △OCA=12×AC×OE                                 =12×(9+x)×3                                 =32(9+x)cm2       Area of △BOA=12×AB×OF                                  =12×(6+x)×3                                  =32(6+x)cm2       Area of △ABC=54cm2                [Given]     ∵Area of △ABC=Area of △OBC+Area of △OCA+Area of △BOA         54=452+32(9+x)+32(6+x) ⇒    54×2=45+27+3x+18+3x ⇒    108−45−27−18=6x ⇒    6x=18 ⇒    x=3 So, AB=AF+FB=x+6=3+6=9 cm and AC=AE+EC=x+9=3+9=12 cm Hence,lengths of AB and AC are 9 cm and 12 cm respectively.  Suggest corrections   