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.
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
Area of △OCA=12×AC×OE
Area of △BOA=12×AB×OF
Area of △ABC=54cm2 [Given] ∵Area of△ABC=Area of△OBC+Area of△OCA+Area of△BOA
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.