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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

    108452718=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.
 

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