Answer :

Given: Area of inscribing circle = 38.5 cm2

Let BCD is a triangle of perimeter 32 cm. AB, AD and AC are the internal bisectors of respective angles of the triangle. The three internal bisectors meet at A. Perpendiculars drawn from A on sides BC, CD and DB are AG, AE and AF respectively.

AG = AE = AF

Legth of inner radius of triangle = AG

Let AG be r units.

Area of a circle = πr2

r = 3.5 cm

Now, Area of ∆BCD = area of ∆BCA + area of ∆DCA +area of ∆BDA

Area ∆BCD = 56cm2

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