Perimeter of a tr

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