Answer :

Given: Area of inscribing circle = 38.5 cm^{2}

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 = πr^{2}

⇒ r = 3.5 cm

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

⇒ Area ∆BCD = 56cm^{2}

Rate this question :

Jaya drew an inciWest Bengal Mathematics

Faruk will draw aWest Bengal Mathematics

The cost of fenciWest Bengal Mathematics

Grandmother of ReWest Bengal Mathematics

Palash and PiyaliWest Bengal Mathematics

The time which toWest Bengal Mathematics

Today the cow of West Bengal Mathematics

A 7 meter wide paWest Bengal Mathematics

Buying a bangle fWest Bengal Mathematics

Let us write by cWest Bengal Mathematics