Q22 of 35 Page 311

Perimeter of a triangle is 32 cm. and area of inscribing circle is 38.5 sq. cm. Let us write by calculating the area of this triangle.

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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