Q. 224.4( 14 Votes )

# Perimeter of a tr

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

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