Finding the Inradius

Inradius of the triangle is Δ/s, ∆ is the area of the triangle and s is semi perimeter.∆ is 84 in this case and s is 21.

Just use the formula below.

$\boxed{\text{inradius}=\dfrac{2A}{P}}$

$\text{where: A=area of the triangle and P=perimeter of the triangle}$

  area1=(1/2)*(14R+15R+13R)=21R        

  S=(13+14+15)/2=21       
  area2=SQRT(21*(21-15)*(21-14)*(21-13))=84     
  area1=area2 --> 21R=84 --> R=4

call intersections points between circle &triangle sides AB,bc&CA x,y&z seriously. ax=az=k ,xb=13-k,by=13-k =xb ,yc=k+1. so cz=k+1,,,,15=k+k+1 so k=7. cos(bac)=13^2+15^2-14^2÷(2.13.15)=99÷195. Put angle bac=f =2t where t =xam ,m is center of the circle so cos(f)=2cos^2 (t)-1 ,cos^2 (t)= .75384615 ,sin^2 t=.24615385 so tan^2 t. =.326530619 , tan t=.5714286 =R\k so k=1.75R k=7 so R=7\1.75=4#####