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The radius of a triangle's inscribed circle is called inradius.
Inradius formula : r=2A/pr=2Ap, where AA = Area of the triangle and pp=perimeter of the triangle.
1) calculate d the distance between the corners of the triangles, use the distance formula :
d=sqrt((x2−x1)^2+(y2−y1)^2d=√(x2−x1)2+(y2−y1)2
Given A(3,3), B(1,2),C(2,1)A(3,3),B(1,2),C(2,1)
=> AB=sqrt((1-3)^2+(2-3)^2)=sqrt(2^2+1^2)=sqrt5⇒AB=√(1−3)2+(2−3)2=√22+12=√5
=> AC=sqrt((1-3)^2+(2-3)^2)=sqrt(2^2+1^2)=sqrt5⇒AC=√(1−3)2+(2−3)2=√22+12=√5
=> BC=sqrt((2-1)^2+(1-2)^2)=sqrt(1^2+1^2)=sqrt2⇒BC=√(2−1)2+(1−2)2=√12+12=√2
perimeter p=AB+AC+BC=sqrt5+sqrt5+sqrt2=2sqrt5+sqrt2p=AB+AC+BC=√5+√5+√2=2√5+√2
Since AB=ACAB=AC, the triangle is isosceles.
Since triangle ABD is right-angled,
=> AD=sqrt((AB^2-(BD)^2)⇒AD=√(AB2−(BD)2); (BD=(1/2)BC)(BD=(12)BC)
=> AD=sqrt((sqrt5)^2-(sqrt(2)/2)^2)=sqrt(5-2/4)=sqrt(9/2)=3/sqrt2⇒AD=
⎷(√5)2−(√22)2=√5−24=√92=3√2
Area of triangle A= (1/2)*AD*BC=(1/2)*(3/sqrt2)*sqrt2=3/2A=(12)⋅AD⋅BC=(12)⋅(3√2)⋅√2=32
Inradius r=(2A)/p=(2*(3/2))/(2sqrt5+sqrt2)=0.5097r=2Ap=2⋅(32)2√5+√2=0.5097
