Question

Circle inscribed in a quadrilateral properties

Answer 1

Answer:

Quadrilateral circumscribing a circle (also called tangential quadrilateral) is a quadrangle whose sides are tangent to a circle inside it.

Quadrilateral circumscribing a circle (also called tangential quadrilateral) is a quadrangle whose sides are tangent to a circle inside it. Where r = radius of inscribed circle and s = semi-perimeter = (a + b + c + d)/2

Where r = radius of inscribed circle and s = semi-perimeter = (a + b + c + d)/2

Where r = radius of inscribed circle and s = semi-perimeter = (a + b + c + d)/2 Derivation for area

Where r = radius of inscribed circle and s = semi-perimeter = (a + b + c + d)/2 Derivation for areaLet O and r be the center and radius of the inscribed circle, respectively

AAOB=12ar

AAOB=12arABOC=12br

AAOB=12arABOC=12brACOD=12cr

AAOB=12arABOC=12brACOD=12crAAOD=12dr

AAOB=12arABOC=12brACOD=12crAAOD=12dr

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total area

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAOD

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12dr

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)r

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)rA=sr (okay!)

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)rA=sr (okay!)

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)rA=sr (okay!) Some known properties

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)rA=sr (okay!) Some known propertiesOpposite sides subtend supplementary angles at the center of inscribed circle. From the figure above, ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°.

AAOB=12arABOC=12brACOD=12crAAOD=12dr Total areaA=AAOB+ABOC+ACOD+AAODA=12ar+12br+12cr+12drA=12(a+b+c+d)rA=sr (okay!) Some known propertiesOpposite sides subtend supplementary angles at the center of inscribed circle. From the figure above, ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°.The area can be divided into four kites. See figure below.If the opposite angles are equal (A = C and B = D), it is a rhombus