Question

Question 3
What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Answer 1

Let ABCD be a convex quadrilateral. The diagonal AC divides the quadrilateral into two triangles.
$\angle A+\angle B+\angle C+\angle D=\angle 1+\angle 6+\angle 5+\angle 4+\angle 3+\angle 2=(\angle 1+\angle 2+\angle 3)+(\angle 4+\angle 5+\angle 6)={180}^{\circ }+{180}^{\circ }\left(anglesumpropertyofatriangle\right)={360}^{\circ }$
Hence, the sum of measures of the angles of a convex quadrilateral is ${360}^{\circ }.$
This property holds good even for a non-convex quadrilateral.

Let ABCD be a non – convex quadrilateral. The diagonal BD divides the quadrilateral into two triangles.
Using angle sum property of triangle,
$In\Delta ABD,\angle 1+\angle 2+\angle 3={180}^{\circ }...\left(i\right)In\Delta BDC,\angle 4+\angle 5+\angle 6={180}^{\circ }...\left(ii\right).Addingeq.\left(i\right)and\left(ii\right),weget,\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6={360}^{\circ }\Rightarrow \angle 1+\angle 6+(\angle 3+\angle 4)+(\angle 2+\angle 5)={360}^{\circ }$
Hence proved.