Question

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.

$\angle A+\angle B+\angle C+\angle D=\angle 1+\angle 6+\angle 5+\angle 4+\angle 3+\angle 2$
= $\angle 1+\angle 2+\angle 3)+(\angle 4+\angle 5+\angle 6)$
= ${180}^{\circ }+{180}^{\circ }$ [Angle sum property of a triangle)
= ${360}^{\circ }$
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 $\Delta BDC,\angle 4+\angle 5+\angle 6={180}^{\circ }...\left(ii\right).$
Adding eq. (i) and (ii),we get,
$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6={360}^{\circ }$
$\Rightarrow \angle 1+\angle 6+(\angle 3+\angle 4)+(\angle 2+\angle 5)={360}^{\circ }$
Therefore, sum of interior angles of a non convex quadrilateral is also ${360}^{\circ }$.
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