Angle Sum Property of a Polygon

In the adjoining figure of a triangle the sum of interior angles of a triangle is always 180°.

i.e. for ∆ABC, ∠BAC + ∠ABC + ∠ACB = 180°

i.e. ∠A + ∠B + ∠C = 180°

In the adjoining figure of a quadrilateral ABCD, if diagonal BD of the quadrilateral is drawn, the quadrilateral will be divided into two triangles i.e. ∆ABD and ∆BDC.

Since, the sum of interior angles of a triangle is 180°.

Therefore, in ∆ABD, ∠ABD+ ∠BDA + ∠DAB = 180°

and, in ∆BDC, ∠BDC + ∠DCB + ∠CBD = 180°

In the adjoining figure of a pentagon ABCDE, on joining AC and AD of the pentagon is divided into three triangles ∆ABC, ∆ACD and ∆ADE.

Since, the sum of the interior angles of the triangles is 180°

Therefore, the sum of interior angles of the pentagon ABCDE = Sum of interior angles of (∆ABC + ∆ACD + ∆ADE)

= 180° + 180° + 180°

= 540°

In the adjoining figure of a hexagon ABCDE, on joining AC, AD and AE, the given hexagon is divided into four triangles i.e. ∆ABC, ∆ACD, ∆ADE and ∆AEF.

The sum of the interior angles of the hexagon ABCDEF = sum of the interior angles of (∆ABC + ∆ACD + ∆ADE + ∆AEF)

= 180° + 180° + 180° + 180°

= 720°