Angle Sum Property of a Polygon

We will learn about the interior angle sum property of a polygon.

The sum of interior angles of a polygon can also be obtained without using the angle sum formula.

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°

$Sum of Interior Angles of a Triangle$

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°

$Sum of Interior Angles of a Quadrilateral$

Adding we get;

∠ABD+ ∠BDA + ∠DAB + ∠BDC + ∠DCB + ∠CBD = 180° + 180°

⇒ (∠ABD + ∠CBD) + ∠DAB + (∠BDC + ∠BDA) + ∠DCB = 360°

⇒ ∠ABC + ∠DAB + ∠ADC + ∠DCB = 360°

⇒ ∠B + ∠A + ∠D + ∠C = 360°

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°

$Sum of Interior Angles of a Pentagon$

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°

$Sum of Interior Angles of a Hexagon$