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°
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°
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°
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°
Polygon and its Classification
Interior and Exterior of the Polygon
Number of Triangles Contained in a Polygon
Angle Sum Property of a Polygon
Problems on Angle Sum Property of a Polygon
Sum of the Interior Angles of a Polygon
Sum of the Exterior Angles of a Polygon
7th Grade Math Problems
8th Grade Math Practice
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