# General Polygons - Angles

#### Contents

## General Polygons

A simple closed figure made up of only line segments is called a polygon. We generally classify polygons according to the number of sides(or vertices) they have. For example, a polygon with 3 sides or vertices is called a Triangle, similarly a polygon with 4 sides or vertices is called a quadrilateral.

A regular polygon is both **equiangular** and **equilateral**. For example, a square has sides of equal length and angles of equal measure, hence its a regular polygon unlike a rectangle which does not have all sides equal. Similarly a regular pentagon is a polygon which has all 5 sides equal **and** has all angles equal.

## Angle sum Property

You must be familiar with the angle sum property of a triangle which states that the sum of the measurements of the three interior angles of a triangle is \(180^\circ\).

Now if we consider a quadrilateral, it can be divided into 2 triangles, thus the sum of the measures of the four angles of a quadrilateral is \(360^\circ\). Similarly any \(n\)-sided polygon can be divided in triangles.

Now since it can be a tedious job to divide a polygon in many triangles and then calculate the sum of all its angles, **We can find the sum \(S\) of all the interior angles of any polygon with \(n\) sides by the formula:**

\(S = (n-2) \times 180^\circ\)

We shall use induction in this proof.

The base case of \(n=3\) is true as the sum of interior angles of a triangle is \(180^\circ\) or \(\pi\).

Now, assume that the statement is true for \(P_k\) (here \(P_k\) denotes a \(k\) sided polygon).

Now, consider \(P_{k+1}\) with vertices \(A_1, A_2 \ldots A_k, A_{k+1}\).

Split this polygon into two parts by joining \(A_1\) and \(A_k\).

Now, \[ \begin{align} \text{Sum of interior angles of } P_{k+1} &= \text{ Sum of interior angles of } A_1 A_2 \ldots A_{k-1} A_k + \text{ Sum of interior angles of } A_1 A_k A_{k+1} \\ &= (k-2)\times 180 ^\circ + 180^\circ \\ &= ((k+1)-2) \times 180^\circ \end{align} \]

This completes the inductive step. \[ \large Q. E. D\]

Sameer was given homework in geometry. He was stuck with a question. The question said that the sum of the interior angles of a polygon did not exceed 2014 degrees. It asked him to find the Max. number of sides the polygon would contain. Enter your answer as the maximum number of sides it could contain.

## Sum of the measures of the exterior angles of a polygon

On many occasions a knowledge of exterior angles may throw light on the nature of interior angles and sides.

**The sum of the measures of the external angles of any polygon is \(360^\circ\).** This is true whatever be the number of sides of the polygon.

Label the interior angles of \(P_n\) as \(Int \angle_1, Int \angle_2, \ldots Int \angle_{n-1}, Int \angle_n\) and the corresponding exterior angles as \(Ext \angle_1, Ext \angle_2, \ldots Ext \angle_{n-1}, Ext \angle_n\).

Note that \(Int \angle_i + Ext \angle_i = 180^\circ \quad i=1,2 \ldots n-1,n \)

Hence, \[ \begin{align} \text{Sum of all exterior angles} &=n \times 180^\circ - \text{Sum of all interior angles} \\ &= n \times 180^\circ - (n-2) \times 180^\circ \\ &= 360^\circ \\ \large Q. E. D \end{align} \]

## Find the number of sides of a polygon whose each exterior angle has a measure of \(20^\circ\).

Total measure of all exterior angles = \(360^\circ\)

Measure of each exterior angle = \(20^\circ\)

Therefore, the number of exterior angles = \( \frac{360^\circ}{20^\circ}\) = \(18\)

Thus the polygon has 18 sides.\( _\square \)

The interior angles of a polygon are in arithmetic progression. The smallest angle is \(120^\circ\) and the common difference is \(5^\circ\). Find the number of sides of the polygon.

**Cite as:**General Polygons - Angles.

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