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CAT 2025 Lesson : Quadrilaterals - Properties of Polygons

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1. Introduction

In the management entrance tests, questions from quadrilaterals are conceptually more advanced than those pertaining to polygons of 5 or more finite sides. Accordingly this lesson covers concepts pertaining to quadrilaterals in detail and the basic properties of other polygons with 5 or more finite sides.

A polygon is a shape enclosed by 3 or more intersecting lines. A triangle is a polygon with 3 sides. We covered triangles in the previous lesson. We will start this lesson with the basic properties that apply to all polygons, the concept of regular polygons and then learn about quadrilaterals.

2. Basics of Polygons

An n-sided polygon will be bound by nn intersecting lines with nn vertices, nn edges and nn interior (and nn exterior) angles. In the 5-sided polygon (pentagon) shown below, the

(a) 5 vertices are A, B, C, D and E

(b) 5 edges are the line segments AB, BC, CD, DE and EA.

(c) 5 interior angles are ∠\angle1, ∠\angle2, ∠\angle3, ∠\angle4 and ∠\angle5.

(d) 5 exterior angles are ∠\angle6, ∠\angle7, ∠\angle8, ∠\angle9 and ∠\angle10.


2.1 Properties of Polygons

In an n-sided polygon (where nn > 3), from any vertex, we can draw diagonals to all vertices other than itself and its two adjacent vertices. So, from 1 vertex we can draw (n−3)(n - 3) diagonals.

(n−3)(n - 3) cuts will divide a polygon into (n−2)(n - 2) triangles. So, sum of interior angles is the sum of angles of the (n−2)(n - 2) triangles formed.

∴ Property 1: Sum of interior angles of an nn-sided polygon =(n−2)×180o= (n - 2) \times 180^{\mathrm{o}}

Example 1

What is the sum of interior angles of the pentagon ABCDE?

Solution

The 2 diagonals from A are AC and AD

This forms 3 triangles – △ \triangle ABC , △ \triangle ACD and △ \triangle ADE

Sum of interior angles of ABCDE = Sum of interior angles of △ \triangle ABC , △ \triangle ACD and △ \triangle ADE

=180o+180o+180o =180^{\mathrm{o}} + 180^{\mathrm{o}} + 180^{\mathrm{o}} =3×180o = 3 \times 180^{\mathrm{o}}

=540o = 540^{\mathrm{o}}


Answer: 540o540^{\mathrm{o}}


If each side is extended in 1-direction across vertices, exterior angles are formed. At each vertex, we have a linear pair (sum of angles is 180o 180^{\mathrm{o}} ) comprising of 1 interior angle and 1 exterior angle.

Sum of exterior angles == Sum of linear pairs - Sum of interior angles

=n×180o−(n−2)×180o = n \times 180^{\mathrm{o}} - (n - 2) \times 180^{\mathrm{o}}

= 180o(n−n+2) 180^{\mathrm{o}} (n - n + 2)

= 360o360^{\mathrm{o}}

∴ Property 2: Sum of exterior angles of an nn-sided polygon = 360o 360^{\mathrm{o}}

Example 2

What is the sum of exterior angles of a pentagon?

Solution

At each vertex, the two angles marked form a linear pair. For instance, ∠1+∠6=180o \angle 1 + \angle 6 = 180^{\mathrm{o}}

Sum of the 5 linear pairs =∠1+∠2+⋯+∠10=5×180o = \angle 1 + \angle 2 + \dots + \angle 10 = 5 \times 180^{\mathrm{o}}

Sum of Interior angles =∠1+∠2+⋯+∠5=3×180o= \angle 1 + \angle 2 + \dots + \angle 5 = 3 \times 180^{\mathrm{o}}

Subtracting the 2 equations above,

Sum of Exterior angles =∠6+∠7+⋯+∠10=2×180o=360o= \angle 6 + \angle 7 + \dots + \angle 10 = 2 \times 180^{\mathrm{o}} = 360^{\mathrm{o}}


OR, directly applying the formula in Property 2,
sum of exterior angles =360o= 360^{\mathrm{o}}

Answer: 360o360^{\mathrm{o}}


Diagonals are lines that connect any 2 vertices of the polygon, that are not adjacent to each other.
(Note: Triangle is the only polygon with 0 diagonals, as all vertices are adjacent to each other.)

As explained at the start of this page, from any 1 vertex we can draw (nn - 3) diagonals.

From nn vertices we can draw n(n−3)n(n - 3) diagonals. But, we are counting each diagonal twice. (e.g., AC and CA are counted as 2 different diagonals).

∴ Property 3: Number of diagonals in an n-sided polygon = n(n−3)2 \dfrac { n (n - 3)}{2}

Example 3

What is the number of diagonals in a pentagon?

Solution

In this figure, 2 diagonals can be drawn from each of the 5 vertices.

As these are double counted,

Number of diagonals =2×52=5 = \dfrac { 2 \times 5}{2} = 5

Directly applying the formula in Property 3,

Number of diagonals =5(5−3)2=5 = \dfrac {5 (5 - 3)}{2} = 5

Answer: 55


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