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CAT 2025 Lesson : Quadrilaterals - Types of Polygons

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2.2 Types of Polygons

There are 2 types of polygons

1) Concave Polygons that have at least one interior angle measuring more than 180o180^{\mathrm{o}}. In the following figures, the angles marked are the concave angles.



2) Convex Polygons are those that have all angles measuring less than 180o180^{\mathrm{o}}



Most questions in the exam pertain to convex polygons. Polygons are also classified or referred to basis the number of the sides they have. You should remember the names of nn-sides figures up to 10 sides.

Number of sides Name of the polygon Number of sides Name of the polygon
4 Quadrilateral 8 Octagon
5 Pentagon 9 Nonagon
6 Hexagon / Sexagon 10 Decagon
7 Heptagon / Septagon

Example 4

In a concave decagon, what is the maximum number of interior angles that could measure more than 180o180^{\mathrm{o}} ?

Solution

A decagon is a 10-sided figure.

Sum of interior angles of a decagon =180×(10−2)=180×8 = 180 \times (10 - 2) = 180 \times 8

Here, we can have a maximum of 7 angles that can be more than (or marginally more than) 180o 180^{\mathrm{o}} , while the last is less than 180o 180^{\mathrm{o}} .

Answer: 77


2.3 Regular Polygons

In any regular polygon, all its sides are of equal length, all its interior angles are equal and all its exterior angles are equal.

In an nn-sided regular polygon, each interior angle is the sum of all angles divided by the number of sides.

Interior Angle =(n−2)×180on = \dfrac{ (n - 2) \times 180^{\mathrm{o}}}{n} ; Exterior Angle =360on = \dfrac {360^{\mathrm{o}}}{n}

Example 5

In an nn-sided figure, each angle measures either 145o 145^{\mathrm{o}} or 210o 210^{\mathrm{o}} . If exactly 18 angles measure 145o 145^{\mathrm{o}} , then n n = ?

Solution

Number of angles measuring 145o=18 145^{\mathrm{o}} = 18 (Given in the question)

Number of angles measuring 210o=n−18 210^{\mathrm{o}} = n - 18

Sum of angles of the nn-sided figure =(n−2)×180o = (n - 2) \times 180^{\mathrm{o}}

⇒ 18×145+(n−18)×210=(n−2)×180 18 \times 145 + ( n - 18 ) \times 210 = (n - 2) \times 180

⇒ 2610+210n−3780=180n−360 2610 + 210n - 3780 = 180n - 360

⇒ 30n=81030n = 810

⇒ n=27n = 27

Answer: 2727


Example 6

If a regular polygon has 27 diagonals, then each of its exterior angles{in degrees) equals?

Solution

Let the regular polygon be an nn-sided figure.

Number of diagonals =n(n−3)2=27⇒n(n−3)=54 = \dfrac{n (n - 3)}{2} = 27⇒ n(n - 3) = 54

⇒ n(n−3)=9(9−6)n (n - 3) = 9(9 - 6)

⇒ n=9n = 9

Each Exterior Angle of the regular nonagon =3609=40o = \dfrac {360}{9} = 40^{\mathrm{o}}

Answer: 40o40^{\mathrm{o}}


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