ICSE Class 7 Maths Chapter 27 Polygons

Chapter 27: Polygons

[Including Quadrilaterals and Their Types]

27.1 Introduction

A closed plane geometrical figure, bounded by atleast three line segments, is called a polygon.

The adjoining figure is a polygon as it is:

(i) closed

(ii) bounded by five line segments AB, BC, CD, DE and AE.

Also, it is clear from the given polygon that:

(i) the line segments AB, BC, CD, DE and AE intersect at their end points.

(ii) two line segments, with a common vertex, are not collinear, i.e., the angle at any vertex is not 180°.

A polygon is named according to the number of sides (line-segments) in it:

27.2 Sum Of Interior Angles Of A Polygon

1. Triangle

Students already know that the sum of interior angles of a triangle is always 180°.

In \(\triangle ABC\), \(\angle BAC + \angle ABC + \angle ACB = 180°\)

\(\angle A + \angle B + \angle C = 180°\)

2. Quadrilateral

Consider a quadrilateral ABCD as shown alongside. If diagonal AC of the quadrilateral is drawn, the quadrilateral will be divided into two triangles ABC and ADC.

Since, the sum of interior angles of a triangle is 180°.

In \(\triangle ABC\), \(\angle ABC + \angle BAC + \angle ACB = 180°\)

And, in \(\triangle ADC\), \(\angle DAC + \angle ADC + \angle ACD = 180°\)

Adding we get:

\(\angle ABC + \angle BAC + \angle ACB + \angle DAC + \angle ADC + \angle ACD = 180° + 180°\)

\((\angle BAC + \angle DAC) + \angle ABC + (\angle ACB + \angle ACD) + \angle ADC = 360°\)

\(\angle BAD + \angle ABC + \angle BCD + \angle ADC = 360°\)

\(\angle A + \angle B + \angle C + \angle D = 360°\)

Alternative method

On drawing the diagonal AC, the given quadrilateral is divided into two triangles. Since, the sum of the interior angles of a triangle is 180°.

Sum of interior angles of the quadrilateral ABCD = Sum of interior angles of \(\triangle ABC\) + sum of interior angles of \(\triangle ADC\) = 180° + 180° = 360°

3. Pentagon

Consider a pentagon ABCDE as shown alongside. On joining CA and CE, the given pentagon is divided into three triangles ABC, CDE and ACE.

Since, the sum of the interior angles of a triangles is 180° Sum of the interior angles of the pentagon ABCDE = Sum of interior angles of (\(\triangle ABC + \triangle CDE + \triangle ACE\)) = 180° + 180° + 180° = 540°

4. Hexagon

It is clear from the given figure that there are four triangles and as such the sum of the interior angles of the hexagon ABCDEF = Sum of interior angles of (\(\triangle ABC + \triangle ACF + \triangle FCE + \triangle ECD\)) = 180° + 180° + 180° + 180° = 720°

27.3 Using Formula

The sum of interior angles of a polygon can also be obtained by using the following formula:

Sum of interior angles of a polygon = (n - 2) \times\) 180°

where, n = number of sides of the polygon.

(i) For a triangle

n = 3 (a triangle has 3 sides)

and, sum of interior angles = (n - 2) \times 180° = (3 - 2) \times 180° = 180°

(ii) For a quadrilateral

n = 4

and, sum of interior angles = (n - 2) \times 180° = (4 - 2) \times 180° = 360°

(iii) For a pentagon

n = 5

and, sum of interior angles = (n - 2) \times 180° = (5 - 2) \times 180° = 3 \times 180° = 540°

(iv) For a hexagon

n = 6

and, sum of interior angles = (n - 2) \times 180° = (6 - 2) \times 180° = 4 \times 180° = 720°

Teacher's Note

Understanding polygon angles helps students calculate measurements in real-world contexts, such as designing roof trusses in construction or planning seating arrangements in auditoriums.

Example 1

Find the sum of the interior angles of a polygon with 8 sides.

Solution

Number of sides in the polygon is 8 \(\Rightarrow\) n = 8

The sum of the interior angles of the 8-sided polygon = (n - 2) \times 180° = (8 - 2) \times 180° = 6 \times 180° = 1080°

Example 2

Find the number of sides of a polygon whose sum of interior angles is 1440°.

Solution

Let the number of sides in the polygon be n.

\((n - 2) \times 180° = 1440°\)

\(n - 2 = \frac{1440°}{180°}\)

\(n = 8 + 2 = 10\)

The number of sides in the polygon = 10

Example 3

Can a polygon have the sum of its interior angles as: (i) 2160° (ii) 2400°

Solution

The number of sides of a polygon is always a whole number which must be greater than or equal to three (3). Thus,

1. A polygon can not have its number of sides in fraction or in decimal.

2. The least number of sides in a polygon is 3.

(i) Let the number of sides of the polygon = n

\((n - 2) \times 180° = 2160°\) \(\Rightarrow\) \(n - 2 = \frac{2160°}{180°} = 12\)

\(\Rightarrow\) \(n = 12 + 2 = 14\)

Since, the number of sides is a whole number which is not less than 3, therefore a polygon can have sum of its interior angles equal to 2160°.

(ii) Let the number of sides of the polygon = n

\((n - 2) \times 180° = 2400°\) \(\Rightarrow\) \(n - 2 = \frac{2400°}{180°} = \frac{40}{3}\)

\(\Rightarrow\) \(n = \frac{40}{3} + 2 = \frac{46}{3}\), which is not a whole number

\(\therefore\) A polygon can not have the sum of its interior angles equal to 2400°.

Teacher's Note

This concept helps students verify whether proposed angle measurements are geometrically possible, similar to checking if data makes sense before accepting survey results.

This is a preview of the first 3 pages. To get the complete book, click below.