Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons Exercise 28(B)

Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons Exercise 28(B) provides students with a clear idea of concepts which are important from the latest syllabus pattern. The solutions created by the experts are comprehensive, according to the current ICSE exam. Finding the number of sides in a regular polygon is the main concept discussed under this exercise. Frequent practice of solutions helps students to solve difficult problems within a short span of time. To ace the annual exam, students can access Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons Exercise 28(B) PDF, from the links provided here with a free download option.

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Exercise 28(A) Solutions

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Exercise 28(B)

1. Fill in the blanks:

In case of regular polygon, with

Solution:

(i) Each exterior angle = 360⁰ / 6

= 60⁰

Each interior angle = 180⁰ – 60⁰

= 120⁰

(ii) Each exterior angle = 360⁰ / 8

= 45⁰

Each interior angle = 180⁰ – 45⁰

= 135⁰

(iii) Given that, each exterior angle = 36⁰

So, number of sides = 360⁰ / 36⁰

= 10 sides

Each interior angle = 180⁰ – 36⁰

= 144⁰

(iv) Given that, each exterior angle = 20⁰

Hence, number of sides = 360⁰ / 20⁰

= 18 sides

Each interior angle = 180⁰ – 20⁰

= 160⁰

(v) Given that, each interior angle = 135⁰

Hence, exterior angle = 180⁰ – 135⁰

= 45⁰

Therefore, number of sides = 360⁰ / 45⁰

= 8 sides

(vi) Given that, each interior angle = 165⁰

Hence, exterior angle = 180⁰ – 165⁰

= 15⁰

Therefore, the number of sides = 360⁰ / 15⁰

= 24 sides

2. Find the number of sides in a regular polygon, if its each interior angle is:

(i) 160⁰

(ii) 150⁰

Solution:

(i) 160⁰

Let the number of sides of a regular polygon = n

Each interior angle = 60⁰

The sum of interior angle of polygon can be calculated as,

(2n – 4) × 90⁰ = 160⁰ × n

180⁰n – 360⁰ = 160⁰n

180⁰n – 160⁰n = 360⁰

20⁰n = 360⁰

n = 360⁰ / 20⁰

We get,

n = 18

Hence, the number of sides = 18

(ii) 150⁰

Let us consider the number of sides of regular polygon be n

The sum of the interior angle of polygon = (2n – 4) × 90⁰

Each interior angle = 150⁰

The sum of the interior angle of polygon can be calculated as,

(2n – 4) × 90⁰ = 150⁰ × n

180⁰n – 360⁰ = 150⁰n

180⁰n – 150⁰n = 360⁰

30⁰n = 360⁰

n = 360⁰ / 30⁰

We get,

n = 12

Hence, the number of sides = 12

3. Find number of sides in a regular polygon, if its each exterior angle is:

(i) 30⁰

(ii) 36⁰

Solution:

(i) 30⁰

Let us assume the number of sides be n

Each exterior angle = 30⁰

Each exterior angle of polygon = 360⁰ / n

Now, we have

360⁰ / n = 30⁰

n = 360⁰ / 30⁰

We get,

n = 12

Hence, the number of sides = 12

(ii) 36⁰

Let us assume the number of sides be n

Each exterior angle = 36⁰

Each exterior angle of polygon = 360⁰ / n

Now, we have

360⁰ / n = 36⁰

n = 360⁰ / 36⁰

We get,

n = 10

Hence, the number of sides = 10

4. Is it possible to have a regular polygon whose each interior angle is:

(i) 135⁰

(ii) 155⁰

Solution:

(i) 135⁰

Let the number of sides of regular polygon be n

The sum of the interior angle of polygon = (2n – 4) × 90⁰

Each interior angle = 135⁰

The sum of interior angle of polygon can be calculated as,

(2n – 4) × 90⁰ = 135⁰ × n

180⁰n – 360⁰ = 135⁰n

180⁰n – 135⁰n = 360⁰

45⁰n = 360⁰

n = 360⁰ / 45⁰

We get,

n = 8

Since, it is a whole number

Therefore, it is possible to have a regular polygon whose interior angle is 135⁰

(ii) 155⁰

Let the number of sides of a regular polygon is n

The sum of the interior angle of polygon is (2n – 4) × 90⁰

Each interior angle = 155⁰

The sum of the interior angle of polygon can be calculated as,

(2n – 4) × 90⁰ = 155⁰ × n

180⁰n – 360⁰ = 155⁰n

180⁰n – 155⁰n = 360⁰

25⁰n = 360⁰

n = 360⁰ / 25⁰

We get,

n = 72 / 5

Since, it is not a whole number

Therefore, it is not possible to form a regular polygon whose interior angle is 155⁰

5. Is it possible to have a regular polygon whose each exterior angle is:

(i) 100⁰

(ii) 36⁰

Solution:

(i) 100⁰

Let the number of sides be n

Each exterior angle = 100⁰

Each exterior angle of a polygon is calculated as,

360⁰ / n

So,

360⁰ / n = 100⁰

n = 360⁰ / 100⁰

We get,

n = 18 / 5

Since, it is not a whole number

Therefore, it is not possible to form a regular polygon

(ii) 36⁰

Let us consider the number of sides be n

Each exterior angle = 36⁰

Each exterior angle of polygon = 360⁰ / n

So,

360⁰ / n = 36⁰

n = 360⁰ / 36⁰

We get,

n = 10

Since, it is a whole number

Therefore, it is possible to form a regular polygon

6. The ratio between the interior angle and the exterior angle of a regular polygon is 2: 1. Find:

(i) each exterior angle of this polygon.

(ii) number of sides in the polygon.

Solution:

(i) Given

Interior angle: exterior angle = 2: 1

Let us assume the interior angle = 2x⁰ and the exterior angle = x⁰

The sum of the interior angle and exterior angle is 180⁰

Hence,

2x⁰ + x⁰ = 180⁰

3x = 180⁰

x = 180⁰ / 3

We get,

x = 60⁰

Therefore, each exterior angle = 60⁰

(ii) Let us assume the number of sides be n

Each exterior angle = 60⁰

Each exterior angle of polygon = 360⁰ / n

So,

360 / n = 60⁰

n = 360⁰ / 60⁰

We get,

n = 6

Hence, the number of sides = 6