Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons

Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons has answers created by experts who have vast knowledge in the education industry. A Polygon is a closed plane figure bounded by at least three line segments. Regular practice of solutions helps students to improve time management skills, which are important in attempting examinations. For in-depth knowledge about the concepts covered, students can use Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons PDF, from the links which are given here.

Chapter 28 Polygons has detailed answers for each and every question relying on Polygons. Students who follow solutions obtain proficiency in solving complex problems effortlessly. This also helps them to build up interest in the respective subjects.

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Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons

Exercise 28(A) Solutions

Exercise 28(B) Solutions

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Exercise 28(A)

1. State, which of the following are polygons:

Solution:

(i) The given figure is not closed.

Hence, the figure is not a polygon.

(ii) The given figure is closed.

Hence, the figure is a polygon.

(iii) The given figure is closed.

Hence, the figure is a polygon.

(iv) In the given figure, one of the sides is an arc.

Hence, the figure is not polygon.

(v) The side intersects each other in the given figure.

Hence, the figure is not polygon.

2. Find the sum of interior angles of a polygon with:

(i) 9 sides

(ii) 13 sides

(iii) 16 sides

Solution:

(i) 9 sides

Number of sides n = 9

The Sum of interior angles of polygon = (2n – 4) × 90⁰

= (2 × 9 – 4) × 90⁰

= (18 – 4) × 90⁰

= 14 × 90⁰

We get,

= 1260⁰

(ii) 13 sides

Number of sides n = 13

The sum of interior angles of polygon = (2n -4) × 90⁰

= (2 × 13 – 4) × 90⁰

= (26 – 4) × 90⁰

= 22 × 90⁰

We get,

= 1980⁰

(iii) 16 sides

Number of sides n = 16

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

= (2 × 16 – 4) × 90⁰

= (32 – 4) × 90⁰

= 28 × 90⁰

We get,

= 2520⁰

3. Find the number of sides of a polygon, if the sum of its interior angles is:

(i) 1440⁰

(ii) 1620⁰

Solution:

(i) 1440⁰

The sum of interior angles of polygon = 1440⁰

Let the number of sides = n

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

The side of polygon can be calculated as,

(2n – 4) × 90⁰ = 1440⁰

2n – 4 = 1440⁰ / 90⁰

2(n – 2) = 1440⁰ / 90⁰

n – 2 = 1440⁰ / (2 × 90⁰)

On further calculation, we get

n – 2 = 16⁰ / 2

We get,

n – 2 = 8

n = 8 + 2

n = 10

Hence, the side of polygon = 10

(ii) 1620⁰

Given

The sum of interior angles of polygon = 1620⁰

Let number of sides = n

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

The side of polygon can be calculated as,

(2n – 4) × 90⁰ = 1620⁰

2(n – 2) × 90⁰ = 1620⁰

n – 2 = 1620⁰ / (2 × 90⁰)

n – 2 = 810⁰ / 90⁰

We get,

n – 2 = 9

n = 9 + 2

n = 11

Hence, the side of polygon = 11

4. Is it possible to have a polygon, whose sum of interior angles is 1030⁰.

Solution:

Given

The sum of interior angles of polygon = 1030⁰

Let us consider the number of sides = n

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

The side of polygon is calculated as,

(2n – 4) × 90⁰ = 1030⁰

2(n – 2)= 1030⁰ / 90⁰

On further calculation, we get

(n – 2) = 1030⁰ / (2 × 90⁰)

(n – 2) = 103⁰ / 18⁰

n = 5.72 + 2

n = 7.72

which is not a whole number.

Therefore, it is not a polygon, whose sum of interior angles is 1030⁰

5. (i) If all the angles of a hexagon arc equal, find the measure of each angle.

(ii) If all the angles of an octagon are equal, find the measure of each angle.

Solution:

(i) Number of sides of polygon n = 6

Let us consider each angle be = x⁰

We know,

The sum of interior angles of hexagon = 6x⁰

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

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

(2n – 4) × 90⁰ = Sum of angles

(2 × 6 – 4) × 90⁰ = 6x⁰

(12 – 4) × 90⁰ = 6x⁰

6x⁰ = 8 × 90⁰

x⁰ = (8 × 90⁰) / 6

We get,

x = 120⁰

Therefore, each angle of hexagon = 120⁰

(ii) Number of sides of octagon n = 8

Let us consider each angle be = x⁰

We know that,

The sum of interior angles of octagon = 8x⁰

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

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

(2n – 4) × 90⁰ = Sum of angles

(2n – 4) × 90⁰ = 8x⁰

(2 × 8 – 4) × 90⁰ = 8x⁰

12 × 90⁰ = 8x⁰

8x⁰ = 12 × 90⁰

x⁰ = (12 × 90⁰) / 8

We get,

x⁰ = 135⁰

Therefore, each angle of octagon = 135⁰

6. One angle of a quadrilateral is 90⁰ and all other angles are equal; find each equal angle

Solution:

Let us consider all the three equal angle of a quadrilateral be x⁰

The sum of angles of a quadrilateral = 360⁰

x + x + x + 90⁰ = 360⁰

3x + 90⁰ = 360⁰

3x = 360⁰ – 90⁰

3x = 270⁰

x = 270⁰ / 3

We get,

x = 90⁰

The measure of each equal angle = 90⁰

7. If angles of quadrilateral are in the ratio 4: 5: 3: 6; find each angle of the quadrilateral.

Solution:

Let us consider the angles of quadrilateral be 4x, 5x, 3x and 6x

We know,

The sum of angles of quadrilateral = 360⁰

4x + 5x + 3x + 6x = 360⁰

18x = 360⁰

x = 360⁰ / 18

We get,

x = 20⁰

Now, all the angles are,

4x = 4 × 20⁰

= 80⁰

5x = 5 × 20⁰

= 100⁰

3x = 3 × 20⁰

= 60⁰

6x = 6 × 20⁰

= 120⁰

Therefore, the angles of the quadrilateral are 80⁰, 100⁰, 60⁰ and 120⁰

8. If one angle of a pentagon is 120⁰ and each of the remaining four angles is x⁰, find the magnitude of x.

Solution:

Given

One angle of a pentagon = 120⁰

Number of sides of pentagon n = 5

Let us consider all other equal angle of pentagon be x

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

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

(2n – 4) × 90⁰ = (2 × 5 – 4) × 90⁰

= 6 × 90⁰

We get,

= 540⁰

Therefore, the sum of interior angles of pentagon is 540⁰

Now,

x + x + x + x + 120⁰ = 540⁰

4x + 120⁰ = 540⁰

4x = 540⁰ – 120⁰

4x = 420⁰

x = 420⁰ / 4

We get,

x = 105⁰

Hence, the value of x = 105⁰

9. The angles of a pentagon are in the ratio 5: 4: 5: 7: 6; find each angle of the pentagon.

Solution:

Let us consider all the angle of pentagon as 5x, 4x, 5x, 7x and 6x

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

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

(2n – 4) × 90⁰ = (2 × 5 – 4) × 90⁰

= 6 × 90⁰

We get,

= 540⁰

The sum of interior angles of pentagon = 540⁰

Hence,

5x + 4x + 5x + 7x + 6x = 540⁰

27x = 540⁰

x = 540⁰ / 27

We get,

x = 20⁰

Thus, each angle,

5x = 5 × 20⁰

= 100⁰

4x = 4 × 20⁰

= 80⁰

5x = 5 × 20⁰

= 100⁰

7x = 7 × 20⁰

= 140⁰

6x = 6 × 20⁰

= 120⁰

Therefore, all the angles of a pentagon are 100⁰, 80⁰, 100⁰, 140⁰ and 120⁰

10. Two angles of a hexagon are 90⁰ and 110⁰. If the remaining four angles arc equal, find each equal angle.

Solution:

Let us consider all the angle of hexagon as x

Number of sides in hexagon n = 6

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

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

(2n – 4) × 90⁰ = (2 × 6 – 4) × 90⁰

= (12 – 4) × 90⁰

= 8 × 90⁰

We get,

= 720⁰

The sum of interior angles of pentagon is 720⁰

Hence,

90⁰ + 110⁰ + x + x + x + x = 720⁰

200⁰ + 4x = 720⁰

4x = 720⁰ – 200⁰

4x = 520⁰

We get,

x = 130⁰

Hence, the measure of each equal angle = 130⁰

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