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

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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 â€“ 4) Ã— 90^{0}

= (2 Ã— 9 â€“ 4) Ã— 90^{0}

= (18 â€“ 4) Ã— 90^{0}

= 14 Ã— 90^{0}

We get,

= 1260^{0}

(ii) 13 sides

Number of sides n = 13

The sum of interior angles of polygon = (2n -4) Ã— 90^{0}

= (2 Ã— 13 â€“ 4) Ã— 90^{0}

= (26 â€“ 4) Ã— 90^{0}

= 22 Ã— 90^{0}

We get,

= 1980^{0}

(iii) 16 sides

Number of sides n = 16

The sum of interior angles of polygon = (2n â€“ 4) Ã— 90^{0}

= (2 Ã— 16 â€“ 4) Ã— 90^{0}

= (32 â€“ 4) Ã— 90^{0}

= 28 Ã— 90^{0}

We get,

= 2520^{0}

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

**(i) 1440 ^{0}**

**(ii) 1620 ^{0}**

**Solution:**

(i) 1440^{0}

The sum of interior angles of polygon = 1440^{0}

Let the number of sides = n

The sum of interior angle of polygon is (2n â€“ 4) Ã— 90^{0}

The side of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = 1440^{0}

2n â€“ 4 = 1440^{0} / 90^{0}

2(n â€“ 2) = 1440^{0} / 90^{0}

n â€“ 2 = 1440^{0} / (2 Ã— 90^{0})

On further calculation, we get

n â€“ 2 = 16^{0} / 2

We get,

n â€“ 2 = 8

n = 8 + 2

n = 10

Hence, the side of polygon = 10

(ii) 1620^{0}

Given

The sum of interior angles of polygon = 1620^{0}

Let number of sides = n

The sum of interior angles of polygon = (2n â€“ 4) **Ã— **90^{0}

The side of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = 1620^{0}

2(n â€“ 2) Ã— 90^{0} = 1620^{0}

n â€“ 2 = 1620^{0} / (2 Ã— 90^{0})

n â€“ 2 = 810^{0} / 90^{0}

We get,

n â€“ 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 1030 ^{0}.**

**Solution:**

Given

The sum of interior angles of polygon = 1030^{0}

Let us consider the number of sides = n

The sum of interior angle of polygon = (2n â€“ 4) Ã— 90^{0}

The side of polygon is calculated as,

(2n â€“ 4) Ã— 90^{0} = 1030^{0}

2(n â€“ 2)= 1030^{0} / 90^{0}

On further calculation, we get

(n â€“ 2) = 1030^{0} / (2 **Ã— **90^{0})

(n â€“ 2) = 103^{0} / 18^{0}

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 1030^{0}

**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 = x^{0}

We know,

The sum of interior angles of hexagon = 6x^{0}

The sum of interior angle of polygon = (2n â€“ 4) Ã— 90^{0}

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

(2n â€“ 4) Ã— 90^{0} = Sum of angles

(2 Ã— 6 â€“ 4) Ã— 90^{0} = 6x^{0}

(12 â€“ 4) Ã— 90^{0} = 6x^{0}

6x^{0} = 8 Ã— 90^{0}

x^{0} = (8 Ã— 90^{0}) / 6

We get,

x = 120^{0}

Therefore, each angle of hexagon = 120^{0}

(ii) Number of sides of octagon n = 8

Let us consider each angle be = x^{0}

We know that,

The sum of interior angles of octagon = 8x^{0}

The sum of interior angles of polygon = (2n â€“ 4) Ã— 90^{0}

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

(2n â€“ 4) Ã— 90^{0} = Sum of angles

(2n â€“ 4) Ã— 90^{0} = 8x^{0}

(2 Ã— 8 â€“ 4) Ã— 90^{0} = 8x^{0}

12 Ã— 90^{0} = 8x^{0}

8x^{0} = 12 Ã— 90^{0}

x^{0} = (12 Ã— 90^{0}) / 8

We get,

x^{0} = 135^{0}

Therefore, each angle of octagon = 135^{0}

**6. One angle of a quadrilateral is 90 ^{0} and all other angles are equal; find each equal angle**

**Solution:**

Let us consider all the three equal angle of a quadrilateral be x^{0}

The sum of angles of a quadrilateral = 360^{0}

x + x + x + 90^{0} = 360^{0}

3x + 90^{0} = 360^{0}

3x = 360^{0} â€“ 90^{0}

3x = 270^{0}

x = 270^{0} / 3

We get,

x = 90^{0}

The measure of each equal angle = 90^{0}

**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 = 360^{0}

4x + 5x + 3x + 6x = 360^{0}

18x = 360^{0}

x = 360^{0} / 18

We get,

x = 20^{0}

Now, all the angles are,

4x = 4 Ã— 20^{0}

= 80^{0}

5x = 5 Ã— 20^{0}

= 100^{0}

3x = 3 Ã— 20^{0}

= 60^{0}

6x = 6 Ã— 20^{0}

= 120^{0}

Therefore, the angles of the quadrilateral are 80^{0}, 100^{0}, 60^{0} and 120^{0}

**8. If one angle of a pentagon is 120 ^{0} and each of the remaining four angles is x^{0}, find the magnitude of x.**

**Solution:**

Given

One angle of a pentagon = 120^{0}

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 â€“ 4) Ã— 90^{0}

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

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 5 â€“ 4) Ã— 90^{0}

= 6 Ã— 90^{0 }

We get,

= 540^{0}

Therefore, the sum of interior angles of pentagon is 540^{0}

Now,

x + x + x + x + 120^{0} = 540^{0}

4x + 120^{0} = 540^{0}

4x = 540^{0} – 120^{0}

4x = 420^{0}

x = 420^{0} / 4

We get,

x = 105^{0}

Hence, the value of x = 105^{0}

**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 â€“ 4) Ã— 90^{0}

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

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 5 â€“ 4) Ã— 90^{0}

= 6 Ã— 90^{0}

We get,

= 540^{0}

The sum of interior angles of pentagon = 540^{0}

Hence,

5x + 4x + 5x + 7x + 6x = 540^{0}

27x = 540^{0}

x = 540^{0} / 27

We get,

x = 20^{0}

Thus, each angle,

5x = 5 Ã— 20^{0}

= 100^{0}

4x = 4 Ã— 20^{0}

= 80^{0}

5x = 5 Ã— 20^{0}

= 100^{0}

7x = 7 Ã— 20^{0}

= 140^{0}

6x = 6 Ã— 20^{0}

= 120^{0}

Therefore, all the angles of a pentagon are 100^{0}, 80^{0}, 100^{0}, 140^{0} and 120^{0}

**10. Two angles of a hexagon are 90 ^{0} and 110^{0}. 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 â€“ 4) Ã— 90^{0}

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

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 6 â€“ 4) Ã— 90^{0}

= (12 â€“ 4) Ã— 90^{0}

= 8 Ã— 90^{0}

We get,

= 720^{0}

The sum of interior angles of pentagon is 720^{0}

Hence,

90^{0} + 110^{0} + x + x + x + x = 720^{0}

200^{0} + 4x = 720^{0}

4x = 720^{0} – 200^{0}

4x = 520^{0}

We get,

x = 130^{0}

Hence, the measure of each equal angle = 130^{0}

**Exercise 28(B)**

**1. Fill in the blanks:**

**In case of regular polygon, with**

Number of sides |
Each exterior angle |
Each interior angle |

(i) 6 |
â€¦â€¦â€¦â€¦. |
â€¦â€¦â€¦â€¦.. |

(ii) 8 |
â€¦â€¦â€¦â€¦. |
â€¦â€¦â€¦â€¦.. |

(iii) â€¦â€¦â€¦â€¦ |
36^{0} |
â€¦â€¦â€¦â€¦.. |

(iv) â€¦â€¦â€¦â€¦ |
20^{0} |
â€¦â€¦â€¦â€¦.. |

(v) â€¦â€¦â€¦â€¦ |
â€¦â€¦â€¦â€¦â€¦ |
135^{0} |

(vi) â€¦â€¦â€¦.. |
â€¦â€¦â€¦â€¦… |
165^{0} |

**Solution:**

Number of sides | Each exterior angle | Each interior angle |

(i) 6 | 60^{0} |
120^{0} |

(ii) 8 | 45^{0} |
135^{0} |

(iii) 10 | 36^{0} |
144^{0} |

(iv) 18 | 20^{0} |
160^{0} |

(v) 8 | 45^{0} |
135^{0} |

(vi) 24 | 15^{0} |
165^{0} |

(i) Each exterior angle = 360^{0} / 6

= 60^{0}

Each interior angle = 180^{0} – 60^{0}

= 120^{0}

(ii) Each exterior angle = 360^{0} / 8

= 45^{0}

Each interior angle = 180^{0} – 45^{0}

= 135^{0}

(iii) Given that, each exterior angle = 36^{0}

So, number of sides = 360^{0} / 36^{0}

= 10 sides

Each interior angle = 180^{0} â€“ 36^{0}

= 144^{0}

(iv) Given that, each exterior angle = 20^{0}

Hence, number of sides = 360^{0} / 20^{0}

= 18 sides

Each interior angle = 180^{0} – 20^{0}

= 160^{0}

(v) Given that, each interior angle = 135^{0}

Hence, exterior angle = 180^{0} – 135^{0}

= 45^{0}

Therefore, number of sides = 360^{0} / 45^{0}

= 8 sides

(vi) Given that, each interior angle = 165^{0}

Hence, exterior angle = 180^{0} – 165^{0}

= 15^{0}

Therefore, the number of sides = 360^{0} / 15^{0}

= 24 sides

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

**(i) 160 ^{0}**

**(ii) 150 ^{0}**

**Solution:**

(i) 160^{0}

Let the number of sides of a regular polygon = n

Each interior angle = 60^{0}

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

(2n â€“ 4) Ã— 90^{0} = 160^{0} Ã— n

180^{0}n â€“ 360^{0} = 160^{0}n

180^{0}n â€“ 160^{0}n = 360^{0}

20^{0}n = 360^{0}

n = 360^{0} / 20^{0}

We get,

n = 18

Hence, the number of sides = 18

(ii) 150^{0}

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

The sum of the interior angle of polygon = (2n â€“ 4) **Ã— **90^{0}

Each interior angle = 150^{0}

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

(2n â€“ 4) Ã— 90^{0} = 150^{0} Ã— n

180^{0}n â€“ 360^{0} = 150^{0}n

180^{0}n â€“ 150^{0}n = 360^{0}

30^{0}n = 360^{0}

n = 360^{0} / 30^{0}

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) 30 ^{0}**

**(ii) 36 ^{0}**

**Solution:**

(i) 30^{0}

Let us assume the number of sides be n

Each exterior angle = 30^{0}

Each exterior angle of polygon = 360^{0} / n

Now, we have

360^{0} / n = 30^{0}

n = 360^{0} / 30^{0}

We get,

n = 12

Hence, the number of sides = 12

(ii) 36^{0}

Let us assume the number of sides be n

Each exterior angle = 36^{0}

Each exterior angle of polygon = 360^{0} / n

Now, we have

360^{0} / n = 36^{0}

n = 360^{0} / 36^{0}

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) 135 ^{0}**

**(ii) 155 ^{0}**

**Solution:**

(i) 135^{0}

Let the number of sides of regular polygon be n

The sum of the interior angle of polygon = (2n â€“ 4) Ã— 90^{0}

Each interior angle = 135^{0}

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

(2n â€“ 4) Ã— 90^{0} = 135^{0} Ã— n

180^{0}n â€“ 360^{0} = 135^{0}n

180^{0}n â€“ 135^{0}n = 360^{0}

45^{0}n = 360^{0}

n = 360^{0} / 45^{0}

We get,

n = 8

Since, it is a whole number

Therefore, it is possible to have a regular polygon whose interior angle is 135^{0}

(ii) 155^{0}

Let the number of sides of a regular polygon is n

The sum of the interior angle of polygon is (2n â€“ 4) **Ã— **90^{0}

Each interior angle = 155^{0}

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

(2n â€“ 4) Ã— 90^{0} = 155^{0} Ã— n

180^{0}n â€“ 360^{0} = 155^{0}n

180^{0}n â€“ 155^{0}n = 360^{0}

25^{0}n = 360^{0}

n = 360^{0} / 25^{0}

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 155^{0}

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

**(i) 100 ^{0}**

**(ii) 36 ^{0}**

**Solution:**

(i) 100^{0}

Let the number of sides be n

Each exterior angle = 100^{0}

Each exterior angle of a polygon is calculated as,

360^{0} / n

So,

360^{0} / n = 100^{0}

n = 360^{0} / 100^{0}

We get,

n = 18 / 5

Since, it is not a whole number

Therefore, it is not possible to form a regular polygon

(ii) 36^{0}

Let us consider the number of sides be n

Each exterior angle = 36^{0}

Each exterior angle of polygon = 360^{0} / n

So,

360^{0} / n = 36^{0}

n = 360^{0} / 36^{0}

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 = 2x^{0} and the exterior angle = x^{0}

The sum of the interior angle and exterior angle is 180^{0}

Hence,

2x^{0} + x^{0} = 180^{0}

3x = 180^{0}

x = 180^{0} / 3

We get,

x = 60^{0}

Therefore, each exterior angle = 60^{0}

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

Each exterior angle = 60^{0}

Each exterior angle of polygon = 360^{0} / n

So,

360 / n = 60^{0}

n = 360^{0} / 60^{0}

We get,

n = 6

Hence, the number of sides = 6

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