# Polygons ICSE Class-6th Concise Selina Maths Solutions Chapter-28

**Polygons ICSE Class-6th Concise** Selina Mathematics Solutions Chapter-28. We provide step by step Solutions of Exercise / lesson-28 **Polygons** for **ICSE Class-6 Concise** Selina Mathematics.

Our Solutions contain all type Questions of Exe-28 A and Exe-28 B with Notes on **Polygons** to develop skill and confidence. Visit official Website **CISCE** for detail information about **ICSE** Board **Class-6**.

**Polygons ICSE Class-6th Concise** Selina Mathematics Solutions Chapter-28

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** **

**What are Polygons?**

- A Polygon is a closed figure made up of lines segments (not curves) in two-dimensions.
- A minimum of three line segments are required for making a closed figure, thus a polygon with a minimum of three sides is known as Triangle.

** Type of Polygons**

#### Regular Polygon

If all the sides and interior angles of the polygon are equal, then it is known as a regular polygon.

#### Irregular Polygon

If all the sides and the interior angles of the polygon are of different measure, then it is known as an irregular polygon.

#### Convex Polygon

If all the interior angles of a polygon are strictly less than 180 degrees, then it is known as a convex polygon. The vertex will point outwards from the centre of the shape.

#### Concave Polygon

If one or more interior angles of a polygon are more than 180 degrees, then it is known as a concave polygon. A concave polygon can have at least four sides. The vertex points towards inside of the polygon.

### Exercise – 28 A **Polygons ICSE Class-6th Concise** Mathematics Selina Solutions

**Question -1.**

State, which of the following are polygons :

**Answer-1**:

Only figure (ii) and (iii) are polygons.

**Question- 2.**

Find the sum of interior angles of a polygon with :

(i) 9 sides

(ii) 13 sides

(iii) 16 sides

**Answer-2**:

**(i)** 9 sides

No. of sides n = 9

∴Sum of interior angles of polygon = (2n – 4) x 90°

= (2 x 9 – 4) x 90°

= 14 x 90°= 1260°

**(ii)** 13 sides

No. of sides n = 13

∴ Sum of interior angles of polygon = (2n – 4) x 90° = (2 x 13 – 4) x 90° = 1980°

**(iii)** 16 sides

No. of sides n = 16

∴ Sum of interior angles of polygon = (2n – 4) x 90°

= (2 x 16 – 4) x 90°

= (32 – 4) x 90° = 28 x 90°

= 2520

**Question -3.**

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

(i) 1440°

(ii) 1620°

**Answer-3**

**(i) 1440°**

1440°

Let no. of. sides = n

∴ Sum of interior angles of polygon = 1440°

∴ (2n – 4) × 90° = 1440°

⇒ 2n – 4 = ^{1440}⁄_{90}

⇒ 2(n – 2) = ^{1440}⁄_{90}

⇒ n – 2 = ^{1440}⁄_{90×2}

⇒ n – 2 = 8

⇒ n = 8 + 2

⇒ n = 10

**(ii) 1620°**

Let no. of sides = n

∴ Sum of angles of polygon = 1620°

∴ (2n – a) × 90° = 1620°

⇒ 2(n – 2) = ^{1620}⁄_{90}

⇒ n – 2 = ^{1620}⁄_{90×2}

⇒ n – 2 = 9

⇒ n = 9 + 2

⇒ n = 11

**Question- 4.**

**Answer-4**:

Let no. of. sides be = n

Sum of interior angles of polygon = 1030°

∴ (2n – 4) × 90° = 1030°

⇒ 2(n – 2) = ^{1030}⁄_{90}

⇒ (n – 2) = ^{1030}⁄_{90×2}

⇒ (n – 2) = ^{103}⁄_{18}

⇒ n = ** ^{103}⁄_{18}**+2

⇒ n = ^{139}⁄_{18}

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 1030°.

**Question -5.**

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

**Answer-5**:

No. of sides of hexagon, n = 6

Let each angle be = x°

Sum of angles = 6x°

(n – 2) x 180° = Sum of angles

(6 – 2) x 180° = 6x°

4 x 180 = 6x

x = 4×^{180}⁄_{6}

x = 120°

∴ Each angle of hexagon = 120°

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

No. of. sides of octagon n = 8

Let each angle be = x°

∴ Sum of angles = 8x°

∴ (2n – 4) × 90° = Sum of angles

(2 × 8 – 4) × 90° = 8x°

12 × 90° = 8x°

⇒ x° =^{90 x 12}⁄_{8}

⇒ x° = 135°

∴ Each angle of octagon = 135°

**Question -6.**

**Answer-6**:

Let the angles of a quadrilateral be x°,

x°, x° and 90°

∴ Sum of interior angles of quadrilateral = 360°

⇒ x° + x° + x° + 90° = 360°

⇒ 3x° = 360° – 90°

⇒ x = ^{270}⁄_{3}

⇒ x = 90°

**Question -7.**

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

**Answer-7**:

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

∴ 4x + 5x + 3x + 6x = 360°

18x = 360°

x = ** ^{360}⁄_{18}**=20°

∴ First angle = 4x = 4 × 20° = 180°

Second angle = 5x = 5 × 20° = 100°

Third angle = 3x = 3 × 20° = 60°

Fourth angle = 6x = 6 × 20° = 120°

**Question -8.**

**Answer-8**:

One angle of a pentagon = 120°

Let remaining four angles be x, x, x and x

Their sum = 4x + 120°

But sum of all the interior angles of a pentagon = (2n – 4) x 90°

= (2 x 5 – 4) x 90° = 540°

= 3 x 180° = 540°

∴ 4x+120o° = 540°

4x = 540° – 120°

4x = 420

x = ** ^{420}⁄_{4}** ⇒ x = 105°

∴Equal angles are 105° (Each)

**Question- 9.**

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

**Answer-9**:

Let the angles of the pentagon be 5x, 4x, 5x, 7x, 6x

Their sum = 5x + 4x + 5x + 7x + 6x = 27x

Sum of interior angles of a polygon

= (2n – 4) × 90°

= (2 × 5 – 4) × 90° = 540°

∴ 27x = 540 ⇒ ^{540}⁄_{27}

⇒ x = 20°

∴ Angles are 5 × 20° = 100°

4 × 20° = 80°

5 × 20° = 100°

7 × 20° = 140°

6 × 20° = 120°

**Question- 10.**

Two angles of a hexagon are 90° and 110°. If the remaining four angles arc equal, find each equal angle.

**Answer-10**:

Two angles of a hexagon are 90°, 110°

Let remaining four angles be x, x, x and x

Their sum = 4x + 200°

But sum of all the interior angles of a hexagon

= (2n – 4) × 90°

= (2 × 6 – 4) × 90° = 8 × 90° = 720°

∴ 4x + 200° = 720°

⇒ 4x = 720° – 200° = 520°

⇒ x = ** ^{520}⁄_{4}**=130°

∴ Equal angles are 130° (each)

**Polygons Exe-28 B for ICSE Class-6th Concise** Selina Mathematics Solved Questions

**Question -1.**

Fill in the blanks :

In case of regular polygon, with

**Answer-1**:

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

(i) 6 | 60° |
120° |

(ii) 8 | 45° |
135° |

(iii) 10 |
36° | 144° |

(iv) 18 |
20° | 160° |

(v) 8 |
45° |
135° |

(vi) 24 |
15° |
165° |

(i) Each exterior angle = ** ^{360}⁄_{6}**=60∘

Each interior angle = 180° – 60° = 120°

(ii) Each exterior angle = ** ^{360}⁄_{8}** = 45°

Each interior angle = 180° – 45° = 135°

(iii) Since each exterior angles = 36°

∴ Number of sides = ** ^{360}⁄_{36}**=10

Also, interior angle = 180°- 20° = 160°

(iv) Since each exterior angles = 20°

∴ Number of sides = ** ^{360}⁄_{20}**=18

Also, interior angle = 180° – 20° = 160°

(v) Since interior angles = 135°

∴ exterior angle = 180° – 135°

∴ Number of sides = ** ^{360}⁄_{45}**°=8

(vi) Since interior angle = 165°

∴ exterior angle = 180° – 165° = 15°

∴ Number of sides = …** ^{360}⁄_{15}**°..= 24…..

**Question -2.**

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

(i) 160°

(ii) 150°

**Answer-2**:

(i) 160°

Let no.of.sides of regular polygon be n.

Each interior angle = 160°

∴** ^{n-2}⁄ n**×180∘=160∘

180n – 360° = 160n

180n – 160n = 360°

20n = 360°

n = 18

(ii) 150°

Let no.of.sides of regular polygon be n.

Each interior angle = 150°

=150∘

180n – 360° = 150n

180n – 150n = 360°

30n = 360°

n = 12

**Question -3.**

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

(i) 30°

(ii) 36°

**Answer-3**:

(i) 30°

Let number of sides = n

∴^{360}⁄_{n }=30∘

n = ^{360}⁄_{30}∘

n = 12

(ii) 36°

Let no. of. sides = n

∴^{360}⁄_{n }=36°

n = ^{360}⁄_{36}∘

n = 10

**Question -4.**

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

(i) 135°

(ii) 155°

**Answer-4**

(i) 135°

No. of. sides = n

Each interior angle = 135°

=135∘

180n – 360° = 135n

180n – 135n = 360°

n = ^{360}⁄_{45}∘

n = 8

Which is a whole number.

Hence, it is possible to have a regular polygon whose interior angle is 135°.

(ii) 155°

No. of. sides = n

Each interior angle = 155°

=155∘

180n – 360° = 155n

180n – 155n = 360°

25n = 360°

n = ^{360}⁄_{25}∘

n = ^{72}⁄_{5}∘

Which is not a whole number.

Hence, it is not possible to have a regular polygon whose interior angle is 155°

**Question -5.**

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

(i) 100°

(ii) 36°

**Answer-5**:

(i) 100°

Let no. of. sides = n

Each exterior angle = 100°

= ^{360}⁄_{n}∘=100∘

∴ n = ^{360}⁄_{100}∘

n = ^{18}⁄5

Which is not a whole number.

Hence, it is not possible to have a regular polygon whose each exterior angle is 100°.

(ii) 36°

et no. of. sides = n

Each exterior angle = 36°

= ^{360}⁄_{n}∘=36∘

∴ n = ^{360}⁄_{36}∘

n = 10

Which is a whole number.

Hence, it is not possible to have a regular polygon whose each exterior angle is 36°.

**Question- 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.

**Answer-6**

**(i) each exterior angle of this polygon.**

Interior angle : exterior angle = 2 : 1

Let interior angle = 2x° & exterior angle = x°

∴ 2x° + x° = 180°

3x = 180°

**x = 60°**

∴ Each exterior angle = 60°

**(ii) number of sides in the polygon.**

Let no.of. sides = n

^{360}⁄n =60°

n = ^{360}⁄_{60}∘

**n = 6**

–: End of** Polygons ICSE Class-6th Concise **Solutions :–

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