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


–: Select Topics :–

 Notes on Polygons

Exe-28 A,

 Exe-28 B,


 

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.

formulas of Polygons


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

Question -1.

State, which of the following are polygons :
Selina Concise Mathematics Class 6 ICSE Solutions - Polygons-a1
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 = 144090

⇒ 2(n – 2) = 144090

⇒ n – 2 = 144090×2

⇒ 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) = 162090

⇒ n – 2 = 162090×2

⇒ n – 2 = 9

⇒ n = 9 + 2

⇒ n = 11

Question- 4.

 Is it possible to have a polygon, whose sum of interior angles is 1030°.

Answer-4:

Let no. of. sides be = n

Sum of interior angles of polygon = 1030°

∴ (2n – 4) × 90° = 1030°

⇒ 2(n – 2) = 103090

⇒ (n – 2) = 103090×2

⇒ (n – 2) = 10318

⇒ n = 10318+2

⇒ n = 13918

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.

 (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,
Answer-5:

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

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 = 4×1806

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 128

⇒ x° = 135°

∴ Each angle of octagon = 135°

Question -6.

 One angle of a quadrilateral is 90° and all other angles are equal ; find each equal angle.

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 = 2703

⇒ 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 = 36018=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.

 If one angle of a pentagon is 120° and each of the remaining four angles is x°, find the magnitude of x.

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 = 4204  ⇒ 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 ⇒ 54027

⇒ 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 = 5204=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
Selina Concise Mathematics Class 6 ICSE Solutions - Polygons-b1
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 =  3606=60∘

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

(ii) Each exterior angle = 3608 = 45°

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

(iii) Since each exterior angles = 36°

∴ Number of sides = 36036=10

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

(iv) Since each exterior angles = 20°

∴ Number of sides = 36020=18

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

(v) Since interior angles = 135°

∴ exterior angle = 180° – 135°

∴  Number of sides = 36045°=8

(vi) Since interior angle = 165°

∴ exterior angle = 180° – 165° = 15°

∴ Number of sides = …36015°..= 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-2n×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°

Polygons Exe-28 B for ICSE Class-6th Concise Selina Mathematics Ans-2.2

=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

360n =30∘

n = 36030

n = 12

(ii) 36°

Let no. of. sides = n

360=36°

n = 36036

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°

Polygons Exe-28 B for ICSE Class-6th Concise Selina Mathematics Ans-4.1

=135∘

180n – 360° = 135n

180n – 135n = 360°

n = 36045

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°

Polygons Exe-28 B for ICSE Class-6th Concise Selina Mathematics Ans-4.2

=155∘

180n – 360° = 155n

180n – 155n = 360°

25n = 360°

n = 36025

n = 725

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°

=  360n∘=100∘

∴ n =  360100

n =  185

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°

= 360n∘=36∘

∴ n =  36036

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

360n =60°

n = 36060

n = 6

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

Return to –  Concise Selina Maths Solutions for ICSE Class -6


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