POLYGONS – EXERCISE 4.1.3 – Class 9

  1. In each of the following polygons find in degrees the sum of the interior angles and the sum of exterior angles.

(i) Hexagon

(ii) Octagon

(iii) pentagon

(iv) nonagon

(v) decagon

Solution:

(i) Hexagon

Numbers of sides, n = 6

Sum of the interior angles = (2n – 4) 90°

= (2 x 6 – 4) x 90°

= (12 – 4) 90°

= 8 x 90°

= 720°

Sum of the exterior angles = 360°

 

(ii) Octagon

Numbers of sides, n = 8

Sum of the interior angles = (2n – 4) 90°

= (2 x 8 – 4) x 90°

= (16 – 4) 90°

= 12 x 90°

= 1080°

Sum of the exterior angles = 360°

 

(iii) Pentagon

Numbers of sides, n = 5

Sum of the interior angles = (2n – 4) 90°

= (2 x 5 – 4) x 90°

= (10 – 4) 90°

= 6 x 90°

= 540°

Sum of the exterior angles = 360°

 

(iv) Nonagon

Numbers of sides, n = 9

Sum of the interior angles = (2n – 4) 90°

= (2 x 9 – 4) x 90°

= (18 – 4) 90°

= 14 x 90°

=1260°

Sum of the exterior angles = 360°

 

(v) Decagon

Numbers of sides, n = 10

Sum of the interior angles = (2n – 4) 90°

= (2 x 10 – 4) x 90°

= (20 – 4) 90°

= 16 x 90°

= 1440°

Sum of the exterior angles = 360°


  1. How many sides dose a polygon have if the sum of the interior angle is

(i) 540°

(ii) 900°

(iii) 1440°

(iv) 7 straight angles

(v) 8 straight angles

Solution:

(i) Sum of the interior angles = (2n – 4) 90° = 540°

2n – 4 = 540/90 = 6

2n = 6 + 4 = 10

n = 10/2 = 5

The polygon has 5 sides

 

(ii) Sum of the interior angles = (2n – 4) 90° = 900°

→ 2n – 4 = 900/90 = 10

→ 2n = 10 + 4 = 14

→ n = 14/2 = 7

The polygon has 7 sides

 

(iii) The Sum of the interior angles = (2n – 4) 90° = 1440°

→ 2n – 4 = 1440/90 = 16

→ 2n = 16 + 4 = 20

→ n = 20/2 = 10

The polygon has 10 sides

 

(iv) Sum of the interior angles = (2n – 4) 90° = 7 straight angles = 7 x 180

→ (2n – 4) = 7 x 180/90

→ 2n – 4 = 7 x 180/90 = 7 x 2 = 14

→ 2n = 14 + 4 = 18

→ n = 18/2 = 9

The polygon has 9 sides

(v) Sum of the interior angles = (2n – 4) 90° = 8 right angles

→ (2n – 4) 90° = 8 x 90°

→ 2n – 4 = 8 x 9090 = 8

→ 2n = 8 + 4 = 12

→ n = 122 = 6

The polygon has 6 sides


  1. Find the measure of each exterior angle of a regular polygon with sides:

(i) 40

(ii) 30

(iii) 20

(iv) 18

(v) 16

(vi) 2x

(vii) (2a +4b)

Solutions:

(i) Number of sides n = 40

Measure of each exterior angle = ( 360/n ) ° = 360/40 =

 

(ii) Number of sides n = 30

Measure of each exterior angle = ( 360/n ) ° = 360/30 = 12°

(iii) Number of sides n = 20

Measure of each exterior angle = ( 360/n ) ° = 360/20 = 18°

(iv) Number of sides n = 18

Measure of each exterior angle = ( 360/n ) ° = 360/18 = 20°

(v) Number of sides n = 16

Measure of each exterior angle = ( 360/n ) ° = 360/16 =  22.5°

 

(vi) Number of sides n = 2x

Measure of each exterior angle = ( 360/n ) ° = 360/2x = ( 𝟏𝟖𝟎/𝐱

 

 

(vii) Number of sides n = 2a + 4b

Measure of each exterior angle = ( 360/n ) ° = 360/2a + 4b = ( 𝟏𝟖𝟎/𝐚 + 𝟐𝐛

 


  1. Find the numbers of sides of regular polygon. If each exterior angle measures.

(i) 10°

(ii) 20°

(iii) 30°

(iv) 40°

(v) 45°

(vi) 60°

(vii) 72°

(viii) 120° 

Solution:

Solution:

(i) Measure of each exterior angle = x° = 10°

 

Number of sides = 360/x = 360/10 = 35

(ii) Measure of each exterior angle = x° = 20°

 

Number of sides = 360/x = 360/20 = 18

 

 

(iii) Measure of each exterior angle = x° = 30°

Number of sides = 360/x = 360/30 = 12

(iv) Measure of each exterior angle = x° = 40°

 

Number of sides = 360/x = 360/40 = 9

 

v) Measure of each exterior angle = x° = 45°

Number of sides = 360/x = 360/45 = 8

 

(vi) Measure of each exterior angle = x° = 60°

Number of sides = 360/x = 360/60 = 6

 

(vii) Measure of each exterior angle = x° = 72°

Number of sides = 360/x = 360/72 = 5

 

(viii) Measure of each exterior angle = x° = 120°

Number of sides = 360/x = 360/120 = 3


  1. Find the numbers of sides of a regular polygon it each exterior is equal to

(i) Its adjacent interior angle.

(ii) Twice its adjacent interior angle.

(iii) Half its adjacent interior angle.

(iv) One-third of its adjacent interior angle.

Solution:

(i) Exterior angle + its adjacent interior = 180°

e + i = 180°

e = i

e + e = 180°

2e = 180°

e = 180/2 = 90°

Number of sides = 360/e = 360/90 = 4

 

(ii) e + i = 180°

e = 2/i

i = e/2

e + e/2 = 180°

2e+e/2 = 180°

3e = 180 x 2 = 360°

e = 360/3 = 120°

Number of sides = 360/e = 360/120 = 3

 

(iii) e + i = 180°

e = i/2

i = 2e

e + 2e= 180°

3e = 180°

e = 180/3 = 60°

Number of sides = 360/e = 360/60 = 6

 

(iv) e + i = 180°

e = i/3

i = 3e

e + 3e= 180°

4e = 180°

e = 180/4 = 45°

Number of sides = 360/e = 360/45 = 8


  1. find the number of sides in a regular polygon if each interior angle is

(i) twice its adjacent exterior angle

(ii) Four times the adjacent exterior angle.

(iii) Eight times the adjacent exterior angle.

(iv) Seventeen times the adjacent exterior angle.

Solution:

(i) e + i = 180°

i = 2e

e + 2e = 180°

3e = 180°

e = 180/3 = 60°

Number of sides = 360/e = 360/60 = 6

 

(ii) e + i = 180°

i = 4e

e + 4e = 180°

5e = 180°

e = 180/5 = 36°

Number of sides = 360/e = 360/36 = 10

 

(iii) e + i = 180°

i = 8e

e + 8e = 180°

9e = 180°

e = 180/9 = 20°

Number of sides = 360/e = 360/20 = 18

 

(iv) Seventeen times the adjacent exterior angle

e + i = 180°

i = 17e

e + 17e = 180°

18e = 180°

e = 180/18 = 10°

Number of sides = 360/e = 360/10 = 36


  1. The angles of a convex polygon are in the ratio 2 : 3 : 5 : 9 :11. Find the measure of each angle.

Solution:

Ratio of the angle = 2 : 3 : 5 : 9 :11.

Let the angle be 2x, 3x, 5x, 9x and 11x.

The polygon has five angles and therefore five sides.

Sum of angle = (2n – 4)90°

= (2 x 5 – 4)90°

= (10 – 4)90

= 6 x 90°

2x + 3x + 5x + 9x +11x = 6 x 90°

30x = 6 x 90°

x = 6 x 90/30 = 6 x 3 = 18°

The angles are 2x = 2 x 18 = 36°

3x = 3 x 18 = 54°

5x = 5 x 18 = 90°

9x = 9 x 18 = 162°

11x = 11 x 18 = 198°


  1. Prove that the opposite sides of a regular hexagon are parallel.

Solution:

POLYGONS - EXERCISE 4.1.3 - Class 9

Construction: in the regular hexagon ABCDEF, join FC. In the quadrilateral ABCF, AF = BC and ∠FAB = ∠CBA = 120°

ABCF is an isosceles trapezium

FC | | AB ……. (1)

Similarly ED | | FC ……. (2)

From (1) and (2) AB | | ED

Similarly BC | | FE and CD | | FA


 

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