# CLASS-8PROBLEM & SOLUTION OF POLYGON

Example.1) Find the magnitude of each interior angle of a regular octagon.

Ans.) The sum of the eight angle of an octagon = (2 X 8 – 4) right angles = 12 X 90⁰

Now, in a regular polygon, all the angles are equal

So, each angle of a regular octagon = (12 X 90⁰) / 8 = 135⁰   (Ans.)

Example.2)  Calculate the number of sides of a regular polygon if each exterior angle is 18⁰

Ans.) The number of sides of the regular polygon

360⁰                    360⁰

= ----------------- =  ----------- =  20       (Ans.)

(an exterior angle)            18⁰

Example.3) The sum of the interior angles of a polygon is 3060⁰. Then specify how many sides does it have?

Ans.) Let the number of sides of the polygon = n

Then the sum of its interior angles = (2n – 4) right angles = 3060⁰

So,    (2n – 4) X 90⁰ =  3060⁰

=>      2n – 4  =  3060⁰ / 90⁰ = 34

=>       2n =  34 + 4 = 38

=>       n  =  38 / 2  = 19

So,  n = 19 , so the polygon has 19 sides.    (Ans.)

Example.4) Each interior angle of a regular polygon is 168⁰. How many sides does it have

Ans.) Each exterior angle of the regular polygon

= 180⁰ - an interior angle

=  180⁰ - 168⁰ =  12⁰

So, number of sides of the regular polygon =

360⁰                  360

----------------- = ----------- = 30

An exterior angle           12

Alternatively –

If the number of sides of the regular polygon be ‘n’ then the sum of the interior angles = 168⁰ X n

=  (2n – 4) X 90⁰  (by property)

=>   360⁰ =  180⁰ X n - 168⁰ X n

=>   360⁰ = 12⁰ X n

=>    n  =   360⁰ / 12⁰ = 30       (Ans.)

Example.5)  Find the measures of each interior angle of a regular 30-gon                                                                                                                                        360⁰

Ans.)  Each exterior angle of a regular 30-gon = --------- =  12⁰

30

So, each interior angle = 180⁰ - each exterior angle

=  180⁰ - 12⁰ = 168⁰        (Ans.)

Example.6) Is it possible to have a polygon in which the sum of the interior angles is 1260⁰ ?

Ans.)  Let the number of sides of the polygon = n,

Given, sum of interior angles = 1260⁰

=>  (2n – 4) right angle  =  1260⁰

=>  (2n – 4) X 90⁰ = 1260⁰

=>  2n – 4  =  1260⁰ / 90⁰ =  14

=>   2n  =  14 + 4  = 18

=>     n = 18 / 2  = 9

So, from the above-obtained result, we can conclude that polygon is possible.    (Ans.)

Example.7) Is it possible to have a polygon in which the sum of the interior angles is 1170⁰ ?

Ans.)  Let the number of sides of the polygon = n,

Given, sum of interior angles = 1170⁰

=>  (2n – 4) right angle  =  1170⁰

=>  (2n – 4) X 90⁰ = 1170⁰

=>  2n – 4  =  1260⁰ / 90⁰ =  13

=>   2n  =  13 + 4  = 17

17                  1

=>     n  =  ----------  =  8 ---------

2                  2

1

A polygon cannot have 8 ----- sides.So, such a polygon cannot be possible

2

(Ans.)

Example.8)  Is it possible to have a regular polygon with exterior angles of 40⁰?

360⁰

Ans.)  The number of sides of the polygon = -------------------

An exterior angles

360⁰

= ----------- =   9

40⁰

A polygon can have 9 sides. So, Such a regular polygon is possible. (Ans.)

Example.9) The angles of a Pentagon are in the ratio 1 : 2 : 3 : 5 : 7. find the angles

Ans.) Let the angles of the pentagon be x, 2x, 3x, 5x, and 7x

Then the sum of the angles = x + 2x + 3x + 5x + 7x  = 18x

Also, the sum of the angles of the pentagon = (2n – 4) right angle

= (2 X 5 – 4) right angle

=  6 X 90⁰

As per the given condition,   18x =  6 X 90⁰

6 X 90⁰

x  = ------------- =  30⁰

18

Now,  x = 30⁰

2x = 2 X 30⁰ = 60⁰

3x = 3 X 30⁰ = 90⁰

5x = 5 X 30⁰ = 150⁰

7x = 7 X 30⁰ = 210⁰

Hence, the angles of the pentagon are 30⁰, 60⁰, 90⁰, 150⁰, & 210⁰  (Ans.)

Example.10)  If each interior angle of a regular polygon is seventeen times an exterior angle, find the number of sides of the polygon.

Ans.) Let an exterior angles = x,  then an interior angle = 17x

So,   x + 17x = 180⁰

Or,   18x = 180⁰

Or,    x  =  10⁰

So, exterior angle 10⁰

360⁰

Number of sides of the polygon = ------------------

An exterior angle

360⁰

=  ------------ =  36     (Ans.)

10⁰

Example.11) Four of the angles of a heptagon are equal and the fifth is 60⁰ greater than each of the equal angles. Find the angles

Ans.)  Let each of the four equal angles = x

Then the fifth angle = x + 60⁰

Now, the sum of the angles of the heptagon = 6x + (x + 60⁰) = 7x + 60⁰

Now, as per the given condition –

7x + 60⁰ = (2n – 4) right angle

Or,  7x  + 60⁰ = (2 X 7 – 4) right angle

Or,  7x + 60⁰ =  (14 – 4) X 90⁰

Or,   7x + 60⁰ = 10 X 90⁰

Or,   7x + 60⁰ =  900⁰

Or,    7x =  900⁰ - 60⁰ = 840⁰

Or,      x  =  840⁰/7 =  120⁰

Thus,  the each of the six equal angles = 120⁰ and the seventh angles

= 120⁰ + 60⁰ = 180⁰    (Ans.)

Example.12) Eight angles of a polygon are 170⁰ each. The remaining angles are 160⁰ each. Calculate the number of sides of the polygon.

Ans.) Let the number of the sides of the polygon = n

Then the sum of the angles of the polygon = 8 X 170⁰ + (n – 8)  X 160⁰

and (2n – 4) right-angle = (2n – 4) X 90⁰

as per the given condition –

(2n – 4) X 90⁰ =  8 X 170⁰ + (n – 8)  X 160⁰

=>     180⁰n - 360⁰ =  1360⁰ + 160⁰n - 1280⁰

=>     180⁰n - 160⁰n  = 80⁰ + 360⁰

=>         20⁰n =  440⁰

=>           n  =  440⁰ / 20⁰ =  22

Hence the polygon has 22 sides.    (Ans.)