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Exterior angles of polygon - class-VII

Description: exterior angles of polygon
Number of Questions: 41
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Tags: lines, angles and shapes angles and pairs of angles shapes and geometric reasoning maths straight line and angles understanding quadrilaterals polygon rectilinear figures polygons geometric proof understanding shapes
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The interior angle of a regular polygon is double the exterior angle. Then the number in the polygon is 

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $6$

  2. $8$

  3. $9$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let the number of sides in the polygon is $n$

Let the measure of exterior angles be $x$ respectively.
$\therefore$   Measure of interior angle $=2x$
So,
$\Rightarrow$  $n\times 2x=(2n-4)\times 90^o$
$\Rightarrow$  $nx=(n-2)\times 90^o$               ----- ( 1 )
Again we know that,
$\Rightarrow$  $nx=360^o$
$\Rightarrow$  $(n-2)\times 90^o=360^o$               [ From ( 1 ) ]
$\Rightarrow$  $n-2=4$
$\Rightarrow$  $n=6$
$\therefore$   The number of sides in the polygon are $6.$

State true or false:
Is it possible to have a regular polygon whose each exterior angle is $\displaystyle 32^{\circ}$
maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: B
Explanation:
Each exterior angle of regular polygon of n sides is given by $\dfrac{360}{n}$ degree
$\text{According to the question}$
$\dfrac{360}{n} degree=32^0$
$\Rightarrow n=\dfrac{360}{32}$
$\Rightarrow n=\dfrac{45}4$
$\text{Which is not possible because number of sides can never be in fraction.}$

State true or false:
Is it possible to have a regular polygon whose each exterior angle is $\displaystyle 20^{\circ}$

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
$\text{Each exterior angle of regular polygon of n sides is given by}\dfrac{360}{n} degree$
$\text{According to question}$
$\dfrac{360}{n} degree=20^0$
$\Rightarrow n=\dfrac{360}{20}$
$\Rightarrow n=18$
$\text{Clearly there is a polygon of sides 18 whose each exterior angle is 20}^0$

Two alternate sides of a regular polygon, when produced, meet at a right angle, then find the value of each exterior angle of the polygon.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $\displaystyle 45^{\circ}$

  2. $\displaystyle 32^{\circ}$

  3. $\displaystyle 62^{\circ}$

  4. $\displaystyle 15^{\circ}$

Show answer
Correct Option: A
Explanation:

$We\quad know\quad External\quad angles\quad of\quad a\quad regular\quad polygon\quad are\quad equal.$


$\ When\quad the\quad two\quad alternate\quad sides\quad of\quad a\quad regular\quad polygon\quad are\quad produced,\quad they\quad meet\quad at\quad right\quad angle.$

$\ These\quad two\quad extended\quad sides\quad form\quad a\quad triangle\quad with\quad the\quad side\quad of\quad the\quad polygon\quad in\quad between.$

$\ Sum\quad of\quad all\quad interior\quad angles\quad of\quad a\quad \triangle ={ 180 }^{ o }$

$\ \Rightarrow 2\times External\quad angle\quad +\quad { 90 }^{ o }\quad =180$

$\ \Rightarrow 2\times External\quad angle=180-90=90$

$\ \Rightarrow External\quad angle=\dfrac { 90 }{ 2 }$

$ \ \Rightarrow External\quad angle={ 45 }^{ o }
$

State true or false: 

Is it possible to have a regular polygon whose each exterior angle is 40% of a right angle.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Given, regular polygon whose each exterior angle is 40% of a right angle = $ 90^o \times \dfrac{40}{100} = 36^o $
Sum of all exterior angle of any polygon is $ 360^o $
Now,
$ 36^o \times$  number    of   angles  = $ 360^o $
The number  of  angles =$ 10 $
Any polygon have equal number of angles and sides.
Therefore the number of side of the polygon is 10.
Since the number of sides is an integer, therefore their exist a polygon whose each exterior angle is 40% of a right angle. 

State true or false.
Is it possible to have a regular polygon whose each exterior angle is $32^{\circ}$

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Given, a regular polygon whose each exterior angle is $32^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 32^o $
$=> n = \dfrac{45}{4} $
Since, n should be an integer, so it is not possible a regular polygon whose each exterior angle is $32^o $

State true or false: 
Is it possible to have a regular polygon whose each exterior angle is $80^o$
80∘

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Let the number of sides of the polygon is n. (which must be an integer)
If each exterior angle is $ 80^o $, then sum of all exterior angle is $ n \times 80^o $.
And Sum of all exterior angles = $ 180^o $
$=>  n \times 80^o = 180^o $
$=> n = 1.25 $
So, it is not possible to have a regular polygon whose each exterior angle is $ 80^o $

State true or false.
Is it possible to have a regular polygon whose each exterior angle is $20^{\circ}$

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Given, a regular polygon whose each exterior angle is $20^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 20^o $
$=> n = 18 $
Since, n should be an integer, so their exist a regular polygon whose each exterior angle is $20^o $

How many sides does a regular polygon have if the measure of an exterior angle is $24^o$?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $15$

  2. $12$

  3. $14$

  4. $16$

Show answer
Correct Option: A

Find the measure of exterior angle of a regular polygon of 15 sides

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $36^o$

  2. $24^o$

  3. $48^o$

  4. none of the above

Show answer
Correct Option: B
Explanation:

The sum of the exterior angles of a regular polygon is $360^o$

Number of sides of polygon $=15$
As each of the exterior angles are equal,

Exterior angle $=\dfrac{360^o}{15}=24^o$

Find the measure of exterior angle of a regular polygon of 9 sides

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $40^o$

  2. $60^o$

  3. $50^o$

  4. $30^o$

Show answer
Correct Option: A
Explanation:

Each exterior angle of a regular polygon of 9 sides
$=\dfrac {360^o}{n}$, where $n=9=\left (\dfrac {360^o}{9}\right )^o=40^o$

If the sum of all interior angles of a convex polygon is 1440, then the number of sides of the polygon is

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. 8

  2. 10

  3. 11

  4. 12

Show answer
Correct Option: B
Explanation:

If n is the number of sides of the polygon, then $(2n -4)\times 90^o = 1440$
or 2n = 20          or          n = 10

An exterior angle of regular polygon is $\displaystyle 12^{\circ}$ the sum of all the interior angles is

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $\displaystyle 4040^{\circ}$

  2. $\displaystyle 5040^{\circ}$

  3. $\displaystyle 6040^{\circ}$

  4. $\displaystyle 7040^{\circ}$

Show answer
Correct Option: B
Explanation:

Given the exterior angle of regular polygon is 12

We know each  exterior angle of regular polygon=$\dfrac{360}{n}$ where n is the sides of polygon
$\dfrac{360}{n}=12\Rightarrow n=30$
we know that interior angle of  regular polygon=$180(n-2)=180(30-2)=5040^{0}$

The measure of the external angle of a regular hexagon is 

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. ${\pi/3}$

  2. ${\pi/4}$,

  3. ${\pi/6}$

  4. None

Show answer
Correct Option: A
Explanation:

$\Rightarrow$ Sum of exterior angles of a regular hexagon $=360^o$

$\Rightarrow$  Number of sides of regular hexagon $=6$
$\Rightarrow$  The measure of the external angle of a regular hexagon $=\dfrac{360^o}{6}=60^o$
In radian $=60^o\times \dfrac{\pi}{180^o}=\dfrac{\pi}{3}$ 

Is it possible to have a regular polygon with measure of each exterior angle as $22^o$?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. not possible

  2. possible

  3. cannot be determined

  4. none of the above

Show answer
Correct Option: A
Explanation:

Since the number of sides of a regular polygon
$=\dfrac {360}{\text {Exterior angle}}$
$\therefore$ The number of sides of a regular polygon
$=\dfrac {360}{22}[\because$ Exterior angle $=22^o$, given]
$=\dfrac {180}{11}$
Which is not a whole number.
$\therefore$ A regular polygon with measure of each exterior angle as $22^o$ is not possible.

The measure of the external angle of a regular octagon is 

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. ${\pi/4}$

  2. ${\pi/6}$

  3. ${\pi/8}$

  4. ${\pi/12}$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  The sum of the exterior angles of regular octagon is $360^o$.

$\Rightarrow$ Number of sides of octagon $=8$
$\Rightarrow$  The measure of the external angles $=\dfrac{360^o}{8}=45^o$
In radian $=45^o\times \dfrac{\pi}{180^o}=\dfrac{\pi}{4}$
$\therefore$  The measure of the external angle of a regular octagon is $\dfrac{\pi}{4}$

Each exterior angle of a regular hexagon is of

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $120^\circ$

  2. $80^\circ$

  3. $100^\circ$

  4. $60^\circ$

Show answer
Correct Option: D

The exterior angle of a regular polygon is one-third of its interior angle. How many sides does the polygon has?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $10$

  2. $8$

  3. $9$

  4. $13$

Show answer
Correct Option: B
Explanation:

Let no of sides of the polygon is $n$ 

Exterior angle will be $\dfrac{360}{n}$
Interior angle will be $\left ( 180-\dfrac{360}{n}\right)$
Exterior angle is $\dfrac{1}{3}$ of the interior angle
$\Rightarrow \dfrac{360}{n}=\dfrac{1}{3} \left (180-\dfrac{360}{n}\right)$
$\Rightarrow n=8$

The number of sides of a regular polygon whose each exterior angle has a measure of $45^o$ is __________.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $4$

  2. $6$

  3. $8$

  4. $10$

Show answer
Correct Option: C
Explanation:

The exterior angle of a regular polygon is $\dfrac{360}{n}$.

Given, $45^\circ$
$\Rightarrow \dfrac{360}{n}=45$

$\Rightarrow n=8$

The measure of each exterior angle of an n-sided regular polygon is $(\dfrac{180^0}{n})$.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A

If the difference between an interior angle of a regular polygon of $\displaystyle \left ( n+1 \right )$ sides and an interior angle of a regular polygon of $n$ sides is $\displaystyle 4^{\circ}$; find the value of $n$. Also, state the difference between their exterior angles.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $\displaystyle n =9$ and difference between exterior angles $\displaystyle 4^{\circ}$

  2. $\displaystyle n =5$ and difference between exterior angles $\displaystyle 22^{\circ}$

  3. $\displaystyle n =11$ and difference between exterior angles $\displaystyle 12^{\circ}$

  4. None of these

Show answer
Correct Option: A
Explanation:
An interior angle of (n + 1) sided regular polygon = $ \dfrac{180^o((n+1) -2)}{(n+1)} $
An interior angle of n sided regular polygon = $ \dfrac{180^o(n-2)}{n} $
Their difference is $ 4^o $
So, $\dfrac{180^o((n+1) -2)}{(n+1)} - \dfrac{180^o(n-2)}{n}= 4^o$
$=> 45 [  \dfrac{(n-1)}{(n+1)} -  \dfrac{(n-2)}{n} ]= 1 $ 
$=> 45 \dfrac{2}{n(n+1)} = 1 $
$=> n^2 + n -90 = 0$
$=> (n-9)(n+10) = 0$
$=> n = 9, -10$ 
Since n should be a positive number. So, $n = 9$

State true or false.
Is it possible to have a regular polygon whose each exterior angle is $\displaystyle \frac{1}{8}$ of a right angle.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Given, a regular polygon whose each exterior angle is $ \dfrac{1}{8}$ of a right angle = $ \dfrac {1}{8} \times 90^o = \dfrac {45^o}{4} $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = \dfrac {45^o}{4}  $
$=> n = 8 $
Since, n should be an integer, so their exist a regular polygon whose each exterior angle is $ \frac{1}{8}$ of a right angle.

Three of the exterior angles of a hexagon are $40^{\circ}$, $51^{\circ}$ and $86^{\circ}$. If each of the remaining exterior angles is $x^{\circ}$, find the value of $x$.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $58$

  2. $61$

  3. $65$

  4. none of the above

Show answer
Correct Option: B
Explanation:

Three of the exterior angles of a hexagon are $ 40^o, 51^o$  and  $86^o $. Each of the remaining exterior angles is $ x^o $.
Sum of all exterior angle of any polygon is $ 360^o $
$ 40^o + 51^o + 86^o + 3 \times x^o = 360^o $
$ => 3 \times x^o = 183^o $
$ => x^o = 61^o $

The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are $\displaystyle (6x-1)^{\circ}, (10x+2)^{\circ}, (8x+2)^{\circ}, (9x-3)^{\circ}, (5x+4)^{\circ}$ and $(12x+6)^{\circ};$. Find each exterior angle.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $41^{\circ}, 62^{\circ}, 58^{\circ}, 60^{\circ}, 39^{\circ} , 90^{\circ}$

  2. $41^{\circ}, 86^{\circ}, 56^{\circ}, 60^{\circ}, 39^{\circ} , 80^{\circ}$

  3. $41^{\circ}, 72^{\circ}, 58^{\circ}, 60^{\circ}, 39^{\circ} , 90^{\circ}$

  4. $41^{\circ}, 82^{\circ}, 60^{\circ}, 60^{\circ}, 36^{\circ} , 100^{\circ}$

Show answer
Correct Option: C
Explanation:

The sum of the exterior angles of any polygon is always equal to 360.
Exterior angles are $\displaystyle (6x-1)^{\circ}, (10x+2)^{\circ}, (8x+2)^{\circ}, (9x-3)^{\circ}, (5x+4)^{\circ}  and  (12x+6)^{\circ} $
Now, 
 $\displaystyle (6x-1)^{\circ}+ (10x+2)^{\circ}+ (8x+2)^{\circ} + (9x-3)^{\circ} + (5x+4)^{\circ} +  (12x+6)^{\circ} = 360^o $
$ => (50x + 10)^o = 360^o $
$ => x = 7 $
Each Exterior angle 
$ => (6x -1)^o = 6 \times 7 -1 =41^o $
$ => (10x +2)^o = 10 \times 7 +2 =72^o $
$ => (8x +2)^o = 8 \times 7 +2 =58^o $
$ => (9x -3)^o = 9 \times 7 -3 =60^o $
$ => (5x +4)^o = 5 \times 7 +4  =39^o $
$ => (12x +6)^o = 12 \times 7 + 6 =90^o $

Two alternate sides of a regular polygon, when produced, meet at a right angle. Find the number of sides of the polygon. 

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $3$

  2. $8$

  3. $2$

  4. $9$

Show answer
Correct Option: B
Explanation:

In a regular polygon all the exterior angles have the same measure. 


When two alternate sides of a polygon are extended a triangle.

If AB, BC and CD are the sides of a regular polygon and AB and CD when produced meet at P forming a right triangle.

Now, in $ \triangle CPB, \angle PCB = \angle PBC = 45^o $

Therefore, exterior angle of the polygon = $ 45^o $
Exterior angle of a regular polygon = $ \dfrac {360^o}{n} $
$=> 45^o = \dfrac {360^o}{n} $
$ => n = 8 $ 
Number of sides of the polygon = $8$

State true or false:
Is it possible to have a regular polygon whose each interior angle is $\displaystyle 175^{\circ}$

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Each interior angle of regular polygon of $n$ sides is given by $\dfrac{180^o(n-2)}{n}$
According to question

$\dfrac{180^o(n-2)}{n} =175^0$
$\Rightarrow 180^o(n-2)=175n$
$\Rightarrow 180n-360^o=175n$
$\Rightarrow 5n=360^o$
$\Rightarrow n=72$
Clearly there is a polygon of sides $72$ whose each interior angle is $175^0$

The sum of the interior angles of a polygon is four times the sum of its exterior angles. Find the number of sides in the polygon.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $10$

  2. $12$

  3. $8$

  4. $7$

Show answer
Correct Option: A
Explanation:

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to $360^o$.
The sum of the interior angles of polygon = $180 (n-2)$
=> $180 (n-2) = 4 \times 360$
=> $n -2 = 8$ 
=> $n =10$ 
Number of sides in the polygon = $10$

There is a regular polygon whose each interior angle is $175^{\circ}$

State true or false.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Given, a polygon whose each interior angles is $ 175^o $
Sum of interior angles of a polygon is =  $ 180^o (n-2) $
Each interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} $
$ \dfrac {180^o (n-2)}{n}  = 175^o $
$  180^o n - 175^o n = 360^o $
$ n = \dfrac {360}{5} $
$ n = 72 $
Since, n (number of sides) is an integer, therefore there exist a polygon whose each interior angles is $ 175^o $

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with 7 sides.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $360^{\circ}$

  2. $340^{\circ}$

  3. $380^{\circ}$

  4. $390^{\circ}$

Show answer
Correct Option: A
Explanation:

No matter what type of polygon, the sum of the exterior angles is always equal to $360^o$.
It does not depends upon number of sides of polygon.  

How many sides does a polygon have if the sum of the measures of its internal angles is five times as large as the sum of the measures of its exterior angles?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $20$

  2. $12$

  3. $15$

  4. $10$

Show answer
Correct Option: B
Explanation:

Sum of measure of Interior angles of a regular polygon is calculated as,
$(n-2)180$ where,
n: Number of sides of a regular polygon.
Sum of exterior angles of a regular polygon always add up to $360^{o}$
$\therefore$ ,$(n-2)180=5(360)$
$\therefore n=12$

Two times the interior angle of a regular polygon is equal to seven times is exterior angle. Find the interior angle of the polygon and the number of sides in it.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $130^{\circ}$ and n $=$ 9

  2. $140^{\circ}$ and n $=$ 9

  3. $160^{\circ}$ and n $=$ 9

  4. $170^{\circ}$ and n $=$ 9

Show answer
Correct Option: B
Explanation:

Two times the interior angle of a regular polygon is equal to seven times is exterior angle.
Each Interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} $
Each Exterior angle of a polygon = $ \dfrac{360^o}{n} $
Now,
$ 2 \times \dfrac {180^o (n-2)}{n} = 7 \times  \dfrac{360^o}{n}  $
$=> n -2 = 7 $
$=> n = 9 $
Number of sides of polygon is 9.
Each Interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} = \dfrac {180^o (9-2)}{9} = 140^o  $

The measurement of each angle of a polygon is $160$$^o$. The number of its sides is ?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $15$

  2. $18$

  3. $20$

  4. $30$

Show answer
Correct Option: B
Explanation:
Given, measure of each angle of a polygon $=160^o$
Exterior angle $= 180^o -$ Interior angle
$= 180^o - 160^o = 20^o$
$\therefore$ Number of sides $= \displaystyle \frac{360^o}{\text{Exterior angle}} = \frac{360}{20} = 18$
Therefore, number of sides of polygon are $18$.

The ratio of the measure of an exterior angle of a regular $7:2$ nonagon to the measure of one of its interior angles is:

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $7:2$

  2. $2:7$

  3. $4:3$

  4. $3:4$

Show answer
Correct Option: B
Explanation:

Let $7a$ be the interior angle

and $2a$ be the exterior angle
Therefore, $ 7a+2a=180^{0}$
$\Rightarrow 9a=180^{0}$
$\Rightarrow a=20^{0}$
So, $2a=2\times 20$
$=40^{0}$
and $7a=7\times 20$
$=140^{0}$
For a regular polygon of $n$ sides, each exterior angle has a measure of $\dfrac{360}{n}$ degrees.

The measure of each interior angle is $140^{0}$.
Since the exterior angle of each angle has measure $40^{0}$, then the number of sides $n$.
$=\dfrac{360}{n}$
$=9$ sides.

A regular polygon is inscribed in a circle. If a side subtends an angle of $30^{\circ}$ at the centre, what is the number of its sides?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $10$

  2. $8$

  3. $6$

  4. $12$

Show answer
Correct Option: D
Explanation:

For a polygon of 'n' sides, the angle subtended at the centre is $ \dfrac {{360}^{o}}{n} $

Given, angle at the centre $ = {30}^{o} $
$ => \dfrac {{360}^{o}}{n}= {30}^{o} $
$ => n = 12 $

Hence, the polygon has $ 12 $ sides.

Exterior angles of a regular polygon is one-third of its interior angle. Find number of sides in polygon.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. 10

  2. 8

  3. 6

  4. 9

Show answer
Correct Option: B
Explanation:
If we take $n$ as the number of sides of polygon and $E$ be the exterior angle and $I$ be the interior angle.

$\Rightarrow$   $E+I=180^\circ$  

According to the given question we get,
$\Rightarrow$  $E=\dfrac{1}{3} I$

$\Rightarrow$  So, $I=3E$

$\therefore$  $E+3E=180^\circ$

$\Rightarrow$  $E = 45^\circ$

$\Rightarrow$  Interior angle  $=135^\circ$

$\Rightarrow$  Interior angle $=\dfrac {(n-2)\times 180}{n}$

$\Rightarrow$  $135n=180n-360$

$\Rightarrow$  $-45=-360$

$\Rightarrow$  $n=8$

$\therefore$  Number of sides in polygon are $8$.

If the interior angle of a regular polygon exceeds the exterior angle by $ \displaystyle 132^{\circ}  $, then the number of sides of the polygon is :

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $15$

  2. $14$

  3. $13$

  4. $12$

Show answer
Correct Option: A
Explanation:

Let the number of sides in the regular polygon be $n$

Thus each interior angle $=$ $\dfrac{(2n-4)\times 90^{\circ}}{n}$
And each exterior angle $=\dfrac{360^{\circ}}{n}$
Lets go according to question:
Therefore, $  \dfrac{(2n-4)\times 90^{\circ}}{n}-\dfrac{360^{\circ}}{n}=132^{\circ}$
$\Rightarrow 180n-360-360=132n$
$\Rightarrow 48n=720$
$\Rightarrow n=\dfrac{720}{48}=15$

Let the  formula relation the exterior angle and number of sides of a polygon be given as $nA = 360$.
The measure $A$, in degrees, of an exterior angle of a regular polygon is related to the number of sides, $n$, of the polygon by the formula above. If the measure of an exterior angle of a regular polygon is greater than $50$, what is the greatest number of sides it can have?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. 5

  2. 6

  3. 7

  4. 8

Show answer
Correct Option: C
Explanation:

Sum of exterior angles for any polynomial is always $360$. 

Since polynomial has $n$ angles, each with exterior angle is $A$, then 
sum of exterior angles will be $nA$ 
Given, $nA = 360$ 
$\therefore A=\dfrac { 360 }{ n }$  
We are given that: $A > 50$ 
$\Rightarrow \dfrac { 360 }{ n } >50$ 
$\Rightarrow 360 > 50n$ 
$\Rightarrow n<\dfrac { 360 }{ 50 }$  
$\Rightarrow n < 7.2$ 
Hence, the greatest number of angles polygon can have is $7$.

If $B$ the exterior angle of a regular polygon of $n-sides$ and $A$ is any constant then $\cos A + \cos (A + B) + \cos (A + 2B) + .... n$ terms is equal to:

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $0$

  2. $\cos A$

  3. $1$

  4. $\dfrac {\sqrt {3}}{2}$

Show answer
Correct Option: A
Explanation:
The sum of exterior angles of a polygon is $ 2\pi$ or $ 360^{\circ}$

If it is a regular polygon

Exterior Angles $ = \dfrac{2\pi}{n} $ 

n is no of sides

According to question

$ B = \dfrac{2\pi}{n}.$

$ \cos A + \cos (A+B) + \cos (A+2B)+ ...n+ terms $

$ = \cos A + \cos\left ( A + \dfrac{2 \pi}{n} \right )+ \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right )+...+ \cos \left ( A+ (n-2) \dfrac{2\pi}{n} \right ) + \cos \left ( A+(n-1) \dfrac{2\pi}{n} \right )$

$ \cos A+ \cos \left ( A+\dfrac{2\pi}{n} \right ) + \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right )+ ...+ \cos \left ( A+n \left ( \dfrac{2\pi}{n} \right )-2\left ( \dfrac{2\pi}{n} \right )\right ) + \cos \left ( A+n \left ( \dfrac{2\pi}{n} \right )-\dfrac{2\pi}{n}\right )$

$ = \cos A + \cos \left ( A +\dfrac{2\pi}{n} \right )+ \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right ) + ...+ \cos \left ( A+2\pi -2\left ( \dfrac{2\pi}{n} \right ) \right ) + \cos \left ( A + 2\pi- \dfrac{2\pi}{n} \right )$

$ = \cos A + \cos \left ( A+\dfrac{2\pi}{n} \right )+ \cos \left ( A+2\left ( \dfrac{2\pi}{n} \right ) \right ) + ...+ \cos\left ( A-2 \left ( \dfrac{2\pi}{n} \right ) \right )+ cos \left ( A - \dfrac{2\pi}{n} \right )$

$ \left \{ \because \cos (2\pi - \theta) = \cos \theta \right \}$

$ = \cos A + \cos (A) \, \cos \left ( \dfrac{2\pi}{n} \right )- \sin(A) . \sin\left ( \dfrac{2\pi}{n} \right )+ \cos A. \cos \dfrac{4\pi}{n}- \sin A $

$ \sin \dfrac{4\pi}{n}+ ...+ \cos A. \cos \dfrac{4\pi}{n}+ \sin A\, \sin \dfrac{4\pi}{n} + \cos A. \cos \dfrac{2\pi}{n} + \sin A . \sin \dfrac{\pi}{n}$

$ = \cos\,A + \cos\,A. \cos\dfrac{2\pi}{n}+ \cos A \,\cos\dfrac{4\pi}{n} + ... \cos A\, \cos \dfrac{4 \pi}{n} + \cos A.\cos \dfrac{2\pi}{n}.$

$ = \cos A \left \{ 1+ \cos\dfrac{2\pi}{n} + \cos \dfrac{4\pi}{n}+ ... \cos\dfrac{4\pi}{n}+ \cos\dfrac{2\pi}{n} \right \}$

$ = \cos A (1-1)$

$ =0 $

Which one of the following statements is not correct?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. if the exterior angle of a regular polygon is $30$ it has $12$ sides

  2. if the interior and exterior angles of a regular polygon are all equal, it is a rectangle

  3. if the exterior angle of a regular polygon is greater than its interior angle, it is an equilateral triangle

  4. in a regular pentagon, the exterior angle is half of the interior angle

Show answer
Correct Option: D
Explanation:

A) Exterior angle of m-gon$=\cfrac { (m180)-(m-2)180 }{ m } $

If $m=12$, 
$\Longrightarrow $ Exterior angle$=30$.
Therefore A is correct.

B) If ABCD is a rectangle,
Interior$=$Exterior angle$={ 90 }^{ 0 }$ .
Therefore B is correct.

C) In equilateral triangle exterior angle ($120$)$>$ interior angle$60$.
Whereas in others it is less than or equal to interior angle.
Therefore C is true.

D) Exterior angle of pentagon$=72$.
Interior angle of pentagon$=108$.
Exterior angle $\neq \cfrac { 1 }{ 2 } $interior angle.
Therefore D is incorrect.

The sum of the exterior angles of a hexagon is?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $360^{\circ}$

  2. $540^{\circ}$

  3. $720^{\circ}$

  4. none of these

Show answer
Correct Option: A
Explanation:

Number of sides in hexagon $=6$
Sum of the interior angles of a polygon$=(n-2)\pi$
$n(Interior\ Angle)=(n-2)\pi$
$\Rightarrow $ Interior Angle $= \dfrac{4}{6}\pi$

Interior Angle $= 120^\circ$
Exterior Angle $=180- $Interior Angle
$\Rightarrow$ Exterior angle $=60^\circ$
Sum of Exterior angle $=6 \times$ Exterior Angle $=360^\circ$

How many sides does a regular polygon have if the measure of an exterior angle is $24^{0}$?

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $14$

  2. $13$

  3. $15$

  4. $18$

Show answer
Correct Option: C
Explanation:

Here, let the number of sides of the polygon be $n$ 

So, the number of exterior angles is $n$ 
Since it is a regular polygon, each exterior angles are equal to one another.
$ \therefore$ The sum of the exterior angles $={ 360 }^{ o }$
$ \therefore$  Each angle $=\theta =\dfrac { { 360 }^{ o } }{ n }$ 
$\Longrightarrow n=\dfrac { { 360 }^{ o } }{ \theta  } $ 
Here $\theta ={ 24 }^{ o } $ 
$ \therefore  n=\dfrac { { 360 }^{ o } }{ { 24 }^{ o } } =15$
Hence, the answer is $15$.

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