Tag: polygons

Questions Related to polygons

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 $=360^\circ$