Mathematics

Polygons and Angles

102 Questions

Polygons and angles focus on calculating interior and exterior angle sums. Questions also cover properties of regular shapes like hexagons and nonagons. This geometry topic is essential for quantitative aptitude sections in major competitive exams.

Interior angle sumsExterior angle sumsRegular polygonsPolygon propertiesTriangle formation

Polygons and Angles Questions

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

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.

  1. $130^{\circ}$ and n $=$ 9
  2. $140^{\circ}$ and n $=$ 9
  3. $160^{\circ}$ and n $=$ 9
  4. $170^{\circ}$ and n $=$ 9
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $

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

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

  1. $15$
  2. $18$
  3. $20$
  4. $30$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $18$.
Multiple choice maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

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:

  1. $7:2$
  2. $2:7$
  3. $4:3$
  4. $3:4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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?

  1. $10$
  2. $8$
  3. $6$
  4. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. 10

  2. 8

  3. 6

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $8$.

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

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 :

  1. $15$
  2. $14$
  3. $13$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$
$\Rightarrow 180n-360-360=132n$
$\Rightarrow 48n=720$
$\Rightarrow n=\dfrac{720}{48}=15$

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

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?

  1. 5

  2. 6

  3. 7

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $7$.

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

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:

  1. $0$
  2. $\cos A$
  3. $1$
  4. $\dfrac {\sqrt {3}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $
Multiple choice maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

Which one of the following statements is not correct?

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

The sum of the exterior angles of a hexagon is?

  1. $360^{\circ}$
  2. $540^{\circ}$
  3. $720^{\circ}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=360^\circ$

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

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

  1. $14$
  2. $13$
  3. $15$
  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $15$.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $z _{1}$ and $\bar {z} _{1}$ represent adjacent of a regular polygon of $n$ sides with centre at the origin & if $\dfrac{Im\ z _{1}}{Re\ z _{1}}=\sqrt{2}-1$ then the value of $n$ is equal to:

  1. $8$
  2. $12$
  3. $16$
  4. $24$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a regular polygon with n sides, the vertices are exp(i * 2pi * k / n). For adjacent vertices z1 and z2, the angle is 2pi/n. If z1 = r(cos theta + i sin theta), then Im z1 / Re z1 = tan theta = sqrt(2)-1. This corresponds to theta = pi/8. Since the angle between adjacent vertices is 2pi/n, and the symmetry relates to the origin, n=8.

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

${ T } _{ m }$ denotes the number of triangles that can be formed with the vertices of a regular polygon of m sides. If ${ { T } _{ m+1 } }-{ { T } _{ m } }=15,$ then $m=$

  1. $3$
  2. $6$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

T_m = mC3. The condition T_{m+1} - T_m = 15 becomes (m+1)C3 - mC3 = 15, which simplifies to mC2 = 15. Solving m(m-1)/2 = 15 gives m^2 - m - 30 = 0, so (m-6)(m+5)=0. Thus m=6.