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 area of complex plane figures 2d and 3d figures

A polygon has 44 diagonals, The number of its sides is

  1. 11

  2. 10

  3. 8

  4. 7
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Number of diagonals in a polygon $=\cfrac{n(n-3)}2$
$\implies 44=\cfrac{n(n-3)}2$
$\implies n^2-3n-88=0$
$\implies (n-11)(n-8)=0$
$\implies n=11$ or $n=-8$
Therefore, number of sides in a polygon $=11.$
Hence, A is the correct option.
Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The measure of maximum possible exterior angle in a regular polygon is 

  1. $70^o$
  2. $60^o$
  3. $90^o$
  4. $120^o$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Exterior angle of regular polygon = 180-interior angle
exterior angle is maximum when interior angle is minimum.
And we have minimum interior angle for regular triangle that is 60 degree..
So maximum exterior angle will be 180-60=120
So correct answer is Option D
Multiple choice statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

The number of rectangles that can be obtained by joining four of the twelve vertices of a $12$ sided regular polygon is

  1. $66$
  2. $30$
  3. $24$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The first vertex can be choosed in $12$ ways and diagonally opposite to it is $1$ vertex. Now for $3rd$ vertex we have $10$ choices and for $4th$ $1.$
However, each rectangle is counted $8$ times.
$\therefore$ No. of ways $=\dfrac{12\times1\times10\times1}{8}$  $=15$ ways.
Hence, the answer is $15.$
Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

Inscribe a regular pentagon in a circle of radius $3\ cm$. The interior angles of the pentagon are:

  1. $54^\circ$
  2. $60^\circ$
  3. $162^\circ$
  4. $108^\circ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that internal angle of regular pentagon is $\cfrac{(n-2)}{n}180^{\circ}$ where n = number of sides.

Here, n = 5.
So, interior angle is $\cfrac{(5-2)}{5}180^{\circ} = 108^{\circ}$

So correct answer is option D

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons
A polygon has $n$ sides. If all the sides and all the angles are same then this polygon is called a regular polygon. Let ${A} _{1},{A} _{2},{A} _{3},...{A} _{n}$ be a regular polygon of $n$ sides. Let $R$ be the radius of the circumscribed circle of a regular polygon and $r$ be the radius of the inscribed circle of a regular polygon.
If ${A} _{1}{A} _{2}={A} _{2}{A} _{3}={A} _{3}{A} _{4}=...={A} _{n}{A} _{1}=a$

Based on the above information, answer the question:

The area of a regular polygon of $n$ sides is

  1. $\dfrac{n{R}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  2. $n{R}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
  3. $\dfrac{n{r}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  4. $n{r}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The area of a regular polygon of n sides is (where r is inradius, R is circumradius, and a is side of the triangle)

  1. $\displaystyle \frac{nR^{2}}{2}\sin \left ( \frac{2\pi }{n} \right )$
  2. $\displaystyle nr^{2}\tan \left( \frac{\pi }{n} \right )$
  3. $\displaystyle \frac{na^{2}}{4}\cot \frac{\pi }{n} $
  4. $\displaystyle nR^{2}\tan(\frac {\pi}{n})$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Area of the regular polygon will be 
$=\dfrac{nR^{2}}{2}sin(\dfrac{2\pi}{n})$.
Now 
$R=\dfrac{s}{2sin(\dfrac{\pi}{n})}$
Hence
$A=\dfrac{ns^{2}}{8sin^{2}\dfrac{\pi}{n}}.2sin(\dfrac{\pi}{n}).cos(\dfrac{\pi}{n})$

$=\dfrac{ns^{2}}{4}.cot(\dfrac{\pi}{n})$. where s is the side of the polygon.

$=nr^{2}.tan(\dfrac{\pi}{n})$ where r is the incentre.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The area of a regular polygon of $2n$ sides inscribed in a circle is given by?

  1. The geometric mean of the areas of the inscribed and circumscribed polygons of $n$ sides.
  2. The arithmetic mean of the areas of the inscribed and circumscribed polygons of $n$ sides.
  3. The harmonic mean of the areas of the inscribed and circumscribed polygons of $n$ sides.

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $a$ be the radius of the circle 


Then,$\displaystyle s _{1}= $ Area of regular polygon of n sides inscribed in the circle $\displaystyle =\frac{1}{2}na^{2}\sin\left ( \frac{2\pi }{n} \right )$

$\displaystyle s _{2}= $  Area of regular polygon of n sides circumscribing in the circle $\displaystyle  = na^{2}\tan \frac{\pi }{n}$

$\displaystyle s _{3}= $ Area of regular polygon of 2n sides inscribed in the circle $\displaystyle  = na^{2}\tan \frac{\pi }{n}$ 

[replacing $n$ by $2n$ is $\displaystyle {(S _{1}}$]

$\displaystyle \therefore $ Geometric mean of $\displaystyle {S _{1}}$ and 

$\displaystyle {S _{2}}$ $\displaystyle = \sqrt{(S _{1}S _{2})}= na^{2}\sin\left ( \frac{\pi }{n}\right ) = S _{3}$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of the radii of inscribed and circumscribed circles of an n sided regular polygon of side $'a'$ is

  1. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/2x} + \cot \frac{\pi}{x} \right )$
  2. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{2x} \right )$
  3. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If ${A} _{1}{A} _{2}{A} _{3}...{A} _{n}$ be a regular polygon of $n$ sides and 
$\dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}},$then

  1. $n=5$
  2. $n=6$
  3. $n=7$

  4. none of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If radius of circle is $r$ then 
${A} _{1}{A} _{2}=2r\sin{\left(\dfrac{\pi}{n}\right)}$
${A} _{1}{A} _{3}=2r\sin{\left(\dfrac{2\pi}{n}\right)}$
${A} _{1}{A} _{4}=2r\sin{\left(\dfrac{3\pi}{n}\right)}$
$\because \dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}}$
$\Rightarrow \dfrac{1}{2r\sin{\left(\dfrac{\pi}{n}\right)}}=\dfrac{1}{2r\sin{\left(\dfrac{2\pi}{n}\right)}}+\dfrac{1}{2r\sin{\left(\dfrac{3\pi}{n}\right)}}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{3\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}+\sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\left[\sin{\left(\dfrac{3\pi}{n}\right)}-\sin{\left(\dfrac{\pi}{n}\right)}\right]=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}.2\cos{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow 2\sin{\left(\dfrac{2\pi}{n}\right)}\cos{\left(\dfrac{2\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
Using multiple angle formula, $2\sin{A}\cos{A}=\sin{2A}$ we get
$\sin{\left(\dfrac{4\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
$\therefore \dfrac{4\pi}{n}=r+{\left(-1\right)}^{r}\dfrac{3}{n}$ for $r=1,n=7$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of inradius and circumradius of incircle and circumcircle of a regular polygon of side $n$ is

  1. $\dfrac {a}{4}\cot \dfrac {\pi}{2n}$
  2. $a\cot \dfrac {\pi}{n}$
  3. $\dfrac {a}{2} \cot \dfrac {\pi}{2n}$
  4. $a\cot \dfrac {\pi}{2n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$r + R = \dfrac {a}{2}\cot \dfrac {\pi}{n} + \dfrac {a}{2}cosec \dfrac {\pi}{n}$
$= \dfrac {a}{2} \left (\dfrac {1 + \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right ) = \dfrac {a}{2} \dfrac {2\cos^{2} \dfrac {\pi}{2n}}{2\sin \dfrac {\pi}{2n}\cdot \cos \dfrac {\pi}{2n}}$
$= \dfrac {a}{2} \cot \dfrac {\pi}{2n}$.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of the radii of inscribed and circumscribed circles of an $n$ -sided regular polygon with side equal to one unit is?

  1. $\displaystyle \frac{1}{2}\cot \frac{\pi }{2n}$
  2. $\displaystyle \cot \frac{\pi }{2n}$
  3. $\displaystyle \cot \frac{\pi }{n}$
  4. $\displaystyle \frac{1}{2}\tan \frac{\pi }{2n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the figure:
Side of polygon $(AB)=1$
$AO=\dfrac { 1 }{ 2 } $
$\angle O=\dfrac { \pi  }{ 2n } $

In right angled $\triangle COA$ :
$\sin { O } =\dfrac { AC }{ AO } $
$\Rightarrow \sin { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2R } $       ..(1)

$\tan { O } =\dfrac { AC }{ CO } $
$\Rightarrow \tan { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2r } $       ...(2)

From (1) and (2)
$R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ \sin { \dfrac { \pi  }{ n }  }  } +\dfrac { 1 }{ \tan { \dfrac { \pi  }{ n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1+\cos { \dfrac { \pi  }{ n }  }  }{ \sin { \dfrac { \pi  }{ n }  }  }  \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 2\cos ^{ 2 }{ \dfrac { \pi  }{ 2n }  }  }{ 2\cos { \dfrac { \pi  }{ 2n }  } \sin { \dfrac { \pi  }{ 2n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \cot { \dfrac { \pi  }{ 2n }  } $

Ans: A

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

Let ${T _n}$ be the number of all possible triangles formed by joining vertices of an $n$-sided regular polygon. If ${T _{n + 1}} - {T _n} = 10$. then the value of $n$ is 

  1. $7$
  2. $5$
  3. $10$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $T _n=\ ^nC _3$

$T _{n+1}=\ ^{n+1}C _3$
therefore,
$T _{n+1}-T _n=\ ^{n+1}C _3-\ ^nC _3=10$
$\Rightarrow\ ^nC _2+\ ^nC _3-\ ^nC _3=10$        $[\because\ ^nC _r+\ ^nC _{r-1}=\ ^{n+1}C _r]$
$\Rightarrow\ ^nC _2=10$
$\Rightarrow\ ^nC _2=\ ^5C _2$
$\Rightarrow n=5$