Tag: maths

Questions Related to maths

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

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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. $12$

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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 $15$.

State true or false
A  triangle ABC exists such that
$\left(b+c+a\right)\left(b+c-a\right)=5bc$
maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. True

  2. False

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Correct Option: B

Which polygon has no diagonals

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

  1. A triangle

  2. A rectangle

  3. A square

  4. A rhombus

Show answer
Correct Option: A
Explanation:

A triangle has no polygons.

Number of diagonals in a polygon $= \cfrac{n \left( n - 3 \right)}{2}$
Here $n$ is the no. of sides.
Since a triangle has $3$ sides, the number of diagonal it has will be $0$.

If in a $\triangle ABC,{a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2},$ where $R=$ circumradius,then the triangle is

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

  1. equilateral

  2. isosceles

  3. right angled

  4. none of these

Show answer
Correct Option: C
Explanation:

Since, ${a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2}$
Using sine rule, we have
${\left(2R\sin{A}\right)}^{2}+{\left(2R\sin{B}\right)}^{2}+{\left(2R\sin{C}\right)}^{2}=8{R}^{2}$
$\Rightarrow 4{R}^{2}\left[{\left(\sin{A}\right)}^{2}+{\left(\sin{B}\right)}^{2}+{\left(\sin{C}\right)}^{2}\right]=8{R}^{2}$
$\Rightarrow {\sin}^{2}A+{\sin}^{2}B+{\sin}^{2}C=2$
$\Rightarrow 1-{\cos}^{2}A+{\sin}^{2}B+1-{\cos}^{2}C=2$
$\Rightarrow -{\cos}^{2}A+{\sin}^{2}B-{\cos}^{2}C=2-2=0$
$\Rightarrow {\cos}^{2}A-{\sin}^{2}B+{\cos}^{2}C=0$
$\Rightarrow \cos{\left(A+B\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
Since, $A+B+C=\pi \Rightarrow A+B=\pi-C$
$\Rightarrow \cos{\left(\pi-C\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  -\cos{C}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  \cos{C}\left[-cos{\left(A-B\right)}+\cos{C}\right]=0$
$\Rightarrow \cos{C}=0, -cos{\left(A-B\right)}+\cos{C}=0$
$\Rightarrow C=\dfrac{\pi}{2}$ or $-cos{\left(A-B\right)}+\cos{\left(\pi-\left(A+B\right)\right)}=0$
$\Rightarrow -cos{\left(A-B\right)}-cos{\left(A+B\right)}=0$
Using transformation angle formula, we have
$\Rightarrow 2\cos{A}\cos{B}=0$
$\therefore \cos{A}=0,\cos{B}=0$
Hence $\angle{A}=\dfrac{\pi}{2} $or $\angle{B}=\dfrac{\pi}{2}, $ or  $\angle{C}=\dfrac{\pi}{2}$