Tag: maths

Questions Related to maths

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. ${\pi/12}$

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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. $60^\circ$

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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

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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 $