Tag: sum of exterior angles of polygons

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Questions Related to sum of exterior angles of polygons

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

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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $n = 9$
Multiple choice maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 $x$.

  1. $58$
  2. $61$
  3. $65$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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.

  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. $41^{\circ}, 82^{\circ}, 60^{\circ}, 60^{\circ}, 36^{\circ} , 100^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

Two alternate sides of a regular polygon, when produced, meet at a right angle. Find the number of sides of the polygon. 

  1. $3$
  2. $8$
  3. $2$
  4. $9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a regular polygon all the exterior angles have the same measure. 


When two alternate sides of a polygon are extended a triangle.

If AB, BC and CD are the sides of a regular polygon and AB and CD when produced meet at P forming a right triangle.

Now, in $ \triangle CPB, \angle PCB = \angle PBC = 45^o $

Therefore, exterior angle of the polygon = $ 45^o $
Exterior angle of a regular polygon = $ \dfrac {360^o}{n} $
$=> 45^o = \dfrac {360^o}{n} $
$ => n = 8 $ 
Number of sides of the polygon = $8$

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

State true or false:
Is it possible to have a regular polygon whose each interior angle is $\displaystyle 175^{\circ}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Each interior angle of regular polygon of $n$ sides is given by $\dfrac{180^o(n-2)}{n}$
According to question

$\dfrac{180^o(n-2)}{n} =175^0$
$\Rightarrow 180^o(n-2)=175n$
$\Rightarrow 180n-360^o=175n$
$\Rightarrow 5n=360^o$
$\Rightarrow n=72$
Clearly there is a polygon of sides $72$ whose each interior angle is $175^0$

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

The sum of the interior angles of a polygon is four times the sum of its exterior angles. Find the number of sides in the polygon.

  1. $10$
  2. $12$
  3. $8$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to $360^o$.
The sum of the interior angles of polygon = $180 (n-2)$
=> $180 (n-2) = 4 \times 360$
=> $n -2 = 8$ 
=> $n =10$ 
Number of sides in the polygon = $10$

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

There is a regular polygon whose each interior angle is $175^{\circ}$

State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, a polygon whose each interior angles is $ 175^o $
Sum of interior angles of a polygon is =  $ 180^o (n-2) $
Each interior angle of a polygon = $ \dfrac {180^o (n-2)}{n} $
$ \dfrac {180^o (n-2)}{n}  = 175^o $
$  180^o n - 175^o n = 360^o $
$ n = \dfrac {360}{5} $
$ n = 72 $
Since, n (number of sides) is an integer, therefore there exist a polygon whose each interior angles is $ 175^o $

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

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with 7 sides.

  1. $360^{\circ}$
  2. $340^{\circ}$
  3. $380^{\circ}$
  4. $390^{\circ}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

No matter what type of polygon, the sum of the exterior angles is always equal to $360^o$.
It does not depends upon number of sides of polygon.  

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

How many sides does a polygon have if the sum of the measures of its internal angles is five times as large as the sum of the measures of its exterior angles?

  1. $20$
  2. $12$
  3. $15$
  4. $10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum of measure of Interior angles of a regular polygon is calculated as,
$(n-2)180$ where,
n: Number of sides of a regular polygon.
Sum of exterior angles of a regular polygon always add up to $360^{o}$
$\therefore$ ,$(n-2)180=5(360)$
$\therefore n=12$