Tag: polygons

Questions Related to polygons

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 $

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

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

  1. $3$

  2. $8$

  3. $2$

  4. $9$

Show answer
Correct Option: B
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 = $8$

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

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:

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 $175^0$

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.

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

  1. $10$

  2. $12$

  3. $8$

  4. $7$

Show answer
Correct Option: A
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 = $10$

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

State true or false.

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 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 $

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

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

  1. $360^{\circ}$

  2. $340^{\circ}$

  3. $380^{\circ}$

  4. $390^{\circ}$

Show answer
Correct Option: A
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.  

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?

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

  1. $20$

  2. $12$

  3. $15$

  4. $10$

Show answer
Correct Option: B
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 $360^{o}$
$\therefore$ ,$(n-2)180=5(360)$
$\therefore n=12$