Q.1. Calculate the sum of all the interior angles of a polygon having:

Sides = n

Sum \ of \ the \ interior \ angles =(2n-4)\times 90^{\circ}

(i)

6 (2\times 6-4)\times 90^{\circ} =720^{\circ} 

(ii)

8

(2\times 8-4)\times 90^{\circ} =1080^{\circ} 

(iii)

14

(2\times 14-4)\times 90^{\circ} =2160^{\circ} 

(iv) 20

(2\times 20-4)\times 90^{\circ} =3240^{\circ} 

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Q.2. Find the number of sides of a polygon, the sum of whose interior angles is:

Sum \ of \ the \ interior \ angles 

No. \ of \ Sides= \frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}+4) 
(i)

540^{\circ} 

\frac{1}{2}(\frac{540}{90}+4)=5 

(ii)

1080^{\circ} 

\frac{1}{2} (\frac{1080}{90}+4)=8 

(iii)

1980^{\circ} 

\frac{1}{2} (\frac{1980}{90}+4)=13 

(iv)

10\ right \ angles 

\frac{1}{2} (\frac{900}{90}+4)=7 

(v)

16 \ right \ angles 

\frac{1}{2} (\frac{1440}{90}+4)=10 

(vi) 20 \ right \ angles 

\frac{1}{2} (\frac{1800}{90}+4)=12 

m9xQ.3. The sides of a hexagon are produced in order. If the measure of the exterior angles so obtained are (3x-5), (8x+3), (7x-2), (4x+1), (6x+4)and (2x-1) . Find the value of x and the measure of each exterior angle of the hexagon.

Answer:

Sum of the interior angles of hexagon = (2n-4)\times 90^{\circ} 

=(2\times 6-4)\times 90^{\circ}=720^{\circ} 

(3x-5)+(8x+3)+(7x-2)+(4x+1)+(6x+4)+ (2x-1)=360^{\circ} 

30x=360^{\circ} 

x=12^{\circ} 

Now substitute the value of x   in the expressions of all the sides we get:

(3x-5)=31^{\circ} 

(8x+3)=99^{\circ} 

(7x-2)=82^{\circ} 

(4x+1)=49^{\circ} 

(6x+4)= 76^{\circ} 

(2x-1)=23^{\circ} 

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Q.4. Is it possible to have a polygon whose sum of interior angles is 840^{\circ}  

Answer:

No.  Let us calculate the number of sides of this polygon.

No. of Sides = \frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}+4)= \frac{1}{2} (\frac{840^{\circ}}{90}+4)    which is not an integer.

Hence not possible.

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Q.5. Is it possible to have a polygon, the sum of whose interior angles is 7 right angles.

Answer:

No.  Let us calculate the number of sides of this polygon.

No. of Sides = \frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}+4)= \frac{1}{2} (\frac{7\times 90^{\circ}}{90}+4)  =5.5 

Hence not possible.

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Q.6. Is it possible to have a polygon, the sum of whose interior angles is 14 right angles. If yes, how many sides does this polygon have?

Answer:

Yes.  Let us calculate the number of sides of this polygon.

No. of Sides = \frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}+4)= \frac{1}{2} (\frac{14\times 90^{\circ}}{90}+4)  =9 

Hence  possible.

The number of side =9 

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Q.7. Find the measure of each angle of a regular polygon

Polygon  Sides \\= n  interior \ angles =  \frac{(2n-4)\times 90^{\circ}}{n}  Exterior Angle = 180- Interior Angle 
(i) Pentagon  5  \frac{(2\times 5-4)\times 90^{\circ}}{5}=108^{\circ}  180- 108=72^{\circ} 
(ii) Hexagon  6  \frac{(2\times 6-4)\times 90^{\circ}}{6}=120^{\circ}  180- 120=60^{\circ} 
(iii) Heptagon  7  \frac{(2\times 7-4)\times 90^{\circ}}{7}=  \frac{900^{\circ}}{7}  180-  \frac{900}{7}=\frac{360^{\circ}}{7} 
(iv) Octagon  8  \frac{(2\times 8-4)\times 90^{\circ}}{8}=135^{\circ}  180- 135=65^{\circ} 

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Q.8. Find the measure of each angle of a regular polygon having

Sides \\= n  interior \ angles =  \frac{(2n-4)\times 90^{\circ}}{n}  Exterior Angle = 180- Interior Angle 
(i) 9  \frac{(2\times 9-4)\times 90^{\circ}}{9}=140^{\circ}  180- 140=40^{\circ} 
(ii) 15  \frac{(2\times 15-4)\times 90^{\circ}}{15}=156^{\circ}  180- 156=24^{\circ} 
(iii) 24  \frac{(2\times 24-4)\times 90^{\circ}}{24}= 165^{\circ}  180- 165=25^{\circ} 
(iv) 30  \frac{(2\times 30-4)\times 90^{\circ}}{30}=168^{\circ}  180- 168=22^{\circ} 
(v) 6  \frac{(2\times 6-4)\times 90^{\circ}}{6}=120^{\circ}  180- 120=60^{\circ} 
(vi) 10  \frac{(2\times 10-4)\times 90^{\circ}}{10}=144^{\circ}  180- 144=36^{\circ} 
(vii)  20  \frac{(2\times 20-4)\times 90^{\circ}}{20}=162^{\circ}  180- 162=28^{\circ} 

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Q.9. Find the number of sides of a regular polygon each of whose exterior angles are:

Exterior \\ angle  Interior\ angles \\= 180^{\circ}-Exterior Angle  n=  \frac{360^{\circ}}{(180^{\circ}-Interior Angle)} 
(i) 30  180-30=150^{\circ}  n=  \frac{360}{180-150}=12 
(ii) 36  180-36=144^{\circ}  n= \frac{360}{180-144}=10 
(iii) 40  180-40=140^{\circ}  n= \frac{360}{180-140}=9 
(iv) 18  180-18=162^{\circ}  n= \frac{360}{180-162}=20 

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Q.10. Is it possible to have a regular polygon whose interior angles measure 130°

Answer:m22x

No.  Let us calculate the number of sides of this polygon.

No. of Sides n=  \frac{360^{\circ}}{(180^{\circ}-Interior Angle)} = \frac{360^{\circ}}{(180^{\circ}-130^{\circ})}=7.2 

Hence  not possible.

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Q.11. Is it possible to have a regular polygon whose interior angles measure measures 1 3/4  of a right angle.

Answer:

Yes.  Let us calculate the number of sides of this polygon.

No. of Sides n=  \frac{360^{\circ}}{(180^{\circ}-Interior Angle)} = \frac{360^{\circ}}{(180^{\circ}-157.5^{\circ})}=16

Hence possible

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Q.12. Find the number of sides of a regular polygon, if its interior angle is equal to exterior angle.

Answer:

 Interior \ angles = 180^{\circ}-Exterior \ Angle

This means that each of the interior and the exterior angles  =90^{\circ}

No. of Sides n=  \frac{360^{\circ}}{(180^{\circ}-Interior Angle)} = \frac{360^{\circ}}{(180^{\circ}-90^{\circ})}=4

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Q.13. The ratio between the exterior angle and the interior angles is 2:7. Find the number of sides of the polygon.

Answer:

 Interior \ angles = 180^{\circ}-Exterior \ Angle

Let the Exterior Angle =2x and the Interior Angle be  7x \Rightarrow 2x+7x=180 \ or\  x=20

Therefore Interior angle =140^{\circ} 

No. of Sides (n)=  \frac{360^{\circ}}{180^{\circ}-140^{\circ}}= 9

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Q.14. The sum of all the interior angles of a regular polygon is twice the sum of exterior angles. Find the number of sides of the polygon.

Answer:

Sum of Interior Angles =(2n-4)\times 90^{\circ}

Sum of Exterior Angles =360^{\circ}

Given Sum of Interior Angles =2\times (Sum \ of \ Exterior \ Angles) 

\Rightarrow (2n-4)\times 90^{\circ} =720^{\circ} or

n=6

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Q.15. Each exterior angle of a regular polygon is 22.5 . Find the number of sides of the polygon.

Answer:

 Interior \ angles = 180^{\circ}-Exterior \ Angle  =180-22.5=157.5^{\circ}

No. of Sides n=  \frac{360^{\circ}}{(180^{\circ}-Interior Angle)} = \frac{360^{\circ}}{(180^{\circ}-157.5^{\circ})}=16

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Q.16. One angle of an Octagon is 100. And all the other seven angles are equal. What is the measure of each one of the equal angles?

Answer:

One angle given =100^{\circ}

Let each of the equal angles =x

Sum of Interior Angles =(2n-4)\times 90^{\circ} =(2\times 8-4)\times 90^{\circ}=1080^{\circ}

\Rightarrow 100+7x=1080 or x= 140^{\circ}

Each of the equal angles =140^{\circ}

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Q.17. The angles of Septagon are in the ratio of 1:2:3:4:5:6:7:8 . Find the smallest angle.

Answer:

Sum of Interior Angles =(2n-4)\times 90^{\circ}= (2\times 8-4)\times 90^{\circ}=1080^{\circ}

Let the angles be 1x, 2x, 3x, 4x, 5x, 6x, 7x, 8x

Therefore 1x+ 2x+ 3x+ 4x+ 5x+ 6x+ 7x+ 8x=1080^{\circ} or x=30

Hence the angles are 30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}, 210^{\circ} and 240^{\circ}.

The smallest angle is 30^{\circ}

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Q.18. Two angles of a polygon are right angles and each of the other angles is 120^{\circ} . Find the number of sides of the polygon.

Answer:

Let the number of sides =n

Sum of Exterior Angles =2\times 90^{\circ}+(n-2)\times 60

\Rightarrow 60+60n=360

or n=5

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Q.19. Each interior angle of a regular polygon is 144. Find the interior angle of a polygon, which has double the number of sides as the first polygon.

Answer:m33x

Let the number of sides of the first polygon   =n

  n=  \frac{360^{\circ}}{180^{\circ}-Interior Angle}=\frac{360^{\circ}}{180^{\circ}-144}=10

Therefore  the number of sides of the second  polygon   =20

Interior angle of the second polygon =  \frac{(2n-4)\times 90^{\circ}}{n}=  \frac{(2\times 20-4)\times 90^{\circ}}{20}=162

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