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

i)  $6$ sides     ii)  $8$ sides     iii)  $14$ sides     iv)  $20$ sides

 Sides $= n$$= n$ Sum of the interior angles $=(2n-4)\times 90^{\circ}$$=(2n-4)\times 90^{\circ}$ i) $6$$6$ $(2\times 6-4)\times 90^{\circ} =720^{\circ}$$(2\times 6-4)\times 90^{\circ} =720^{\circ}$ ii) $8$$8$ $(2\times 8-4)\times 90^{\circ} =1080^{\circ}$$(2\times 8-4)\times 90^{\circ} =1080^{\circ}$ iii) $14$$14$ $(2\times 14-4)\times 90^{\circ} =2160^{\circ}$$(2\times 14-4)\times 90^{\circ} =2160^{\circ}$ iv) $20$$20$ $(2\times 20-4)\times 90^{\circ} =3240^{\circ}$$(2\times 20-4)\times 90^{\circ} =3240^{\circ}$

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

i) $540^{\circ}$            ii)  $1080^{\circ}$            iii)  $1980^{\circ}$

iv)   $10$ right angles              v)  $16$ right angles             vi) $20$ right angles

 Sum of the interior angles $\displaystyle \text{No. of Sides} = \frac{1}{2} \Big( \frac{\text{sum of interior angles}}{90} +4 \Big)$$\displaystyle \text{No. of Sides} = \frac{1}{2} \Big( \frac{\text{sum of interior angles}}{90} +4 \Big)$ i) $540^{\circ}$$540^{\circ}$ $\frac{1}{2}$$\frac{1}{2}$ $\Big($$\Big($ $\frac{540}{90}$$\frac{540}{90}$ $+4 \Big)=5$$+4 \Big)=5$ ii) $1080^{\circ}$$1080^{\circ}$ $\frac{1}{2}$$\frac{1}{2}$ $\Big ($$\Big ($ $\frac{1080}{90}$$\frac{1080}{90}$ $+4 \Big)=8$$+4 \Big)=8$ iii) $1980^{\circ}$$1980^{\circ}$ $\frac{1}{2}$$\frac{1}{2}$ $\Big ($$\Big ($ $\frac{1980}{90}$$\frac{1980}{90}$ $+4 \Big)=13$$+4 \Big)=13$ iv) $10$$10$ right angles $\frac{1}{2}$$\frac{1}{2}$ $\Big ($$\Big ($ $\frac{900}{90}$$\frac{900}{90}$ $+4 \Big)=7$$+4 \Big)=7$ v) $16$$16$ right angles $\frac{1}{2}$$\frac{1}{2}$ $\Big ($$\Big ($ $\frac{1440}{90}$$\frac{1440}{90}$ $+4 \Big)=10$$+4 \Big)=10$ vi) $20$$20$ right angles $\frac{1}{2}$$\frac{1}{2}$ $\Big ($$\Big ($ $\frac{1800}{90}$$\frac{1800}{90}$ $+4 \Big)=12$$+4 \Big)=12$

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

Sum of the interior angles of hexagon $= (2n-4)\times 90^{\circ} =(2\times 6-4)\times 90^{\circ}=720^{\circ}$

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

$\Rightarrow 30x=360^{\circ}$

$\Rightarrow x=12^{\circ}$

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

$(3x-5)^{\circ}=31^{\circ}$               $(8x+3)^{\circ}=99^{\circ}$

$(7x-2)^{\circ}=82^{\circ}$               $(4x+1)^{\circ}=49^{\circ}$

$(6x+4)^{\circ}= 76^{\circ}$               $(2x-1)^{\circ}=23^{\circ}$

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

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

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

Hence not possible.

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

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

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

Hence not possible.

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Question 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?

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

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

Hence  possible.

The number of side $=9$

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Question 7: Find the measure of each angle of a regular polygon:

i)  Pentagon       ii)  Hexagon        iii)  Heptagon       iv)  Octagon

 Polygon Sides $= n$$= n$ Interior angles $\displaystyle = \frac{(2n-4)\times 90^{\circ}}{n}$$\displaystyle = \frac{(2n-4)\times 90^{\circ}}{n}$ Exterior Angle $= 180-$$= 180-$ Interior Angle i) Pentagon $5$$5$ $\displaystyle \frac{(2\times 5-4)\times 90^{\circ}}{5} =108^{\circ}$$\displaystyle \frac{(2\times 5-4)\times 90^{\circ}}{5} =108^{\circ}$ $\displaystyle 180- 108=72^{\circ}$$\displaystyle 180- 108=72^{\circ}$ ii) Hexagon $6$$6$ $\displaystyle \frac{(2\times 6-4)\times 90^{\circ}}{6} =120^{\circ}$$\displaystyle \frac{(2\times 6-4)\times 90^{\circ}}{6} =120^{\circ}$ $\displaystyle 180- 120=60^{\circ}$$\displaystyle 180- 120=60^{\circ}$ iii) Heptagon $7$$7$ $\displaystyle \frac{(2\times 7-4)\times 90^{\circ}}{7} = \frac{900^{\circ}}{7}$$\displaystyle \frac{(2\times 7-4)\times 90^{\circ}}{7} = \frac{900^{\circ}}{7}$ $1\displaystyle 80- \frac{900}{7} = \frac{360^{\circ}}{7}$$1\displaystyle 80- \frac{900}{7} = \frac{360^{\circ}}{7}$ iv) Octagon $8$$8$ $\displaystyle \frac{(2\times 8-4)\times 90^{\circ}}{8} =135^{\circ}$$\displaystyle \frac{(2\times 8-4)\times 90^{\circ}}{8} =135^{\circ}$ $\displaystyle 180- 135=65^{\circ}$$\displaystyle 180- 135=65^{\circ}$

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Question 8: Find the measure of each angle of a regular polygon having:

i)  $9$ sides     ii)  $15$ sides     iii)  $24$ sides     iv)  $30$ sides

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

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

i)  $6$ sides     ii)  $10$ sides     iii)  $15$ sides     iv)  $20$ sides

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

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Question 11: Is it possible to have a regular polygon whose interior angles measure $130^{\circ}$

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

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

Hence  not possible.

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Question 12: Is it possible to have a regular polygon whose interior angles measure measures $1$ $\frac{3}{4}$  of a right angle.

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

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

Hence possible

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Question 13: Find the number of sides of a regular polygon, if its interior angle is equal to exterior angle.

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

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

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

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Question 14: The ratio between the exterior angle and the interior angles is $2:7$. Find the number of sides of the polygon.

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

Let the Exterior Angle $=2x$ and the Interior Angle be  $7x$

$\Rightarrow 2x+7x=180 \ or\ x=20$

Therefore Interior angle $=140^{\circ}$

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

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Question 15: 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.

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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Question 16: Each exterior angle of a regular polygon is $(22.5)^{\circ}$. Find the number of sides of the polygon.

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

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

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Question 17: One angle of an Octagon is $100^{\circ}$. And all the other seven angles are equal. What is the measure of each one of the equal angles?

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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Question 18: The angles of Septagon are in the ratio of $1:2:3:4:5:6:7:8$. Find the smallest angle.

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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Question 19: 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.

Let the number of sides $=n$

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

$\Rightarrow 60+60n=360$

$n=5$

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Question 20: 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.

Let the number of sides of the first polygon   $=n$
$\displaystyle n= \frac{360^{\circ}}{180^{\circ}-Interior Angle}=\frac{360^{\circ}}{180^{\circ}-144} =10$
Therefore  the number of sides of the second  polygon   $=20$
$\displaystyle \text{Interior angle of the second polygon }= \frac{(2n-4)\times 90^{\circ}}{n}= \frac{(2\times 20-4)\times 90^{\circ}}{20} =162$