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

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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.

No. of Sides $=$ $\frac{1}{2} \Big($ $\frac{sum \ of \ interior \ angles}{90}$ $+4 \Big)=$ $\frac{1}{2}$ $\Big($ $\frac{840^{\circ}}{90}$ $+4 \Big)$ 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.

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

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.

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

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 $= n$ 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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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 $= n$ $interior \ angles =$ $\frac{(2n-4)\times 90^{\circ}}{n}$ 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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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 $= 180^{\circ}-$ 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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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.

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.

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  $=90^{\circ}$

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 $=140^{\circ}$

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.

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

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