Question 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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Question 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) $20 \ right \ angles$ $\frac{1}{2} (\frac{1800}{90}+4)=12$

Question 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.

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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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 $=$ $\frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}$ $+4)=$ $\frac{1}{2} (\frac{840^{\circ}}{90}$ $+4)$ which is not an integer.

Hence not possible.

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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 $=$ $\frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}$ $+4)=$ $\frac{1}{2} (\frac{7\times 90^{\circ}}{90}$ $+4) =5.5$

Hence not possible.

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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 $=$ $\frac{1}{2} (\frac{sum \ of \ interior \ angles}{90}$ $+4)=$ $\frac{1}{2} (\frac{14\times 90^{\circ}}{90}$ $+4) =9$

Hence  possible.

The number of side $=9$

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Question 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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Question 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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Question 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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Question 10: Is it possible to have a regular polygon whose interior angles measure 130°

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 11: Is it possible to have a regular polygon whose interior angles measure measures 1 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 12: 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 13: 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 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.

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 15: Each exterior angle of a regular polygon is $22.5$. 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 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?

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 17: 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 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.

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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Question 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.

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 $=$ $\frac{(2n-4)\times 90^{\circ}}{n}= \frac{(2\times 20-4)\times 90^{\circ}}{20}$ $=162$