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.

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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Question 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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Question 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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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?

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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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°

Answer:

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.

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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Question 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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Question 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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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.

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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Question 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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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?

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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Question 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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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.

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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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.

Answer:

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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