Class 8: Polygons – Exercise 29 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

	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

Answer:

	Sum of the interior angles	$\displaystyle \text{No. of Sides} = \frac{1}{2} \Big( \frac{\text{sum of interior angles}}{90} +4 \Big)$
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.

Answer:

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 $840^{\circ}$

Answer:

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.

Answer:

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?

Answer:

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

Answer:

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

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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

Answer:

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

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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

Answer:

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

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

Answer:

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.

Answer:

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.

Answer:

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

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.

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 $=140^{\circ}$

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

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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.

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

Answer:

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

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

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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?

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 18: 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 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.

Answer:

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.