Class 8: Polygons – Exercise 29

$\displaystyle \textbf{Question 1: }\text{Calculate the sum of all the interior angles}$
$\displaystyle \text{of a polygon having:}$
$\displaystyle \text{(i) }6\text{ sides}\qquad \text{(ii) }8\text{ sides}$
$\displaystyle \text{(iii) }14\text{ sides}\qquad \text{(iv) }20\text{ sides}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$

$\displaystyle \text{(i) For }n=6,$
$\displaystyle (6-2)\times180^\circ=4\times180^\circ=720^\circ.$
$\displaystyle \therefore \text{Sum of the interior angles}=720^\circ.$

$\displaystyle \text{(ii) For }n=8,$
$\displaystyle (8-2)\times180^\circ=6\times180^\circ=1080^\circ.$
$\displaystyle \therefore \text{Sum of the interior angles}=1080^\circ.$

$\displaystyle \text{(iii) For }n=14,$
$\displaystyle (14-2)\times180^\circ=12\times180^\circ=2160^\circ.$
$\displaystyle \therefore \text{Sum of the interior angles}=2160^\circ.$

$\displaystyle \text{(iv) For }n=20,$
$\displaystyle (20-2)\times180^\circ=18\times180^\circ=3240^\circ.$
$\displaystyle \therefore \text{Sum of the interior angles}=3240^\circ.$

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$\displaystyle \textbf{Question 2: }\text{Find the number of sides of a polygon,}$
$\displaystyle \text{the sum of whose interior angles is:}$
$\displaystyle \text{(i) }540^\circ\qquad \text{(ii) }1080^\circ\qquad\text{(iii) }1980^\circ$
$\displaystyle \text{(iv) }10\text{ right angles}\qquad \text{(v) }16\text{ right angles}$
$\displaystyle \text{(vi) }20\text{ right angles}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle \therefore n=\frac{\text{sum of interior angles}}{180^\circ}+2.$

$\displaystyle \text{(i) Sum of interior angles}=540^\circ.$
$\displaystyle n=\frac{540}{180}+2=3+2=5.$
$\displaystyle \therefore \text{The polygon has }5\text{ sides.}$

$\displaystyle \text{(ii) Sum of interior angles}=1080^\circ.$
$\displaystyle n=\frac{1080}{180}+2=6+2=8.$
$\displaystyle \therefore \text{The polygon has }8\text{ sides.}$

$\displaystyle \text{(iii) Sum of interior angles}=1980^\circ.$
$\displaystyle n=\frac{1980}{180}+2=11+2=13.$
$\displaystyle \therefore \text{The polygon has }13\text{ sides.}$

$\displaystyle \text{(iv) Sum of interior angles}=10\times90^\circ=900^\circ.$
$\displaystyle n=\frac{900}{180}+2=5+2=7.$
$\displaystyle \therefore \text{The polygon has }7\text{ sides.}$

$\displaystyle \text{(v) Sum of interior angles}=16\times90^\circ=1440^\circ.$
$\displaystyle n=\frac{1440}{180}+2=8+2=10.$
$\displaystyle \therefore \text{The polygon has }10\text{ sides.}$

$\displaystyle \text{(vi) Sum of interior angles}=20\times90^\circ=1800^\circ.$
$\displaystyle n=\frac{1800}{180}+2=10+2=12.$
$\displaystyle \therefore \text{The polygon has }12\text{ sides.}$

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$\displaystyle \textbf{Question 3: }\text{The sides of a hexagon are produced in order.}$
$\displaystyle \text{If the exterior angles so obtained are }(3x-5)^\circ,\, (8x+3)^\circ,$
$\displaystyle (7x-2)^\circ,\,(4x+1)^\circ,\,(6x+4)^\circ \text{ and }(2x-1)^\circ,$
$\displaystyle \text{find the value of }x\text{ and the measure of each exterior angle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The sum of the exterior angles of any polygon is }360^\circ.$
$\displaystyle (3x-5)+(8x+3)+(7x-2)+(4x+1)+(6x+4)+(2x-1)=360$
$\displaystyle 30x+0=360$
$\displaystyle \therefore x=12.$
$\displaystyle \text{Substituting }x=12\text{ in the expressions for the angles:}$
$\displaystyle (3x-5)^\circ=(3\times12-5)^\circ=31^\circ.$
$\displaystyle (8x+3)^\circ=(8\times12+3)^\circ=99^\circ.$
$\displaystyle (7x-2)^\circ=(7\times12-2)^\circ=82^\circ.$
$\displaystyle (4x+1)^\circ=(4\times12+1)^\circ=49^\circ.$
$\displaystyle (6x+4)^\circ=(6\times12+4)^\circ=76^\circ.$
$\displaystyle (2x-1)^\circ=(2\times12-1)^\circ=23^\circ.$
$\displaystyle \therefore \text{The exterior angles are }31^\circ,\ 99^\circ,\, 82^\circ,\ 49^\circ,\ 76^\circ\text{ and }23^\circ.$

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$\displaystyle \textbf{Question 4: }\text{Is it possible to have a polygon whose sum}$
$\displaystyle \text{of interior angles is }840^\circ?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle (n-2)\times180^\circ=840^\circ$
$\displaystyle n-2=\frac{840}{180}=\frac{14}{3}$
$\displaystyle n=\frac{14}{3}+2=\frac{20}{3}.$
$\displaystyle \text{Since the number of sides must be a whole number,}$
$\displaystyle \frac{20}{3}\text{ cannot represent the number of sides of a polygon.}$
$\displaystyle \therefore \text{No, such a polygon is not possible.}$

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$\displaystyle \textbf{Question 5: }\text{Is it possible to have a polygon, the sum}$
$\displaystyle \text{of whose interior angles is }7\text{ right angles?}$
$\displaystyle \text{Answer:}$
$\displaystyle 7\text{ right angles}=7\times90^\circ=630^\circ.$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle (n-2)\times180^\circ=630^\circ$
$\displaystyle n-2=\frac{630}{180}=\frac{7}{2}$
$\displaystyle n=\frac{7}{2}+2=\frac{11}{2}=5.5.$
$\displaystyle \text{Since the number of sides must be a whole number,}$
$\displaystyle 5.5\text{ cannot represent the number of sides of a polygon.}$
$\displaystyle \therefore \text{No, such a polygon is not possible.}$

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$\displaystyle \textbf{Question 6: }\text{Is it possible to have a polygon, the sum}$
$\displaystyle \text{of whose interior angles is }14\text{ right angles? If yes,}$
$\displaystyle \text{how many sides does this polygon have?}$
$\displaystyle \text{Answer:}$
$\displaystyle 14\text{ right angles}=14\times90^\circ=1260^\circ.$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle (n-2)\times180^\circ=1260^\circ$
$\displaystyle n-2=\frac{1260}{180}=7$
$\displaystyle n=9.$
$\displaystyle \text{Since }9\text{ is a whole number, such a polygon is possible.}$
$\displaystyle \therefore \text{The polygon has }9\text{ sides.}$

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$\displaystyle \textbf{Question 7: }\text{Find the measure of each interior angle}$
$\displaystyle \text{and each exterior angle of the following regular polygons:}$
$\displaystyle \text{(i) Pentagon}\qquad\text{(ii) Hexagon}\qquad \text{(iii) Heptagon}\qquad\text{(iv) Octagon}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a regular polygon with }n\text{ sides,}$
$\displaystyle \text{each interior angle} =\frac{(n-2)\times180^\circ}{n}.$
$\displaystyle \text{Each exterior angle} =\frac{360^\circ}{n}=180^\circ-\text{each interior angle}.$

$\displaystyle \text{(i) Pentagon: }n=5.$
$\displaystyle \text{Each interior angle} =\frac{(5-2)\times180^\circ}{5}=108^\circ.$
$\displaystyle \text{Each exterior angle} =\frac{360^\circ}{5}=72^\circ.$

$\displaystyle \text{(ii) Hexagon: }n=6.$
$\displaystyle \text{Each interior angle} =\frac{(6-2)\times180^\circ}{6}=120^\circ.$
$\displaystyle \text{Each exterior angle} =\frac{360^\circ}{6}=60^\circ.$

$\displaystyle \text{(iii) Heptagon: }n=7.$
$\displaystyle \text{Each interior angle} =\frac{(7-2)\times180^\circ}{7}=\frac{900^\circ}{7}.$
$\displaystyle \text{Each exterior angle} =\frac{360^\circ}{7}.$

$\displaystyle \text{(iv) Octagon: }n=8.$
$\displaystyle \text{Each interior angle} =\frac{(8-2)\times180^\circ}{8}=135^\circ.$
$\displaystyle \text{Each exterior angle} =\frac{360^\circ}{8}=45^\circ.$

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$\displaystyle \textbf{Question 8: }\text{Find the measure of each interior angle}$
$\displaystyle \text{and each exterior angle of a regular polygon having:}$
$\displaystyle \text{(i) }9\text{ sides}\qquad \text{(ii) }15\text{ sides}\qquad\text{(iii) }24\text{ sides}$
$\displaystyle \text{(iv) }30\text{ sides}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a regular polygon with }n\text{ sides,}$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{n}.$
$\displaystyle \text{Each interior angle} =180^\circ-\text{each exterior angle}.$

$\displaystyle \text{(i) For }n=9,$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{9}=40^\circ.$
$\displaystyle \text{Each interior angle}=180^\circ-40^\circ=140^\circ.$
$\displaystyle \therefore \text{Interior angle}=140^\circ,\quad \text{exterior angle}=40^\circ.$

$\displaystyle \text{(ii) For }n=15,$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{15}=24^\circ.$
$\displaystyle \text{Each interior angle}=180^\circ-24^\circ=156^\circ.$
$\displaystyle \therefore \text{Interior angle}=156^\circ,\quad \text{exterior angle}=24^\circ.$

$\displaystyle \text{(iii) For }n=24,$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{24}=15^\circ.$
$\displaystyle \text{Each interior angle}=180^\circ-15^\circ=165^\circ.$
$\displaystyle \therefore \text{Interior angle}=165^\circ,\quad \text{exterior angle}=15^\circ.$

$\displaystyle \text{(iv) For }n=30,$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{30}=12^\circ.$
$\displaystyle \text{Each interior angle}=180^\circ-12^\circ=168^\circ.$
$\displaystyle \therefore \text{Interior angle}=168^\circ,\quad \text{exterior angle}=12^\circ.$

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$\displaystyle \textbf{Question 9: }\text{Find the number of sides of a regular polygon}$
$\displaystyle \text{each of whose exterior angles is:}$
$\displaystyle \text{(i) }30^\circ\qquad \text{(ii) }36^\circ\qquad\text{(iii) }40^\circ\qquad \text{(iv) }18^\circ$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle \text{number of sides} =\frac{360^\circ}{\text{each exterior angle}}.$

$\displaystyle \text{(i) Each exterior angle}=30^\circ.$
$\displaystyle n=\frac{360^\circ}{30^\circ}=12.$
$\displaystyle \therefore \text{The polygon has }12\text{ sides.}$

$\displaystyle \text{(ii) Each exterior angle}=36^\circ.$
$\displaystyle n=\frac{360^\circ}{36^\circ}=10.$
$\displaystyle \therefore \text{The polygon has }10\text{ sides.}$

$\displaystyle \text{(iii) Each exterior angle}=40^\circ.$
$\displaystyle n=\frac{360^\circ}{40^\circ}=9.$
$\displaystyle \therefore \text{The polygon has }9\text{ sides.}$

$\displaystyle \text{(iv) Each exterior angle}=18^\circ.$
$\displaystyle n=\frac{360^\circ}{18^\circ}=20.$
$\displaystyle \therefore \text{The polygon has }20\text{ sides.}$

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$\displaystyle \textbf{Question 11: }\text{Is it possible to have a regular polygon}$
$\displaystyle \text{whose interior angles measure }130^\circ?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Each exterior angle}=180^\circ-130^\circ=50^\circ.$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle n=\frac{360^\circ}{\text{each exterior angle}}.$
$\displaystyle n=\frac{360^\circ}{50^\circ}=7.2.$
$\displaystyle \text{Since the number of sides must be a whole number,}$
$\displaystyle 7.2\text{ cannot represent the number of sides of a polygon.}$
$\displaystyle \therefore \text{No, such a regular polygon is not possible.}$

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$\displaystyle \textbf{Question 12: }\text{Is it possible to have a regular polygon}$
$\displaystyle \text{whose interior angles measure }1\frac{3}{4} \text{ of a right angle?}$
$\displaystyle \text{Answer:}$
$\displaystyle 1\frac{3}{4}\text{ of a right angle} =\frac{7}{4}\times90^\circ=157.5^\circ.$
$\displaystyle \text{Each exterior angle}=180^\circ-157.5^\circ=22.5^\circ.$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle n=\frac{360^\circ}{\text{each exterior angle}}.$
$\displaystyle n=\frac{360^\circ}{22.5^\circ}=16.$
$\displaystyle \text{Since }16\text{ is a whole number, such a polygon is possible.}$
$\displaystyle \therefore \text{Yes, the regular polygon has }16\text{ sides.}$

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$\displaystyle \textbf{Question 13: }\text{Find the number of sides of a regular polygon}$
$\displaystyle \text{if each interior angle is equal to each exterior angle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Each interior angle}+\text{each exterior angle}=180^\circ.$
$\displaystyle \text{Since the interior angle is equal to the exterior angle,}$
$\displaystyle 2\times\text{each exterior angle}=180^\circ.$
$\displaystyle \therefore \text{Each exterior angle}=90^\circ.$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle n=\frac{360^\circ}{\text{each exterior angle}}.$
$\displaystyle n=\frac{360^\circ}{90^\circ}=4.$
$\displaystyle \therefore \text{The regular polygon has }4\text{ sides.}$

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$\displaystyle \textbf{Question 14: }\text{The ratio between the exterior angle and}$
$\displaystyle \text{the interior angle of a regular polygon is }2:7. \text{ Find the number}$
$\displaystyle \text{of sides of the polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the exterior angle be }2x^\circ \text{ and the interior angle be }7x^\circ.$
$\displaystyle \text{An interior angle and its corresponding exterior angle are supplementary.}$
$\displaystyle 2x+7x=180$
$\displaystyle 9x=180$
$\displaystyle \therefore x=20.$
$\displaystyle \therefore \text{Each exterior angle}=2x^\circ=40^\circ.$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle n=\frac{360^\circ}{\text{each exterior angle}}.$
$\displaystyle n=\frac{360^\circ}{40^\circ}=9.$
$\displaystyle \therefore \text{The polygon has }9\text{ sides.}$

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$\displaystyle \textbf{Question 15: }\text{The sum of all the interior angles of a regular}$
$\displaystyle \text{polygon is twice the sum of its exterior angles. Find the number}$
$\displaystyle \text{of sides of the polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a polygon with }n\text{ sides,}$
$\displaystyle \text{sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle \text{The sum of the exterior angles of any polygon is }360^\circ.$
$\displaystyle \text{Given, sum of interior angles} =2\times\text{sum of exterior angles}.$
$\displaystyle (n-2)\times180^\circ=2\times360^\circ$
$\displaystyle (n-2)\times180=720$
$\displaystyle n-2=4$
$\displaystyle \therefore n=6.$
$\displaystyle \therefore \text{The polygon has }6\text{ sides.}$

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$\displaystyle \textbf{Question 16: }\text{Each exterior angle of a regular polygon is}$
$\displaystyle 22.5^\circ.\text{ Find the number of sides of the polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a regular polygon,}$
$\displaystyle \text{each exterior angle}=\frac{360^\circ}{n}.$
$\displaystyle 22.5^\circ=\frac{360^\circ}{n}$
$\displaystyle n=\frac{360^\circ}{22.5^\circ}$
$\displaystyle \therefore n=16.$
$\displaystyle \therefore \text{The polygon has }16\text{ sides.}$

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$\displaystyle \textbf{Question 17: }\text{One angle of an octagon is }100^\circ \text{ and all}$
$\displaystyle \text{the other seven angles are equal. What is the measure of each}$
$\displaystyle \text{of the equal angles?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let each of the seven equal angles be }x^\circ.$
$\displaystyle \text{Sum of interior angles of an octagon} =(8-2)\times180^\circ=1080^\circ.$
$\displaystyle 100^\circ+7x^\circ=1080^\circ$
$\displaystyle 7x=980$
$\displaystyle \therefore x=140^\circ.$
$\displaystyle \therefore \text{Each of the seven equal angles is }140^\circ.$

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$\displaystyle \textbf{Question 18: }\text{The angles of an octagon are in the ratio}$
$\displaystyle 1:2:3:4:5:6:7:8.\text{ Find the smallest angle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Sum of interior angles of an octagon} =(8-2)\times180^\circ=1080^\circ.$
$\displaystyle \text{Let the angles be }x,\ 2x,\ 3x,\ 4x,\ 5x,\ 6x,\ 7x,\ 8x.$
$\displaystyle x+2x+3x+4x+5x+6x+7x+8x=1080^\circ$
$\displaystyle 36x=1080^\circ$
$\displaystyle \therefore x=30^\circ.$
$\displaystyle \therefore \text{The smallest angle is }30^\circ.$

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$\displaystyle \textbf{Question 19: }\text{Two angles of a polygon are right angles}$
$\displaystyle \text{and each of the other angles is }120^\circ. \text{ Find the number}$
$\displaystyle \text{of sides of the polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of sides of the polygon be }n.$
$\displaystyle \text{Two interior angles are }90^\circ\text{ each and the remaining}$
$\displaystyle (n-2)\text{ interior angles are }120^\circ\text{ each}.$
$\displaystyle \text{Sum of interior angles}=(n-2)\times180^\circ.$
$\displaystyle 2(90^\circ)+(n-2)(120^\circ)=(n-2)(180^\circ)$
$\displaystyle 180+120n-240=180n-360$
$\displaystyle 120n-60=180n-360$
$\displaystyle 300=60n$
$\displaystyle \therefore n=5.$
$\displaystyle \therefore \text{The polygon has }5\text{ sides.}$

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$\displaystyle \textbf{Question 20: }\text{Each interior angle of a regular polygon is}$
$\displaystyle 144^\circ.\text{ Find the interior angle of a regular polygon which has}$
$\displaystyle \text{double the number of sides of the first polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Each exterior angle of the first polygon} =180^\circ-144^\circ=36^\circ.$
$\displaystyle \text{For a regular polygon, }n =\frac{360^\circ}{\text{each exterior angle}}.$
$\displaystyle n=\frac{360^\circ}{36^\circ}=10.$
$\displaystyle \therefore \text{The first polygon has }10\text{ sides.}$
$\displaystyle \text{The second polygon has }2\times10=20\text{ sides.}$
$\displaystyle \text{Each exterior angle of the second polygon} =\frac{360^\circ}{20}=18^\circ.$
$\displaystyle \text{Each interior angle of the second polygon} =180^\circ-18^\circ=162^\circ.$
$\displaystyle \therefore \text{The required interior angle is }162^\circ.$