\displaystyle \textbf{Question 1: }\text{Write in degrees the sum of all interior angles of a:}
\displaystyle \text{(i) Hexagon \hspace{1cm} (ii) Septagon}
\displaystyle \text{(iii) Nonagon \hspace{0.8cm} (iv) 15-gon}
\displaystyle \textbf{Answer:}
\displaystyle \textbf{(i) Hexagon}
\displaystyle \text{Sum of interior angles of a hexagon}=(2n-4)\text{ right angles}.
\displaystyle =(2\times6-4)\times90^\circ=(12-4)\times90^\circ.
\displaystyle =8\times90^\circ=720^\circ.
\displaystyle \therefore \text{The sum of the interior angles of a hexagon is }720^\circ.
\displaystyle \textbf{(ii) Septagon}
\displaystyle \text{Sum of interior angles of a septagon}=(2n-4)\text{ right angles}.
\displaystyle =(2\times7-4)\times90^\circ=(14-4)\times90^\circ.
\displaystyle =10\times90^\circ=900^\circ.
\displaystyle \therefore \text{The sum of the interior angles of a septagon is }900^\circ.
\displaystyle \textbf{(iii) Nonagon}
\displaystyle \text{Sum of interior angles of a nonagon}=(2n-4)\text{ right angles}.
\displaystyle =(2\times9-4)\times90^\circ=(18-4)\times90^\circ.
\displaystyle =14\times90^\circ=1260^\circ.
\displaystyle \therefore \text{The sum of the interior angles of a nonagon is }1260^\circ.
\displaystyle \textbf{(iv) 15-gon}
\displaystyle \text{Sum of interior angles of a 15-gon}=(2n-4)\text{ right angles}.
\displaystyle =(2\times15-4)\times90^\circ=(30-4)\times90^\circ.
\displaystyle =26\times90^\circ=2340^\circ.
\displaystyle \therefore \text{The sum of the interior angles of a 15-gon is }2340^\circ.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the measure, in degrees, of each interior angle of a regular:}
\displaystyle \text{(i) Pentagon \hspace{1cm} (ii) Octagon}
\displaystyle \text{(iii) Decagon \hspace{0.8cm} (iv) 16-gon}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each interior angle of a regular polygon of }n\text{ sides is}
\displaystyle \frac{2n-4}{n}\text{ right angles.}
\displaystyle \textbf{(i) Pentagon}
\displaystyle \text{Each interior angle of a pentagon}=\frac{2n-4}{n}\text{ right angles.}
\displaystyle =\frac{2\times5-4}{5}\times90^\circ=\frac{10-4}{5}\times90^\circ.
\displaystyle =\frac{6}{5}\times90^\circ=108^\circ.
\displaystyle \therefore \text{Each interior angle of a regular pentagon is }108^\circ.
\displaystyle \textbf{(ii) Octagon}
\displaystyle \text{Each interior angle of an octagon}=\frac{2n-4}{n}\text{ right angles.}
\displaystyle =\frac{2\times8-4}{8}\times90^\circ=\frac{16-4}{8}\times90^\circ.
\displaystyle =\frac{12}{8}\times90^\circ=135^\circ.
\displaystyle \therefore \text{Each interior angle of a regular octagon is }135^\circ.
\displaystyle \textbf{(iii) Decagon}
\displaystyle \text{Each interior angle of a decagon}=\frac{2n-4}{n}\text{ right angles.}
\displaystyle =\frac{2\times10-4}{10}\times90^\circ=\frac{20-4}{10}\times90^\circ.
\displaystyle =\frac{16}{10}\times90^\circ=144^\circ.
\displaystyle \therefore \text{Each interior angle of a regular decagon is }144^\circ.
\displaystyle \textbf{(iv) 16-gon}
\displaystyle \text{Each interior angle of a 16-gon}=\frac{2n-4}{n}\text{ right angles.}
\displaystyle =\frac{2\times16-4}{16}\times90^\circ=\frac{32-4}{16}\times90^\circ.
\displaystyle =\frac{28}{16}\times90^\circ=\frac{315}{2}^\circ=157.5^\circ.
\displaystyle \therefore \text{Each interior angle of a regular} 16\text{-}gon\text{ is }157.5^\circ.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the measure, in degrees, of each exterior angle of a regular polygon}
\displaystyle \text{containing:}
\displaystyle \text{(i) 6 sides \hspace{1cm} (ii) 8 sides}
\displaystyle \text{(iii) 15 sides \hspace{0.6cm} (iv) 20 sides}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each exterior angle of a polygon of }n\text{ sides is }
\displaystyle \frac{360^\circ}{n}.
\displaystyle \textbf{(i) 6-sided polygon}
\displaystyle \text{Each exterior angle}=\frac{360^\circ}{6}=60^\circ.
\displaystyle \therefore \text{Each exterior angle of a regular 6-sided polygon is }60^\circ.
\displaystyle \textbf{(ii) 8-sided polygon}
\displaystyle \text{Each exterior angle}=\frac{360^\circ}{8}=45^\circ.
\displaystyle \therefore \text{Each exterior angle of a regular 8-sided polygon is }45^\circ.
\displaystyle \textbf{(iii) 15-sided polygon}
\displaystyle \text{Each exterior angle}=\frac{360^\circ}{15}=24^\circ.
\displaystyle \therefore \text{Each exterior angle of a regular 15-sided polygon is }24^\circ.
\displaystyle \textbf{(iv) 20-sided polygon}
\displaystyle \text{Each exterior angle}=\frac{360^\circ}{20}=18^\circ.
\displaystyle \therefore \text{Each exterior angle of a regular 20-sided polygon is }18^\circ.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the number of sides of a polygon, the sum of whose interior angles is:}
\displaystyle \text{(i) 24 right angles \hspace{0.6cm} (ii) }1620^\circ
\displaystyle \text{(iii) }2880^\circ
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that the sum of interior angles of a polygon of }n\text{ sides is}
\displaystyle (2n-4)\text{ right angles.}
\displaystyle \textbf{(i) Sum}=24\text{ right angles}.
\displaystyle (2n-4)=24\Rightarrow2n=24+4=28.
\displaystyle \therefore n=\frac{28}{2}=14.
\displaystyle \therefore \text{The polygon has }14\text{ sides.}
\displaystyle \textbf{(ii) Sum}=1620^\circ.
\displaystyle (2n-4)\times90^\circ=1620^\circ.
\displaystyle \therefore 2n-4=\frac{1620}{90}=18.
\displaystyle \therefore 2n=18+4=22.
\displaystyle \therefore n=\frac{22}{2}=11.
\displaystyle \therefore \text{The polygon has }11\text{ sides.}
\displaystyle \textbf{(iii) Sum}=2880^\circ.
\displaystyle (2n-4)\times90^\circ=2880^\circ.
\displaystyle \therefore 2n-4=\frac{2880}{90}=32.
\displaystyle \therefore 2n=32+4=36.
\displaystyle \therefore n=\frac{36}{2}=18.
\displaystyle \therefore \text{The polygon has }18\text{ sides.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the number of sides in a regular polygon, if each of its exterior angles is:}
\displaystyle \text{(i) }72^\circ \hspace{0.8cm}\text{(ii) }24^\circ
\displaystyle \text{(iii) }22.5^\circ \hspace{0.4cm}\text{(iv) }15^\circ
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each exterior angle of a regular polygon of }n\text{ sides is}
\displaystyle \frac{360^\circ}{n}.
\displaystyle \textbf{(i) Exterior angle}=72^\circ.
\displaystyle \frac{360^\circ}{n}=72^\circ\Rightarrow n=\frac{360}{72}=5.
\displaystyle \therefore \text{The number of sides of the polygon is }5.
\displaystyle \textbf{(ii) Exterior angle}=24^\circ.
\displaystyle \frac{360^\circ}{n}=24^\circ\Rightarrow n=\frac{360}{24}=15.
\displaystyle \therefore \text{The number of sides of the polygon is }15.
\displaystyle \textbf{(iii) Exterior angle}=22.5^\circ.
\displaystyle \frac{360^\circ}{n}=22.5^\circ\Rightarrow n=\frac{360}{22.5}=\frac{360\times10}{225}=16.
\displaystyle \therefore \text{The number of sides of the polygon is }16.
\displaystyle \textbf{(iv) Exterior angle}=15^\circ.
\displaystyle \frac{360^\circ}{n}=15^\circ\Rightarrow n=\frac{360}{15}=24.
\displaystyle \therefore \text{The number of sides of the polygon is }24.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the number of sides in a regular polygon, if each of its interior angles is:}
\displaystyle \text{(i) }120^\circ \hspace{0.8cm}\text{(ii) }150^\circ
\displaystyle \text{(iii) }160^\circ \hspace{0.6cm}\text{(iv) }165^\circ
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each interior angle of a regular polygon of }n\text{ sides is}
\displaystyle \frac{2n-4}{n}\text{ right angles.}
\displaystyle \textbf{(i) Each interior angle}=120^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=120^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{120}{90}=\frac{4}{3}.
\displaystyle \therefore 6n-12=4n.
\displaystyle \therefore 2n=12\Rightarrow n=6.
\displaystyle \therefore \text{The number of sides of the polygon is }6.
\displaystyle \textbf{(ii) Each interior angle}=150^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=150^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{150}{90}=\frac{5}{3}.
\displaystyle \therefore 6n-12=5n.
\displaystyle \therefore n=12.
\displaystyle \therefore \text{The number of sides of the polygon is }12.
\displaystyle \textbf{(iii) Each interior angle}=160^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=160^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{160}{90}=\frac{16}{9}.
\displaystyle \therefore 18n-36=16n.
\displaystyle \therefore 2n=36\Rightarrow n=18.
\displaystyle \therefore \text{The number of sides of the polygon is }18.
\displaystyle \textbf{(iv) Each interior angle}=165^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=165^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{165}{90}=\frac{11}{6}.
\displaystyle \therefore 12n-24=11n.
\displaystyle \therefore n=24.
\displaystyle \therefore \text{The number of sides of the polygon is }24.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Is it possible to describe a polygon, the sum of whose interior angles is:}
\displaystyle \text{(i) }320^\circ \hspace{0.8cm}\text{(ii) }540^\circ
\displaystyle \text{(iii) }11\text{ right angles} \hspace{0.4cm}\text{(iv) }14\text{ right angles?}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that the sum of the interior angles of a polygon of }n\text{ sides is}
\displaystyle (2n-4)\text{ right angles.}
\displaystyle \textbf{(i) Sum of interior angles}=320^\circ.
\displaystyle (2n-4)\times90^\circ=320^\circ.
\displaystyle \therefore 2n-4=\frac{320}{90}=\frac{32}{9}.
\displaystyle \therefore 2n=\frac{32}{9}+4=\frac{68}{9}.
\displaystyle \therefore n=\frac{68}{18}=\frac{34}{9}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to describe a polygon.}
\displaystyle \textbf{(ii) Sum of interior angles}=540^\circ.
\displaystyle (2n-4)\times90^\circ=540^\circ.
\displaystyle \therefore 2n-4=\frac{540}{90}=6.
\displaystyle \therefore 2n=6+4=10.
\displaystyle \therefore n=\frac{10}{2}=5.
\displaystyle \therefore \text{Yes, it is possible to describe a polygon.}
\displaystyle \textbf{(iii) Sum of interior angles}=11\text{ right angles}.
\displaystyle (2n-4)=11.
\displaystyle \therefore 2n=11+4=15.
\displaystyle \therefore n=\frac{15}{2}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to describe a polygon.}
\displaystyle \textbf{(iv) Sum of interior angles}=14\text{ right angles}.
\displaystyle (2n-4)=14.
\displaystyle \therefore 2n=14+4=18.
\displaystyle \therefore n=\frac{18}{2}=9.
\displaystyle \therefore \text{Yes, it is possible to describe a polygon.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Is it possible to have a regular polygon, each of whose exterior angle is:}
\displaystyle \text{(i) }32^\circ \hspace{0.8cm}\text{(ii) }18^\circ
\displaystyle \text{(iii) }\frac{1}{8}\text{ of a right angle} \hspace{0.4cm}\text{(iv) }80^\circ ?
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each exterior angle of a regular polygon of }n\text{ sides is}
\displaystyle \frac{360^\circ}{n}.
\displaystyle \textbf{(i) Exterior angle}=32^\circ.
\displaystyle \frac{360^\circ}{n}=32^\circ\Rightarrow n=\frac{360}{32}=\frac{45}{4}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to have a regular polygon.}
\displaystyle \textbf{(ii) Exterior angle}=18^\circ.
\displaystyle \frac{360^\circ}{n}=18^\circ\Rightarrow n=\frac{360}{18}=20.
\displaystyle \therefore \text{Yes, it is possible to have a regular polygon.}
\displaystyle \textbf{(iii) Exterior angle}=\frac{1}{8}\text{ of a right angle}=\frac{1}{8}\times90^\circ=\frac{45}{4}^\circ.
\displaystyle \frac{360^\circ}{n}=\frac{45}{4}^\circ\Rightarrow n=\frac{360\times4}{45}=32.
\displaystyle \therefore \text{Yes, it is possible to have a regular polygon.}
\displaystyle \textbf{(iv) Exterior angle}=80^\circ.
\displaystyle \frac{360^\circ}{n}=80^\circ\Rightarrow n=\frac{360}{80}=\frac{9}{2}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to have a regular polygon.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Is it possible to have a regular polygon, each of whose interior angles is:}
\displaystyle \text{(i) }120^\circ \hspace{0.8cm}\text{(ii) }105^\circ
\displaystyle \text{(iii) }175^\circ \hspace{0.6cm}\text{(iv) }130^\circ ?
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each interior angle of a regular polygon of }n\text{ sides is}
\displaystyle \frac{2n-4}{n}\text{ right angles.}
\displaystyle \textbf{(i) Interior angle}=120^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=120^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{120}{90}=\frac{4}{3}.
\displaystyle \therefore 6n-12=4n.
\displaystyle \therefore 2n=12\Rightarrow n=6.
\displaystyle \therefore \text{It is possible to have a regular polygon.}
\displaystyle \textbf{(ii) Interior angle}=105^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=105^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{105}{90}=\frac{7}{6}.
\displaystyle \therefore 12n-24=7n.
\displaystyle \therefore 5n=24.
\displaystyle \therefore n=\frac{24}{5}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to have a regular polygon.}
\displaystyle \textbf{(iii) Interior angle}=175^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=175^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{175}{90}=\frac{35}{18}.
\displaystyle \therefore 36n-72=35n.
\displaystyle \therefore n=72.
\displaystyle \therefore \text{It is possible to have a regular polygon.}
\displaystyle \textbf{(iv) Interior angle}=130^\circ.
\displaystyle \frac{2n-4}{n}\times90^\circ=130^\circ.
\displaystyle \therefore \frac{2n-4}{n}=\frac{130}{90}=\frac{13}{9}.
\displaystyle \therefore 18n-36=13n.
\displaystyle \therefore 5n=36.
\displaystyle \therefore n=\frac{36}{5}.
\displaystyle \text{Since }n\text{ is not an integer, it is not possible to have a regular polygon.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The angles of a quadrilateral are in the ratio }6:3:2:4.
\displaystyle \text{Find the angles.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the four angles of a quadrilateral is }360^\circ.
\displaystyle \therefore \angle A+\angle B+\angle C+\angle D=360^\circ.
\displaystyle \text{Also, }\angle A:\angle B:\angle C:\angle D=6:3:2:4.
\displaystyle \text{Let }\angle A=6x,\ \angle B=3x,\ \angle C=2x\text{ and }\angle D=4x.
\displaystyle \therefore 6x+3x+2x+4x=360^\circ.
\displaystyle \Rightarrow 15x=360^\circ\Rightarrow x=\frac{360^\circ}{15}=24^\circ.
\displaystyle \therefore \angle A=6x=6\times24^\circ=144^\circ.
\displaystyle \angle B=3x=3\times24^\circ=72^\circ.
\displaystyle \angle C=2x=2\times24^\circ=48^\circ.
\displaystyle \angle D=4x=4\times24^\circ=96^\circ.
\displaystyle {\therefore \text{The required angles are }144^\circ,\ 72^\circ,\ 48^\circ\text{ and }96^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The angles of a pentagon are in the ratio }3:4:5:2:4.
\displaystyle \text{Find the angles.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the five angles of a pentagon is}
\displaystyle (2n-4)\text{ right angles}.
\displaystyle =(2\times5-4)\times90^\circ=(10-4)\times90^\circ.
\displaystyle =6\times90^\circ=540^\circ.
\displaystyle \text{Let }\angle A=3x,\ \angle B=4x,\ \angle C=5x,\ \angle D=2x\text{ and }\angle E=4x.
\displaystyle \therefore 3x+4x+5x+2x+4x=540^\circ.
\displaystyle \Rightarrow 18x=540^\circ\Rightarrow x=\frac{540^\circ}{18}=30^\circ.
\displaystyle \therefore \angle A=3x=3\times30^\circ=90^\circ.
\displaystyle \angle B=4x=4\times30^\circ=120^\circ.
\displaystyle \angle C=5x=5\times30^\circ=150^\circ.
\displaystyle \angle D=2x=2\times30^\circ=60^\circ.
\displaystyle \angle E=4x=4\times30^\circ=120^\circ.
\displaystyle {\therefore \text{The required angles are }90^\circ,\ 120^\circ,\ 150^\circ,\ 60^\circ\text{ and }120^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The angles of a pentagon are }(3x+5)^\circ,\ (x+16)^\circ,
\displaystyle (2x+9)^\circ,\ (3x-8)^\circ\text{ and }(4x-15)^\circ\text{ respectively. Find the value}
\displaystyle \text{of }x\text{ and hence find the measures of all the angles of the pentagon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the five angles of a pentagon is}
\displaystyle (2n-4)\text{ right angles}.
\displaystyle =(2\times5-4)\times90^\circ=(10-4)\times90^\circ=540^\circ.
\displaystyle \therefore (3x+5)+(x+16)+(2x+9)+(3x-8)+(4x-15)=540.
\displaystyle \Rightarrow 13x+7=540.
\displaystyle \Rightarrow 13x=540-7=533.
\displaystyle \therefore x=\frac{533}{13}=41.
\displaystyle \text{First angle}=3x+5=3\times41+5=128^\circ.
\displaystyle \text{Second angle}=x+16=41+16=57^\circ.
\displaystyle \text{Third angle}=2x+9=2\times41+9=91^\circ.
\displaystyle \text{Fourth angle}=3x-8=3\times41-8=115^\circ.
\displaystyle \text{Fifth angle}=4x-15=4\times41-15=149^\circ.
\displaystyle {\therefore x=41\text{ and the angles are }128^\circ,\ 57^\circ,\ 91^\circ,\ 115^\circ\text{ and }149^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The angles of a hexagon are }2x^\circ,\ (2x+25)^\circ,\ 3(x-15)^\circ,
\displaystyle (3x-20)^\circ,\ 2(x+5)^\circ\text{ and }3(x-5)^\circ\text{ respectively. Find the value}
\displaystyle \text{of }x\text{ and hence find the measures of all the angles of the hexagon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the angles of a hexagon is}
\displaystyle (2n-4)\text{ right angles}.
\displaystyle =(2\times6-4)\times90^\circ=(12-4)\times90^\circ.
\displaystyle =8\times90^\circ=720^\circ.
\displaystyle \therefore 2x+(2x+25)+3(x-15)+(3x-20)+2(x+5)+3(x-5)=720.
\displaystyle \Rightarrow 2x+2x+25+3x-45+3x-20+2x+10+3x-15=720.
\displaystyle \Rightarrow 15x-45=720.
\displaystyle \Rightarrow 15x=720+45=765.
\displaystyle \therefore x=\frac{765}{15}=51.
\displaystyle \text{First angle}=2x=2\times51^\circ=102^\circ.
\displaystyle \text{Second angle}=2x+25=2\times51^\circ+25^\circ=127^\circ.
\displaystyle \text{Third angle}=3(x-15)=3(51^\circ-15^\circ)=108^\circ.
\displaystyle \text{Fourth angle}=3x-20=3\times51^\circ-20^\circ=133^\circ.
\displaystyle \text{Fifth angle}=2(x+5)=2(51^\circ+5^\circ)=112^\circ.
\displaystyle \text{Sixth angle}=3(x-5)=3(51^\circ-5^\circ)=138^\circ.
\displaystyle {\therefore x=51\text{ and the angles are }102^\circ,\ 127^\circ,\ 108^\circ,\ 133^\circ,\ 112^\circ\text{ and }138^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Three of the exterior angles of a hexagon are }40^\circ,\ 52^\circ\text{ and }85^\circ
\displaystyle \text{respectively and each of the remaining exterior angles is }x^\circ.
\displaystyle \text{Calculate the value of }x.
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the exterior angles of a hexagon is }360^\circ.
\displaystyle \therefore 40^\circ+52^\circ+85^\circ+x^\circ+x^\circ+x^\circ=360^\circ.
\displaystyle \Rightarrow 177^\circ+3x=360^\circ.
\displaystyle \Rightarrow 3x=360^\circ-177^\circ=183^\circ.
\displaystyle \therefore x=\frac{183^\circ}{3}=61^\circ.
\displaystyle {\therefore \text{The value of }x\text{ is }61^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{One angle of an octagon is }100^\circ\text{ and the other angles are all equal.}
\displaystyle \text{Find the measure of each of the equal angles.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let each of the remaining seven angles be }x^\circ.
\displaystyle \text{The sum of the interior angles of an octagon is}
\displaystyle (2\times8-4)\times90^\circ=(16-4)\times90^\circ=1080^\circ.
\displaystyle \therefore 100^\circ+7x=1080^\circ.
\displaystyle \Rightarrow 7x=980^\circ.
\displaystyle \therefore x=\frac{980^\circ}{7}=140^\circ.
\displaystyle {\therefore \text{Each of the remaining angles is }140^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The interior angle of a regular polygon is double the exterior angle.}
\displaystyle \text{Find the number of sides in the polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let each exterior angle be }x^\circ.
\displaystyle \text{Then each interior angle}=2x^\circ.
\displaystyle \text{Since interior angle + exterior angle}=180^\circ,
\displaystyle x+2x=180^\circ.
\displaystyle \Rightarrow 3x=180^\circ.
\displaystyle \therefore x=60^\circ.
\displaystyle \text{Also, }n\times\text{(each exterior angle)}=360^\circ.
\displaystyle \therefore n\times60^\circ=360^\circ.
\displaystyle \Rightarrow n=\frac{360}{60}=6.
\displaystyle {\therefore \text{The polygon has }6\text{ sides.}}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The ratio of each interior angle to each exterior angle of a regular}
\displaystyle \text{polygon is }7:2.\text{ Find the number of sides in the polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let each interior angle}=7x^\circ\text{ and each exterior angle}=2x^\circ.
\displaystyle \text{Since interior angle + exterior angle}=180^\circ,
\displaystyle 7x+2x=180^\circ.
\displaystyle \Rightarrow 9x=180^\circ.
\displaystyle \therefore x=20^\circ.
\displaystyle \therefore \text{Each exterior angle}=2x=40^\circ.
\displaystyle \text{Also, }n\times40^\circ=360^\circ.
\displaystyle \Rightarrow n=\frac{360}{40}=9.
\displaystyle {\therefore \text{The number of sides of the polygon is }9.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The sum of the interior angles of a polygon is }6\text{ times the sum of its}
\displaystyle \text{exterior angles. Find the number of sides in the polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The sum of the exterior angles of a polygon is }360^\circ.
\displaystyle \therefore \text{Sum of its interior angles}=6\times360^\circ=2160^\circ.
\displaystyle \text{Let the number of sides of the polygon be }n.
\displaystyle \text{The sum of its interior angles}=(2n-4)\text{ right angles.}
\displaystyle \therefore (2n-4)\times90^\circ=2160^\circ.
\displaystyle \Rightarrow 2n-4=\frac{2160}{90}=24.
\displaystyle \Rightarrow 2n=24+4=28.
\displaystyle \therefore n=\frac{28}{2}=14.
\displaystyle {\therefore \text{The number of sides of the polygon is }14.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Two angles of a convex polygon are right angles and each of the other}
\displaystyle \text{angles is }120^\circ.\text{ Find the number of sides of the polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Two interior angles of the convex polygon are }90^\circ\text{ each.}
\displaystyle \therefore \text{Their corresponding exterior angles are }180^\circ-90^\circ=90^\circ\text{ each.}
\displaystyle \text{Each of the other interior angles is }120^\circ.
\displaystyle \therefore \text{Each corresponding exterior angle is }180^\circ-120^\circ=60^\circ.
\displaystyle \text{Let the number of sides of the polygon be }n.
\displaystyle \text{The sum of its exterior angles is }360^\circ.
\displaystyle \therefore 90^\circ+90^\circ+(n-2)\times60^\circ=360^\circ.
\displaystyle \Rightarrow 180^\circ+60^\circ(n-2)=360^\circ.
\displaystyle \Rightarrow 60^\circ(n-2)=180^\circ.
\displaystyle \Rightarrow n-2=\frac{180}{60}=3.
\displaystyle \therefore n=3+2=5.
\displaystyle {\therefore \text{The number of sides of the polygon is }5.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The ratio between the number of sides of two regular polygons is }3:4
\displaystyle \text{and the ratio between the sums of their interior angles is }2:3.
\displaystyle \text{Find the number of sides in each polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the number of sides of the first and second polygons be }3x\text{ and }4x.
\displaystyle \text{The sum of the interior angles of the first polygon is}
\displaystyle (2\times3x-4)\text{ right angles}=(6x-4)\text{ right angles.}
\displaystyle \text{The sum of the interior angles of the second polygon is}
\displaystyle (2\times4x-4)\text{ right angles}=(8x-4)\text{ right angles.}
\displaystyle \therefore (6x-4):(8x-4)=2:3.
\displaystyle \Rightarrow \frac{6x-4}{8x-4}=\frac{2}{3}.
\displaystyle \Rightarrow 3(6x-4)=2(8x-4).
\displaystyle \Rightarrow 18x-12=16x-8.
\displaystyle \Rightarrow 18x-16x=-8+12.
\displaystyle \Rightarrow 2x=4\Rightarrow x=2.
\displaystyle \therefore \text{Number of sides of the first polygon}=3x=3\times2=6.
\displaystyle \text{Number of sides of the second polygon}=4x=4\times2=8.
\displaystyle {\therefore \text{The two polygons have }6\text{ sides and }8\text{ sides respectively.}}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The number of sides of two regular polygons are in the ratio }4:5\text{ and their}
\displaystyle \text{interior angles are in the ratio }15:16.\text{ Find the number of sides in each polygon.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the number of sides of the first polygon be }4x\text{ and of the second polygon be }5x.
\displaystyle \text{Interior angle of the first polygon}=\frac{2(4x)-4}{4x}\text{ right angles}
\displaystyle =\frac{8x-4}{4x}=\frac{2x-1}{x}\text{ right angles.}
\displaystyle \text{Interior angle of the second polygon}=\frac{2(5x)-4}{5x}\text{ right angles}
\displaystyle =\frac{10x-4}{5x}\text{ right angles.}
\displaystyle \therefore \frac{2x-1}{x}:\frac{10x-4}{5x}=15:16.
\displaystyle \Rightarrow \frac{5(2x-1)}{10x-4}=\frac{15}{16}.
\displaystyle \Rightarrow 16(10x-5)=15(10x-4).
\displaystyle \Rightarrow 160x-80=150x-60.
\displaystyle \Rightarrow 10x=20\Rightarrow x=2.
\displaystyle \therefore \text{Number of sides of the first polygon}=4x=8.
\displaystyle \therefore \text{Number of sides of the second polygon}=5x=10.
\displaystyle {\therefore \text{The two polygons have }8\text{ sides and }10\text{ sides respectively.}}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{How many diagonals are there in a:}
\displaystyle \text{(i) Pentagon \hspace{1cm} (ii) Hexagon}
\displaystyle \text{(iii) Octagon?}
\displaystyle \textbf{Answer:}
\displaystyle \text{Number of diagonals of a polygon of }n\text{ sides}
\displaystyle =\frac{1}{2}n(n-1)-n.
\displaystyle \textbf{(i) Pentagon }(n=5)
\displaystyle \text{Number of diagonals}=\frac{1}{2}\times5(5-1)-5.
\displaystyle =\frac{1}{2}\times5\times4-5=10-5=5.
\displaystyle \therefore \text{A pentagon has }5\text{ diagonals.}
\displaystyle \textbf{(ii) Hexagon }(n=6)
\displaystyle \text{Number of diagonals}=\frac{1}{2}\times6(6-1)-6.
\displaystyle =\frac{1}{2}\times6\times5-6=15-6=9.
\displaystyle \therefore \text{A hexagon has }9\text{ diagonals.}
\displaystyle \textbf{(iii) Octagon }(n=8)
\displaystyle \text{Number of diagonals}=\frac{1}{2}\times8(8-1)-8.
\displaystyle =\frac{1}{2}\times8\times7-8=28-8=20.
\displaystyle \therefore \text{An octagon has }20\text{ diagonals.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The alternate sides of any pentagon are produced to meet, so as to form a}
\displaystyle \text{star-shaped figure, shown in the figure. Prove that the sum of the measures}
\displaystyle \text{of the angles at the vertices of the star is }180^\circ. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{The alternate sides of pentagon }ABCDE\text{ are produced to meet at }P,Q,R,S
\displaystyle \text{and }T\text{ so as to form a star-shaped figure.}
\displaystyle \textbf{To prove: }\angle P+\angle Q+\angle R+\angle S+\angle T=180^\circ.
\displaystyle \text{That is, }\angle a+\angle b+\angle c+\angle d+\angle e=180^\circ.
\displaystyle \textbf{Proof:}
\displaystyle \angle1+\angle2+\angle3+\angle4+\angle5=360^\circ. \qquad \cdots(i)
\displaystyle \text{This is the sum of the exterior angles of pentagon }ABCDE.
\displaystyle \text{Similarly, }\angle6+\angle7+\angle8+\angle9+\angle10=360^\circ. \qquad \cdots(ii)
\displaystyle \text{In }\triangle BCP,\ \angle1+\angle b+\angle a=180^\circ. \qquad \cdots(iii)
\displaystyle \text{In }\triangle CDQ,\ \angle2+\angle7+\angle b=180^\circ. \qquad \cdots(iv)
\displaystyle \text{In }\triangle DER,\ \angle3+\angle8+\angle c=180^\circ. \qquad \cdots(v)
\displaystyle \text{In }\triangle EAS,\ \angle4+\angle9+\angle d=180^\circ. \qquad \cdots(vi)
\displaystyle \text{In }\triangle ABT,\ \angle5+\angle10+\angle e=180^\circ. \qquad \cdots(vii)
\displaystyle \text{Adding equations }(iii)\text{ to }(vii),
\displaystyle \angle1+\angle6+\angle a+\angle2+\angle7+\angle b+\angle3+\angle8+\angle c
\displaystyle +\angle4+\angle9+\angle d+\angle5+\angle10+\angle e=5\times180^\circ.
\displaystyle \Rightarrow (\angle1+\angle2+\angle3+\angle4+\angle5)
\displaystyle +(\angle6+\angle7+\angle8+\angle9+\angle10)
\displaystyle +(\angle a+\angle b+\angle c+\angle d+\angle e)=900^\circ.
\displaystyle \Rightarrow 360^\circ+360^\circ+(\angle a+\angle b+\angle c+\angle d+\angle e)=900^\circ.
\displaystyle \Rightarrow 720^\circ+(\angle a+\angle b+\angle c+\angle d+\angle e)=900^\circ.
\displaystyle \therefore \angle a+\angle b+\angle c+\angle d+\angle e=900^\circ-720^\circ=180^\circ.
\displaystyle \therefore \angle P+\angle Q+\angle R+\angle S+\angle T=180^\circ.
\displaystyle {\therefore \text{The sum of the angles at the vertices of the star is }180^\circ.}
\displaystyle \\

\displaystyle \textbf{Exercise - 14(B)}

\displaystyle \textbf{Question 1: }\text{Construct a regular hexagon of side }2\text{ cm, using ruler and}
\displaystyle \text{compasses only.}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each angle of a regular hexagon is }120^\circ.
\displaystyle \textbf{Steps of Construction:}\displaystyle \text{(i) Draw }AB=2\text{ cm.}
\displaystyle \text{(ii) At }A\text{ and }B,\text{ draw rays making an angle of }120^\circ\text{ each.}
\displaystyle \text{(iii) Cut off }AF=BC=2\text{ cm.}
\displaystyle \text{(iv) Again at }F\text{ and }C,\text{ draw rays making an angle of }120^\circ\text{ each.}
\displaystyle \text{(v) Cut off }FE=CD=2\text{ cm.}
\displaystyle \text{(vi) Join }ED.
\displaystyle \therefore \text{ABCDEF is the required regular hexagon.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Construct a regular hexagon of side }3.5\text{ cm, using ruler and}
\displaystyle \text{compasses only.}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each angle of a regular hexagon is }120^\circ.
\displaystyle \textbf{Steps of Construction:}\displaystyle \text{(i) Draw }AB=3.5\text{ cm.}
\displaystyle \text{(ii) At }A\text{ and }B,\text{ draw rays making an angle of }120^\circ\text{ each.}
\displaystyle \text{(iii) Cut off }AF=BC=3.5\text{ cm.}
\displaystyle \text{(iv) Again at }F\text{ and }C,\text{ draw rays making an angle of }120^\circ\text{ each.}
\displaystyle \text{(v) Cut off }FE=CD=3.5\text{ cm.}
\displaystyle \text{(vi) Join }ED.
\displaystyle \therefore \text{ABCDEF is the required regular hexagon.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Construct a regular pentagon of side }2.5\text{ cm using ruler,}
\displaystyle \text{compasses and protractor.}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each angle of a regular pentagon is }108^\circ.
\displaystyle \textbf{Steps of Construction:}\displaystyle \text{(i) Draw a line segment }AB=2.5\text{ cm.}
\displaystyle \text{(ii) At }A\text{ and }B,\text{ draw rays making an angle of }108^\circ\text{ each.}
\displaystyle \text{(iii) Cut off }AE=BC=2.5\text{ cm.}
\displaystyle \text{(iv) At }E\text{ and }C,\text{ draw rays making an angle of }108^\circ\text{ meeting each}
\displaystyle \text{other at }D.
\displaystyle \therefore \text{ABCDE is the required regular pentagon.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Construct a regular pentagon of side }3.2\text{ cm using ruler,}
\displaystyle \text{compasses and protractor.}
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that each angle of a regular pentagon is }108^\circ.
\displaystyle \textbf{Steps of Construction:}\displaystyle \text{(i) Draw a line segment }AB=3.2\text{ cm.}
\displaystyle \text{(ii) At }A\text{ and }B,\text{ draw rays making an angle of }108^\circ\text{ each.}
\displaystyle \text{(iii) Cut off }AE=BC=3.2\text{ cm.}
\displaystyle \text{(iv) At }E\text{ and }C,\text{ draw rays making an angle of }108^\circ\text{ meeting each}
\displaystyle \text{other at }D.
\displaystyle \therefore \text{ABCDE is the required regular pentagon.}
\displaystyle \\

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