\displaystyle \textbf{Question 1: }\text{Find, in terms of }\pi,\text{ the length of the arc that subtends an angle of }
\displaystyle 30^\circ\text{ at the centre} \ \text{of a circle of radius }4\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\theta=30^\circ\text{ and }r=4\text{ cm.}
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{30^\circ}{360^\circ}\times2\pi\times4
\displaystyle =\frac{1}{12}\times8\pi=\frac{2\pi}{3}\text{ cm.}
\displaystyle \therefore \text{The length of the arc is }\frac{2\pi}{3}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the angle subtended at the centre of a circle of radius }
\displaystyle 5\text{ cm by an arc of length} \ \frac{5\pi}{3}\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=5\text{ cm and }l=\frac{5\pi}{3}\text{ cm.}
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle \frac{5\pi}{3}=\frac{\theta}{360^\circ}\times2\pi\times5
\displaystyle \frac{5\pi}{3}=\frac{10\pi\theta}{360^\circ}
\displaystyle \therefore \theta=\frac{5\pi}{3}\times\frac{360^\circ}{10\pi}=60^\circ.
\displaystyle \therefore \text{The angle subtended at the centre is }60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{An arc of length }20\pi\text{ cm subtends an angle of }144^\circ
\displaystyle \text{ at the centre of a circle. Find the} \ \text{radius of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }l=20\pi\text{ cm and }\theta=144^\circ.
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle 20\pi=\frac{144^\circ}{360^\circ}\times2\pi r
\displaystyle 20\pi=\frac{2}{5}\times2\pi r=\frac{4\pi r}{5}
\displaystyle \therefore r=20\pi\times\frac{5}{4\pi}=25\text{ cm.}
\displaystyle \therefore \text{The radius of the circle is }25\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{An arc of length }15\text{ cm subtends an angle of }45^\circ\text{ at the centre }
\displaystyle \text{of a circle. Find, in terms of }\pi,\text{ the radius of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }l=15\text{ cm and }\theta=45^\circ.
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle 15=\frac{45^\circ}{360^\circ}\times2\pi r
\displaystyle 15=\frac{1}{8}\times2\pi r=\frac{\pi r}{4}
\displaystyle \therefore r=\frac{60}{\pi}\text{ cm.}
\displaystyle \therefore \text{The radius of the circle is }\frac{60}{\pi}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A sector of a circle of radius }4\text{ cm contains an angle of }30^\circ. \\ \text{Find the area of the sector.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=4\text{ cm and }\theta=30^\circ.
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{30^\circ}{360^\circ}\times\pi\times4^2
\displaystyle =\frac{1}{12}\times16\pi=\frac{4\pi}{3}\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector is }\frac{4\pi}{3}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The area of a sector is one-twelfth that of the complete circle. Find the} \\ \text{angle of the sector.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the angle of the sector be }\theta.
\displaystyle \frac{\text{Area of the sector}}{\text{Area of the circle}}=\frac{\theta}{360^\circ}
\displaystyle \therefore \frac{1}{12}=\frac{\theta}{360^\circ}
\displaystyle \therefore \theta=\frac{360^\circ}{12}=30^\circ.
\displaystyle \therefore \text{The angle of the sector is }30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In a circle of radius }35\text{ cm, an arc subtends an angle of }72^\circ
\displaystyle \text{ at the centre. Find the length} \ \text{of the arc and area of the sector.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=35\text{ cm and }\theta=72^\circ.
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{72^\circ}{360^\circ}\times2\times\frac{22}{7}\times35
\displaystyle =\frac{1}{5}\times220=44\text{ cm.}
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{72^\circ}{360^\circ}\times\frac{22}{7}\times35\times35
\displaystyle =\frac{1}{5}\times3850=770\text{ cm}^2.
\displaystyle \therefore \text{The length of the arc is }44\text{ cm and the area of the sector is }770\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The length of the minute hand of a clock is }14\text{ cm. Find the area}
\displaystyle \text{swept by the minute hand in }5\text{ minutes.}\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Length of the minute hand }r=14\text{ cm.}
\displaystyle \text{Angle swept by the minute hand in }1\text{ minute}=6^\circ.
\displaystyle \therefore \text{Angle swept in }5\text{ minutes}=5\times6^\circ=30^\circ.
\displaystyle \text{Area swept by the minute hand}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{30^\circ}{360^\circ}\times\frac{22}{7}\times14\times14
\displaystyle =\frac{1}{12}\times616=\frac{154}{3}\text{ cm}^2
\displaystyle =51.33\text{ cm}^2\text{ (approx.).}
\displaystyle \therefore \text{The area swept by the minute hand is }51.33\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A sector is cut-off from a circle of radius }21\text{ cm. The angle of the sector is }
\displaystyle 120^\circ.\text{ Find} \ \text{the length of its arc and the area.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=21\text{ cm and }\theta=120^\circ.
\displaystyle \text{Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{120^\circ}{360^\circ}\times2\times\frac{22}{7}\times21
\displaystyle =\frac{1}{3}\times132=44\text{ cm.}
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{120^\circ}{360^\circ}\times\frac{22}{7}\times21\times21
\displaystyle =\frac{1}{3}\times1386=462\text{ cm}^2.
\displaystyle \therefore \text{The length of the arc is }44\text{ cm and the area of the sector is }462\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In a circle of radius }21\text{ cm, an arc subtends an angle of }60^\circ\text{ at the centre.}
\displaystyle \text{Find (i) the length of the arc (ii) area of the sector formed by the arc. }\left(\text{Use }\pi=\frac{22}{7}\right)
\displaystyle \hfill\text{[CBSE 2013, 17, 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=21\text{ cm and }\theta=60^\circ.
\displaystyle \text{(i) Length of the arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{60^\circ}{360^\circ}\times2\times\frac{22}{7}\times21
\displaystyle =\frac{1}{6}\times132=22\text{ cm.}
\displaystyle \therefore \text{The length of the arc is }22\text{ cm.}

\displaystyle \text{(ii) Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times21\times21
\displaystyle =\frac{1}{6}\times1386=231\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector is }231\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The length of the minute hand of a clock is }5\text{ cm. Find the area swept}
\displaystyle \text{by the minute hand during the time period between }6{:}05\text{ a.m. and }6{:}40\text{ a.m.} 
\displaystyle \text{Answer:}
\displaystyle \text{Length of the minute hand }r=5\text{ cm.}
\displaystyle \text{Time elapsed from }6{:}05\text{ a.m. to }6{:}40\text{ a.m.}=35\text{ minutes.}
\displaystyle \text{Angle swept by the minute hand in }1\text{ minute}=6^\circ.
\displaystyle \therefore \text{Angle swept in }35\text{ minutes}=35\times6^\circ=210^\circ.
\displaystyle \text{Area swept by the minute hand}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{210^\circ}{360^\circ}\times\pi\times5^2
\displaystyle =\frac{7}{12}\times25\pi=\frac{175\pi}{12}\text{ cm}^2.
\displaystyle \text{Using }\pi=\frac{22}{7},\quad \frac{175\pi}{12}=\frac{275}{6}=45.83\text{ cm}^2\text{ (approx.).}
\displaystyle \therefore \text{The area swept by the minute hand is }45.83\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The perimeter of a sector of a circle of radius }5.6\text{ m is }20.0\text{ m.}
\displaystyle \text{Find the area of the sector.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, radius of the sector }r=5.6\text{ m.}
\displaystyle \text{Perimeter of the sector}=20\text{ m.}
\displaystyle \text{Perimeter of a sector}=\text{arc length}+2r
\displaystyle \therefore \text{Arc length }l=20-2(5.6)
\displaystyle =20-11.2=8.8\text{ m.}
\displaystyle \text{Area of a sector}=\frac{1}{2}lr
\displaystyle =\frac{1}{2}\times8.8\times5.6
\displaystyle =24.64\text{ m}^2.
\displaystyle \therefore \text{The area of the sector is }24.64\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{AB is a chord of a circle with centre }O\text{ and radius }4\text{ cm. }AB
\displaystyle \text{ is of length }4\text{ cm. Find} \ \text{the area of the sector of the circle formed by chord }AB.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }OA=OB=4\text{ cm and }AB=4\text{ cm.}
\displaystyle \therefore OA=OB=AB.
\displaystyle \therefore \triangle OAB\text{ is an equilateral triangle.}
\displaystyle \therefore \angle AOB=60^\circ.
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{60^\circ}{360^\circ}\times\pi\times4^2
\displaystyle =\frac{1}{6}\times16\pi=\frac{8\pi}{3}\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector formed by chord }AB\text{ is }\frac{8\pi}{3}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In a circle of radius }6\text{ cm, a chord of length }
\displaystyle 10\text{ cm makes an angle of }110^\circ \ \text{at the centre of the circle. Find:}
\displaystyle \text{(i) the circumference of the circle}\qquad\text{(ii) the area of the circle}
\displaystyle \text{(iii) the length of the arc }AB\qquad\text{(iv) the area of the sector }OAB.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=6\text{ cm and }\angle AOB=110^\circ.
\displaystyle \text{(i) Circumference of the circle}=2\pi r
\displaystyle =2\pi\times6=12\pi\text{ cm.}
\displaystyle \therefore \text{The circumference of the circle is }12\pi\text{ cm.}

\displaystyle \text{(ii) Area of the circle}=\pi r^2
\displaystyle =\pi\times6^2=36\pi\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }36\pi\text{ cm}^2.

\displaystyle \text{(iii) Length of arc }AB=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{110^\circ}{360^\circ}\times2\pi\times6
\displaystyle =\frac{11}{36}\times12\pi=\frac{11\pi}{3}\text{ cm.}
\displaystyle \therefore \text{The length of arc }AB\text{ is }\frac{11\pi}{3}\text{ cm.}

\displaystyle \text{(iv) Area of sector }OAB=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{110^\circ}{360^\circ}\times\pi\times6^2
\displaystyle =\frac{11}{36}\times36\pi=11\pi\text{ cm}^2.
\displaystyle \therefore \text{The area of sector }OAB\text{ is }11\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A horse is tied to a peg at one corner of a square shaped grass}
\displaystyle \text{field of side }15\text{ m by} \ \text{means of a }5\text{ m long rope. Find the area of that part}
\displaystyle \text{of the field in which the horse can graze. Also, find the increase in grazing area}
\displaystyle \text{if length of rope is increased to } 10\text{ m. }\left(\text{Use }\pi=3.14\right) \hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the horse is tied at a corner, the grazing region is a quadrant of a circle.}
\displaystyle \text{When the length of the rope is }5\text{ m, radius }r=5\text{ m.}
\displaystyle \text{Area in which the horse can graze}=\frac{1}{4}\pi r^2
\displaystyle =\frac{1}{4}\times3.14\times5^2
\displaystyle =19.625\text{ m}^2.
\displaystyle \therefore \text{The area in which the horse can graze is }19.625\text{ m}^2.
\displaystyle \text{When the length of the rope is increased to }10\text{ m, radius }R=10\text{ m.}
\displaystyle \text{New grazing area}=\frac{1}{4}\pi R^2
\displaystyle =\frac{1}{4}\times3.14\times10^2
\displaystyle =78.5\text{ m}^2.
\displaystyle \text{Increase in grazing area}=78.5-19.625
\displaystyle =58.875\text{ m}^2.
\displaystyle \therefore \text{The increase in grazing area is }58.875\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In Figure }\text{ shows a sector of a circle, centre }O,\text{ containing an angle }\theta^\circ.
\displaystyle \text{Prove that:}
\displaystyle \text{(i) Perimeter of the shaded region is }r\left(\tan\theta+\sec\theta+\frac{\pi\theta}{180}-1\right).
\displaystyle \text{(ii) Area of the shaded region is }\frac{r^2}{2}\left(\tan\theta-\frac{\pi\theta}{180}\right). \displaystyle \text{Answer:}
\displaystyle \text{Since }AB\text{ is tangent to the circle at }A,\ OA\perp AB.
\displaystyle \text{Also, }OA=OC=r\text{ and }\angle AOB=\theta^\circ.
\displaystyle \text{In right-angled }\triangle OAB,
\displaystyle \tan\theta=\frac{AB}{OA}=\frac{AB}{r}
\displaystyle \therefore AB=r\tan\theta.
\displaystyle \sec\theta=\frac{OB}{OA}=\frac{OB}{r}
\displaystyle \therefore OB=r\sec\theta.
\displaystyle \text{Since }O,C,B\text{ are collinear,}
\displaystyle BC=OB-OC=r\sec\theta-r=r(\sec\theta-1).
\displaystyle \text{Length of minor arc }CA=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{\pi r\theta}{180}.
\displaystyle \text{(i) Perimeter of the shaded region}=AB+BC+\text{arc }CA
\displaystyle =r\tan\theta+r(\sec\theta-1)+\frac{\pi r\theta}{180}
\displaystyle =r\left(\tan\theta+\sec\theta+\frac{\pi\theta}{180}-1\right).
\displaystyle \therefore \text{Perimeter of the shaded region is }r\left(\tan\theta+\sec\theta+\frac{\pi\theta}{180}-1\right).

\displaystyle \text{(ii) Area of the shaded region}=\text{area of }\triangle OAB-\text{area of sector }OAC
\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}\times OA\times AB
\displaystyle =\frac{1}{2}\times r\times r\tan\theta=\frac{r^2}{2}\tan\theta.
\displaystyle \text{Area of sector }OAC=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{\pi r^2\theta}{360}.
\displaystyle \therefore \text{Area of the shaded region}=\frac{r^2}{2}\tan\theta-\frac{\pi r^2\theta}{360}
\displaystyle =\frac{r^2}{2}\left(\tan\theta-\frac{\pi\theta}{180}\right).
\displaystyle \therefore \text{Area of the shaded region is }\frac{r^2}{2}\left(\tan\theta-\frac{\pi\theta}{180}\right).
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In Figure }\text{ shows a sector of a circle of radius }r\text{ cm containing an angle }\theta^\circ.
\displaystyle \text{The area of the sector is }A\text{ cm}^2\text{ and perimeter of the sector is }50\text{ cm. Prove that}
\displaystyle \text{(i) }\theta=\frac{360}{\pi}\left(\frac{25}{r}-1\right)\qquad\text{(ii) }A=25r-r^2. \displaystyle \text{Answer:}
\displaystyle \text{Let the length of arc }AB=l\text{ cm.}
\displaystyle \text{Perimeter of the sector}=OA+OB+\text{arc }AB
\displaystyle \therefore r+r+l=50
\displaystyle \therefore l=50-2r.
\displaystyle \text{(i) Length of an arc }l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle 50-2r=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle \theta=\frac{360^\circ(50-2r)}{2\pi r}
\displaystyle =\frac{360^\circ(25-r)}{\pi r}
\displaystyle =\frac{360^\circ}{\pi}\left(\frac{25}{r}-1\right).
\displaystyle \therefore \theta=\frac{360}{\pi}\left(\frac{25}{r}-1\right).

\displaystyle \text{(ii) Area of a sector }A=\frac{1}{2}lr
\displaystyle =\frac{1}{2}(50-2r)r
\displaystyle =25r-r^2.
\displaystyle \therefore A=25r-r^2.
\displaystyle \\

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