\displaystyle \textbf{Question 1: }\text{A chord }PQ\text{ of length }12\text{ cm subtends an angle of }120^\circ\text{ at the}
\displaystyle \text{centre of a circle. Find the area of the minor segment cut off by the chord }PQ.
\displaystyle \text{Answer:}
\displaystyle \text{Let }O\text{ be the centre of the circle and }OM\perp PQ.
\displaystyle \therefore PM=MQ=\frac{12}{2}=6\text{ cm and }\angle POM=\frac{120^\circ}{2}=60^\circ.
\displaystyle \text{In right-angled }\triangle OPM,
\displaystyle \sin60^\circ=\frac{PM}{OP}=\frac{6}{r}
\displaystyle \frac{\sqrt3}{2}=\frac{6}{r}
\displaystyle \therefore r=\frac{12}{\sqrt3}=4\sqrt3\text{ cm.}
\displaystyle \text{Area of sector }POQ=\frac{120^\circ}{360^\circ}\times\pi r^2
\displaystyle =\frac{1}{3}\times\pi\times(4\sqrt3)^2=16\pi\text{ cm}^2.
\displaystyle \text{Area of }\triangle POQ=\frac{1}{2}r^2\sin120^\circ
\displaystyle =\frac{1}{2}\times48\times\frac{\sqrt3}{2}=12\sqrt3\text{ cm}^2.
\displaystyle \text{Area of the minor segment}=\text{area of sector }POQ-\text{area of }\triangle POQ
\displaystyle =16\pi-12\sqrt3\text{ cm}^2.
\displaystyle \therefore \text{The area of the minor segment is }(16\pi-12\sqrt3)\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A chord }AB\text{ of a circle, of radius }14\text{ cm makes an angle of }60^\circ
\displaystyle \text{ at the centre} \ \text{of the circle. Find the area of the minor segment of the circle. Also, find}
\displaystyle \text{the area of the major segment of the circle. }\left(\text{Use }\pi=\frac{22}{7}\right)\hfill\text{[CBSE 2019, 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=14\text{ cm and }\angle AOB=60^\circ.
\displaystyle \text{Since }OA=OB=14\text{ cm and }\angle AOB=60^\circ,\ \triangle AOB\text{ is equilateral.}
\displaystyle \therefore AB=14\text{ cm.}
\displaystyle \text{Area of sector }AOB=\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times14^2
\displaystyle =\frac{1}{6}\times616=\frac{308}{3}\text{ cm}^2.
\displaystyle \text{Area of }\triangle AOB=\frac{\sqrt3}{4}(14)^2=49\sqrt3\text{ cm}^2.
\displaystyle \text{Area of the minor segment}=\frac{308}{3}-49\sqrt3\text{ cm}^2.
\displaystyle \therefore \text{The area of the minor segment is }\left(\frac{308}{3}-49\sqrt3\right)\text{ cm}^2.

\displaystyle \text{Area of the circle}=\pi r^2=\frac{22}{7}\times14^2=616\text{ cm}^2.
\displaystyle \text{Area of the major segment}=\text{area of the circle}-\text{area of the minor segment}
\displaystyle =616-\left(\frac{308}{3}-49\sqrt3\right)
\displaystyle =\frac{1540}{3}+49\sqrt3\text{ cm}^2.
\displaystyle \therefore \text{The area of the major segment is }\left(\frac{1540}{3}+49\sqrt3\right)\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A chord of a circle of radius }20\text{ cm subtends an angle of }90^\circ\text{ at the}
\displaystyle \text{centre. Find the area of the corresponding major segment of the circle. }\left(\text{Use }\pi=3.14\right) 
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=20\text{ cm and }\theta=90^\circ.
\displaystyle \text{Area of the }90^\circ\text{ sector}=\frac{90^\circ}{360^\circ}\times3.14\times20^2
\displaystyle =\frac{1}{4}\times3.14\times400=314\text{ cm}^2.
\displaystyle \text{Area of the triangle}=\frac{1}{2}\times20\times20=200\text{ cm}^2.
\displaystyle \text{Area of the minor segment}=314-200=114\text{ cm}^2.
\displaystyle \text{Area of the circle}=3.14\times20^2=1256\text{ cm}^2.
\displaystyle \text{Area of the major segment}=1256-114=1142\text{ cm}^2.
\displaystyle \therefore \text{The area of the major segment is }1142\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A chord }10\text{ cm long is drawn in a circle whose radius is }5\sqrt2\text{ cm. }
\displaystyle \text{Find the area of both the segments. }\left(\text{Take }\pi=3.14\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }O\text{ be the centre of the circle and }AB\text{ be the chord.}
\displaystyle OA=OB=5\sqrt2\text{ cm and }AB=10\text{ cm.}
\displaystyle OA^2+OB^2=(5\sqrt2)^2+(5\sqrt2)^2=50+50=100=AB^2.
\displaystyle \therefore \triangle AOB\text{ is right-angled at }O,\text{ so }\angle AOB=90^\circ.
\displaystyle \text{Area of sector }AOB=\frac{90^\circ}{360^\circ}\times\pi r^2
\displaystyle =\frac{1}{4}\times3.14\times(5\sqrt2)^2
\displaystyle =\frac{1}{4}\times3.14\times50=39.25\text{ cm}^2.
\displaystyle \text{Area of }\triangle AOB=\frac{1}{2}\times OA\times OB
\displaystyle =\frac{1}{2}\times5\sqrt2\times5\sqrt2=25\text{ cm}^2.
\displaystyle \text{Area of the minor segment}=39.25-25=14.25\text{ cm}^2.
\displaystyle \therefore \text{The area of the minor segment is }14.25\text{ cm}^2.

\displaystyle \text{Area of the circle}=\pi r^2=3.14\times50=157\text{ cm}^2.
\displaystyle \text{Area of the major segment}=157-14.25=142.75\text{ cm}^2.
\displaystyle \therefore \text{The area of the major segment is }142.75\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The radius of a circle with centre }O\text{ is }5\text{ cm (in adjoining figure. }
\displaystyle \text{Two radii }OA\text{ and }OB  \ \text{are drawn at right angles to each other. Find the areas of the}
\displaystyle \text{segments made by the chord }AB. \ \left(\text{Take }\pi=3.14\right). \displaystyle \text{Answer:}
\displaystyle \text{Given, }OA=OB=5\text{ cm and }\angle AOB=90^\circ.
\displaystyle \text{Area of sector }AOB=\frac{90^\circ}{360^\circ}\times\pi r^2
\displaystyle =\frac{1}{4}\times3.14\times5^2=19.625\text{ cm}^2.
\displaystyle \text{Area of }\triangle AOB=\frac{1}{2}\times OA\times OB
\displaystyle =\frac{1}{2}\times5\times5=12.5\text{ cm}^2.
\displaystyle \text{Area of the minor segment}=\text{area of sector }AOB-\text{area of }\triangle AOB
\displaystyle =19.625-12.5=7.125\text{ cm}^2.
\displaystyle \therefore \text{The area of the minor segment is }7.125\text{ cm}^2.

\displaystyle \text{Area of the circle}=\pi r^2=3.14\times5^2=78.5\text{ cm}^2.
\displaystyle \text{Area of the major segment}=\text{area of the circle}-\text{area of the minor segment}
\displaystyle =78.5-7.125=71.375\text{ cm}^2.
\displaystyle \therefore \text{The area of the major segment is }71.375\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In adjoining figure. }AB\text{ is the diameter of a circle, centre }O.\ C\text{ is a}
\displaystyle \text{point on the circumference such that }\angle COB=\theta.\text{ The area of the minor segment}
\displaystyle \text{cut off by } AC \ \text{ is equal to twice the area} \ \text{of the sector }BOC. \\ \text{ Prove that }\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\pi\left(\frac{1}{2}-\frac{\theta}{120}\right). \displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r.
\displaystyle \text{Since }AB\text{ is a diameter, }\angle AOB=180^\circ.
\displaystyle \text{Given, }\angle COB=\theta^\circ.
\displaystyle \therefore \angle AOC=(180-\theta)^\circ.
\displaystyle \text{Area of the minor segment cut off by }AC
\displaystyle =\text{area of sector }AOC-\text{area of }\triangle AOC
\displaystyle =\frac{180-\theta}{360}\pi r^2-\frac{1}{2}r^2\sin(180^\circ-\theta)
\displaystyle =\frac{180-\theta}{360}\pi r^2-\frac{1}{2}r^2\sin\theta.
\displaystyle \text{Area of sector }BOC=\frac{\theta}{360}\pi r^2.
\displaystyle \text{According to the given condition,}
\displaystyle \frac{180-\theta}{360}\pi r^2-\frac{1}{2}r^2\sin\theta
\displaystyle =2\times\frac{\theta}{360}\pi r^2.
\displaystyle \frac{180-\theta}{360}\pi-\frac{1}{2}\sin\theta=\frac{\theta\pi}{180}.
\displaystyle \text{Multiplying by }2,
\displaystyle \frac{180-\theta}{180}\pi-\sin\theta=\frac{\theta\pi}{90}.
\displaystyle \sin\theta=\pi\left(1-\frac{\theta}{180}-\frac{\theta}{90}\right)
\displaystyle =\pi\left(1-\frac{\theta}{60}\right).
\displaystyle \text{Since }\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2},
\displaystyle 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\pi\left(1-\frac{\theta}{60}\right).
\displaystyle \therefore \sin\frac{\theta}{2}\cos\frac{\theta}{2}=\pi\left(\frac{1}{2}-\frac{\theta}{120}\right).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A chord of a circle subtends an angle of }\theta\text{ at the centre of the circle.}
\displaystyle \text{The area of the minor segment cut off by the chord is one eighth of the area of the} \\ \text{circle. Prove that}
\displaystyle 8\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\pi=\frac{\pi\theta}{45}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r.
\displaystyle \text{Area of the sector}=\frac{\theta}{360}\pi r^2.
\displaystyle \text{Area of the triangle}=r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.
\displaystyle \therefore \text{Area of the minor segment}
\displaystyle =\frac{\theta}{360}\pi r^2-r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.
\displaystyle \text{Given, area of the minor segment}=\frac{1}{8}\times\text{area of the circle}.
\displaystyle \therefore \frac{\theta}{360}\pi r^2-r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\frac{1}{8}\pi r^2.
\displaystyle \frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\frac{\pi}{8}.
\displaystyle \sin\frac{\theta}{2}\cos\frac{\theta}{2}=\frac{\pi\theta}{360}-\frac{\pi}{8}.
\displaystyle \text{Multiplying both sides by }8,
\displaystyle 8\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\frac{\pi\theta}{45}-\pi.
\displaystyle \therefore 8\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\pi=\frac{\pi\theta}{45}.
\displaystyle \text{Hence proved.}
\displaystyle \\

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