\displaystyle \textbf{Question 1: }\text{What is the ratio of the areas of a circle and an equilateral triangle whose}
\displaystyle \text{diameter and a side are respectively equal?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the diameter of the circle and side of the equilateral triangle be }a.
\displaystyle \therefore \text{Radius of the circle}=\frac{a}{2}.
\displaystyle \text{Area of the circle}=\pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{4}.
\displaystyle \text{Area of the equilateral triangle}=\frac{\sqrt3}{4}a^2.
\displaystyle \therefore \text{Required ratio}=\frac{\pi a^2}{4}:\frac{\sqrt3 a^2}{4}
\displaystyle =\pi:\sqrt3.
\displaystyle \therefore \text{The ratio of the areas is }\pi:\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If the circumferences of two circles are in the ratio }2:3,\text{ what is the} \\ \text{ratio of their areas?}
\displaystyle \text{Answer:}
\displaystyle \text{Since circumference is proportional to radius,}
\displaystyle r_1:r_2=2:3.
\displaystyle \therefore \text{Ratio of their areas}=r_1^2:r_2^2
\displaystyle =2^2:3^2=4:9.
\displaystyle \therefore \text{The ratio of their areas is }4:9.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the area of the sector of a circle whose radius is }r\text{ and length} \\ \text{of the arc is }l.
\displaystyle \text{Answer:}
\displaystyle \text{Area of a sector}=\frac{1}{2}\times\text{arc length}\times\text{radius}
\displaystyle =\frac{1}{2}lr.
\displaystyle \therefore \text{The area of the sector is }\frac{1}{2}lr.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the length, in terms of }\pi,\text{ of the arc that subtends an angle of }
\displaystyle 36^\circ\text{ at the centre} \ \text{of a circle of radius }5\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\theta=36^\circ\text{ and }r=5\text{ cm.}
\displaystyle \text{Length of the arc}=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{36^\circ}{360^\circ}\times2\pi\times5
\displaystyle =\frac{1}{10}\times10\pi=\pi\text{ cm.}
\displaystyle \therefore \text{The length of the arc is }\pi\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the angle subtended at the centre of a circle of radius }6\text{ cm} \\ \text{by an arc of length }3\pi\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=6\text{ cm and arc length }l=3\pi\text{ cm.}
\displaystyle l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle 3\pi=\frac{\theta}{360^\circ}\times2\pi\times6
\displaystyle 3=\frac{\theta}{30}
\displaystyle \therefore \theta=90^\circ.
\displaystyle \therefore \text{The angle subtended at the centre is }90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{What is the area of a sector of a circle of radius }5\text{ cm formed by an arc} \\ \text{of length }3.5\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=5\text{ cm and arc length }l=3.5\text{ cm.}
\displaystyle \text{Area of the sector}=\frac{1}{2}lr
\displaystyle =\frac{1}{2}\times3.5\times5
\displaystyle =8.75\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector is }8.75\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In a circle of radius }10\text{ cm, an arc subtends an angle of }108^\circ
\displaystyle \text{ at the centre. What is the area} \ \text{of the sector in terms of }\pi\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=10\text{ cm and }\theta=108^\circ.
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{108^\circ}{360^\circ}\times\pi\times10^2
\displaystyle =\frac{3}{10}\times100\pi
\displaystyle =30\pi\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector is }30\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If a square is inscribed in a circle, what is the ratio of the areas of the circle} \\ \text{and the square?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a.
\displaystyle \text{Diameter of the circle}=\text{diagonal of the square}=a\sqrt2.
\displaystyle \therefore \text{Radius of the circle}=\frac{a}{\sqrt2}.
\displaystyle \text{Area of circle}:\text{area of square}=\pi\left(\frac{a}{\sqrt2}\right)^2:a^2
\displaystyle =\frac{\pi a^2}{2}:a^2=\pi:2.
\displaystyle \therefore \text{The required ratio is }\pi:2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the formula for the area of a sector of angle }\theta\text{ (in degrees) of} \\ \text{a circle of radius }r.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the sector}=\frac{\theta}{360}\times\pi r^2.
\displaystyle \therefore \text{The required formula is }\frac{\theta}{360}\pi r^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the formula for the area of a segment in a circle of radius }r,\text{ given that the}
\displaystyle \text{sector angle is }\theta\text{ (in degrees).}
\displaystyle \text{Answer:}
\displaystyle \text{Area of minor segment}=\text{area of sector}-\text{area of triangle}
\displaystyle =\frac{\theta}{360}\pi r^2-r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.
\displaystyle \therefore \text{Area of minor segment}=\left(\frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)r^2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If the adjoining figure is a sector of a circle of radius }10.5\text{ cm, what is the perimeter}
\displaystyle \text{of the sector? }\left(\text{Take }\pi=\frac{22}{7}\right) \displaystyle \text{Answer:}
\displaystyle \text{Given, }r=10.5\text{ cm and }\theta=60^\circ.
\displaystyle \text{Length of the arc}=\frac{60^\circ}{360^\circ}\times2\pi r
\displaystyle =\frac{1}{6}\times2\times\frac{22}{7}\times10.5=11\text{ cm.}
\displaystyle \text{Perimeter of the sector}=2r+\text{arc length}
\displaystyle =2(10.5)+11=32\text{ cm.}
\displaystyle \therefore \text{The perimeter of the sector is }32\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the diameter of a semi-circular protractor is }14\text{ cm, then find its perimeter.}
\displaystyle \hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=14\text{ cm.}
\displaystyle \therefore \text{Radius }r=7\text{ cm.}
\displaystyle \text{Perimeter of a semicircle}=\pi r+2r
\displaystyle =\frac{22}{7}\times7+14
\displaystyle =22+14=36\text{ cm.}
\displaystyle \therefore \text{The perimeter of the semi-circular protractor is }36\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A piece of wire }22\text{ cm long is bent into the form of an arc of a circle}
\displaystyle \text{subtending an angle of }60^\circ\text{ at its centre. Find the radius of the circle. }\left(\text{Use }\pi=\frac{22}{7}\right)
\displaystyle \hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, arc length }l=22\text{ cm and }\theta=60^\circ.
\displaystyle l=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle 22=\frac{60^\circ}{360^\circ}\times2\times\frac{22}{7}\times r
\displaystyle 22=\frac{22r}{21}
\displaystyle \therefore r=21\text{ cm.}
\displaystyle \therefore \text{The radius of the circle is }21\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the area of the largest triangle that can be inscribed in a semi-circle}
\displaystyle \text{of radius} \ r\text{ units.} \ \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{For the largest triangle, take the diameter }2r\text{ as its base.}
\displaystyle \text{The maximum possible height of the triangle is }r.
\displaystyle \text{Maximum area}=\frac{1}{2}\times2r\times r
\displaystyle =r^2\text{ square units.}
\displaystyle \therefore \text{The area of the largest triangle is }r^2\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{An arc subtends an angle of }90^\circ\text{ at the centre of the circle of radius }
\displaystyle 14\text{ cm. Write the area of} \ \text{minor sector thus formed in terms of }\pi.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=14\text{ cm and }\theta=90^\circ.
\displaystyle \text{Area of the minor sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{90^\circ}{360^\circ}\times\pi\times14^2
\displaystyle =\frac{1}{4}\times196\pi=49\pi\text{ cm}^2.
\displaystyle \therefore \text{The area of the minor sector is }49\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the area of a sector of circle of radius }21\text{ cm and central angle }120^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=21\text{ cm and }\theta=120^\circ.
\displaystyle \text{Area of the sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{120^\circ}{360^\circ}\times\pi\times21^2
\displaystyle =\frac{1}{3}\times441\pi=147\pi\text{ cm}^2.
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 147\pi=147\times\frac{22}{7}=462\text{ cm}^2.
\displaystyle \therefore \text{The area of the sector is }462\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{What is the area of a square inscribed in a circle of diameter }p\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle \text{Diagonal of the square}=\text{diameter of the circle}=p\text{ cm.}
\displaystyle \therefore a\sqrt2=p
\displaystyle \therefore a=\frac{p}{\sqrt2}.
\displaystyle \text{Area of the square}=a^2
\displaystyle =\left(\frac{p}{\sqrt2}\right)^2=\frac{p^2}{2}\text{ cm}^2.
\displaystyle \therefore \text{The area of the square is }\frac{p^2}{2}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Is it true to say that area of a segment of a circle is less than the area of}
\displaystyle \text{its corresponding sector? Why?}
\displaystyle \text{Answer:}
\displaystyle \text{No, it is not always true.}
\displaystyle \text{Area of a minor segment}=\text{area of corresponding minor sector}-\text{area of triangle.}
\displaystyle \therefore \text{A minor segment has area less than that of its corresponding minor sector.}
\displaystyle \text{However, a major segment has area greater than that of its corresponding major sector.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If the numerical value of the area of a circle is equal to the numerical value}
\displaystyle \text{of its circumference, find its radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of circle}=\text{circumference of circle}
\displaystyle \pi r^2=2\pi r
\displaystyle r=2,\quad \text{since }r>0.
\displaystyle \therefore \text{The radius of the circle is }2\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{How many revolutions does a circular wheel of radius }r\text{ metres make in }
\displaystyle \text{covering a distance of metres?}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference of the wheel}=2\pi r\text{ metres.}
\displaystyle \text{Number of revolutions}=\frac{\text{distance covered}}{\text{circumference of the wheel}}
\displaystyle =\frac{s}{2\pi r}.
\displaystyle \therefore \text{The wheel makes }\frac{s}{2\pi r}\text{ revolutions.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the ratio of the area of the circle circumscribing a square to the}
\displaystyle \text{area of the circle inscribed in the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square be }a.
\displaystyle \text{Radius of the circumscribed circle}=\frac{a\sqrt2}{2}=\frac{a}{\sqrt2}.
\displaystyle \text{Radius of the inscribed circle}=\frac{a}{2}.
\displaystyle \text{Required ratio}=\pi\left(\frac{a}{\sqrt2}\right)^2:\pi\left(\frac{a}{2}\right)^2
\displaystyle =\frac{\pi a^2}{2}:\frac{\pi a^2}{4}
\displaystyle =2:1.
\displaystyle \therefore \text{The required ratio is }2:1.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the length of the arc of a circle which subtends an angle of }60^\circ
\displaystyle \text{ at the centre of the circle} \ \text{of radius }42\text{ cm.}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }r=42\text{ cm and }\theta=60^\circ.
\displaystyle \text{Length of the arc}=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{60^\circ}{360^\circ}\times2\pi\times42
\displaystyle =14\pi\text{ cm.}
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 14\pi=44\text{ cm.}
\displaystyle \therefore \text{The length of the arc is }44\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In the figure, } \ ABCD\text{ is a square of side }10\text{ cm. A sector of radius }
\displaystyle 5\text{ cm is cut out} \ \text{from one of the corners. Find the area of the shaded region. }
\displaystyle \left(\text{Take }\pi=3.14\right) \ \hfill\text{[CBSE 2024]} \displaystyle \text{Answer:}
\displaystyle \text{Area of square }ABCD=10^2=100\text{ cm}^2.
\displaystyle \text{The sector cut out at corner }A\text{ is a quadrant of radius }5\text{ cm.}
\displaystyle \text{Area of the quadrant}=\frac{1}{4}\pi r^2
\displaystyle =\frac{1}{4}\times3.14\times5^2
\displaystyle =19.625\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=100-19.625
\displaystyle =80.375\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }80.375\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the figure, } \text{ the shape of the top of a table is that of a sector of a}
\displaystyle \text{circle with centre }O\text{ and } \angle AOB=90^\circ.\text{ If }AO=OB=42\text{ cm, then}
\displaystyle \text{find the perimeter of the top of the table.} \ \hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Given, }r=42\text{ cm and }\angle AOB=90^\circ.
\displaystyle \text{Angle of the major sector}=360^\circ-90^\circ=270^\circ.
\displaystyle \text{Length of the major arc }AB=\frac{270^\circ}{360^\circ}\times2\pi r
\displaystyle =\frac{3}{4}\times2\times\frac{22}{7}\times42=198\text{ cm.}
\displaystyle \text{Perimeter of the table top}=\text{major arc }AB+AO+OB
\displaystyle =198+42+42=282\text{ cm.}
\displaystyle \therefore \text{The perimeter of the top of the table is }282\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In the figure, } \text{ three sectors of a circle of radius }5\text{ cm, making angles }35^\circ,
\displaystyle 50^\circ\text{ and }95^\circ \ \text{at the centre are shaded. Find the area of the shaded region. }\left(\text{Use }\pi=\frac{22}{7}\right)
\displaystyle \hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Sum of the angles of the shaded sectors}=35^\circ+50^\circ+95^\circ=180^\circ.
\displaystyle \text{Area of the shaded region}=\frac{180^\circ}{360^\circ}\times\pi r^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times5^2
\displaystyle =\frac{275}{7}\text{ cm}^2
\displaystyle =39.29\text{ cm}^2\text{ (approx.).}
\displaystyle \therefore \text{The area of the shaded region is }39.29\text{ cm}^2\text{ (approx.).}
\displaystyle \\

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