\displaystyle \textbf{Question 1: }\text{If the area of a circle is }154\text{ cm}^2,\text{ then its perimeter is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \pi r^2=154
\displaystyle \frac{22}{7}r^2=154
\displaystyle r^2=49\Rightarrow r=7\text{ cm.}
\displaystyle \text{Perimeter}=2\pi r=2\times\frac{22}{7}\times7=44\text{ cm.}
\displaystyle \therefore \text{The blank is }44\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If the perimeter of a circle is equal to that of a square, then the ratio}
\displaystyle \text{of their areas is } \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r\text{ and side of the square be }a.
\displaystyle 2\pi r=4a\Rightarrow a=\frac{\pi r}{2}.
\displaystyle \text{Area of circle}:\text{area of square}=\pi r^2:a^2
\displaystyle =\pi r^2:\frac{\pi^2r^2}{4}=4:\pi.
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 4:\frac{22}{7}=14:11.
\displaystyle \therefore \text{The blank is }14:11.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The area of the circle that can be inscribed in a square of side }6\text{ cm is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the inscribed circle}=\text{side of the square}=6\text{ cm.}
\displaystyle \therefore r=3\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2=\pi(3)^2=9\pi\text{ cm}^2.
\displaystyle \therefore \text{The blank is }9\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The diameter of a circle whose area is equal to the sum of the areas of}
\displaystyle \text{two circles of radii } 24\text{ cm and }7\text{ cm is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}
\displaystyle \pi R^2=\pi(24)^2+\pi(7)^2
\displaystyle R^2=576+49=625
\displaystyle \therefore R=25\text{ cm.}
\displaystyle \therefore \text{Diameter}=2R=50\text{ cm.}
\displaystyle \therefore \text{The blank is }50\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The area of the square that can be inscribed in a circle of radius }8\text{ cm is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the circle}=2\times8=16\text{ cm.}
\displaystyle \text{Diagonal of the inscribed square}=\text{diameter of the circle}=16\text{ cm.}
\displaystyle \text{If the side of the square is }a,\text{ then }a\sqrt2=16.
\displaystyle \therefore a=8\sqrt2\text{ cm.}
\displaystyle \text{Area of the square}=a^2=(8\sqrt2)^2=128\text{ cm}^2.
\displaystyle \therefore \text{The blank is }128\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{It is proposed to build a single park equal in area to the sum of areas of two}
\displaystyle \text{circular parks of diameters }16\text{ m and }12\text{ m in a locality. The radius of the new} \\ \text{park would be }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Radii of the two circular parks are }8\text{ m and }6\text{ m.}
\displaystyle \text{Let the radius of the new park be }R\text{ m.}
\displaystyle \pi R^2=\pi(8)^2+\pi(6)^2
\displaystyle R^2=64+36=100
\displaystyle \therefore R=10\text{ m.}
\displaystyle \therefore \text{The blank is }10\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The radius of a circle whose circumference is equal to the sum of the}
\displaystyle \text{circumferences of the two circles of diameters }36\text{ cm and }20\text{ cm, is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Radii of the two circles are }18\text{ cm and }10\text{ cm.}
\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}
\displaystyle 2\pi R=2\pi(18)+2\pi(10)
\displaystyle R=18+10=28\text{ cm.}
\displaystyle \therefore \text{The blank is }28\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the difference between outer and inner radii of a circular ring is }14\text{ cm,}
\displaystyle \text{then the difference between outer and inner circumferences is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer and inner radii be }R\text{ and }r\text{ respectively.}
\displaystyle R-r=14\text{ cm.}
\displaystyle \text{Difference of circumferences}=2\pi R-2\pi r
\displaystyle =2\pi(R-r)=2\pi\times14=28\pi\text{ cm.}
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 28\pi=88\text{ cm.}
\displaystyle \therefore \text{The blank is }88\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The area of a sector whose perimeter is four times its radius }r,\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the arc of the sector be }l.
\displaystyle \text{Perimeter of the sector}=l+2r=4r
\displaystyle \therefore l=2r.
\displaystyle \text{Area of the sector}=\frac{1}{2}lr
\displaystyle =\frac{1}{2}\times2r\times r=r^2.
\displaystyle \therefore \text{The blank is }r^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Two circles touch each other externally. The sum of their areas is }490\pi\text{ cm}^2.
\displaystyle \text{Their centres are } \text{separated by }28\text{ cm. The difference of their perimeters is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the circles be }r_1\text{ and }r_2,\text{ where }r_1>r_2.
\displaystyle \text{Since the circles touch externally, }r_1+r_2=28.
\displaystyle \pi r_1^2+\pi r_2^2=490\pi
\displaystyle \therefore r_1^2+r_2^2=490.
\displaystyle (r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2
\displaystyle 28^2=490+2r_1r_2
\displaystyle \therefore r_1r_2=147.
\displaystyle (r_1-r_2)^2=r_1^2+r_2^2-2r_1r_2
\displaystyle =490-294=196.
\displaystyle \therefore r_1-r_2=14\text{ cm.}
\displaystyle \text{Difference of their perimeters}=2\pi(r_1-r_2)
\displaystyle =2\pi\times14=28\pi=88\text{ cm.}
\displaystyle \therefore \text{The blank is }88\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }l\text{ is the arc length of a sector of a circle of radius }r\text{ and }A
\displaystyle \text{ is the area of the sector, then }A:l= \ \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the sector }A=\frac{1}{2}lr.
\displaystyle \therefore A:l=\frac{1}{2}lr:l
\displaystyle =\frac{r}{2}:1=r:2.
\displaystyle \therefore \text{The blank is }r:2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the circumferences of two circles are in the ratio }c_1:c_2,
\displaystyle \text{ then the ratio of their areas is} \ \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since circumference is proportional to radius,}
\displaystyle r_1:r_2=c_1:c_2.
\displaystyle \text{Ratio of their areas}=r_1^2:r_2^2
\displaystyle =c_1^2:c_2^2.
\displaystyle \therefore \text{The blank is }c_1^2:c_2^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The area of a sector of angle }\theta^\circ\text{ of a circle of radius }r\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of a sector}=\frac{\theta}{360}\times\pi r^2.
\displaystyle \therefore \text{The blank is }\frac{\theta}{360}\pi r^2.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The length of the arc of a sector of angle }\theta^\circ\text{ of a circle of radius }r\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Length of the arc}=\frac{\theta}{360}\times2\pi r
\displaystyle =\frac{\pi r\theta}{180}.
\displaystyle \therefore \text{The blank is }\frac{\pi r\theta}{180}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The area of the minor segment of angle }\theta^\circ\text{ of a circle of radius }r\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of minor segment}=\text{area of sector}-\text{area of triangle}.
\displaystyle \text{Area of sector}=\frac{\theta}{360}\pi r^2.
\displaystyle \text{Area of triangle}=r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.
\displaystyle \therefore \text{Area of minor segment}=\frac{\theta}{360}\pi r^2-r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
\displaystyle =\left(\frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)r^2.
\displaystyle \therefore \text{The blank is }\left(\frac{\pi\theta}{360}-\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)r^2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The difference in the areas of the regular hexagon circumscribing a circle of radius }
\displaystyle 10\text{ cm and the}  \ \text{regular hexagon inscribed in the circle is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For the circumscribing regular hexagon, apothem}=10\text{ cm.}
\displaystyle \text{If its side is }a,\text{ then }\frac{a}{2}=10\tan30^\circ=\frac{10}{\sqrt3}.
\displaystyle \therefore a=\frac{20}{\sqrt3}\text{ cm.}
\displaystyle \text{Area of circumscribing hexagon}=\frac{1}{2}\times\text{perimeter}\times\text{apothem}
\displaystyle =\frac{1}{2}\times6\times\frac{20}{\sqrt3}\times10=200\sqrt3\text{ cm}^2.
\displaystyle \text{For the inscribed regular hexagon, side}=\text{radius}=10\text{ cm.}
\displaystyle \text{Area of inscribed hexagon}=6\times\frac{\sqrt3}{4}\times10^2=150\sqrt3\text{ cm}^2.
\displaystyle \text{Required difference}=200\sqrt3-150\sqrt3=50\sqrt3\text{ cm}^2.
\displaystyle \therefore \text{The blank is }50\sqrt3\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the figure, } \ ABCD\text{ is a square of side }10\text{ cm and a circle is inscribed in it.}
\displaystyle \text{The area of the shaded part is }\underline{\hspace{1.5cm}}. \displaystyle \text{Answer:}
\displaystyle \text{Radius of the inscribed circle}=\frac{10}{2}=5\text{ cm.}
\displaystyle \text{Area of square}-\text{area of circle}=10^2-\pi(5)^2=100-25\pi.
\displaystyle \text{The four corner regions outside the circle are equal.}
\displaystyle \therefore \text{Area of the shaded part}=\frac{1}{4}(100-25\pi)
\displaystyle =25-\frac{25\pi}{4}\text{ cm}^2.
\displaystyle \therefore \text{The blank is }\left(25-\frac{25\pi}{4}\right)\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the figure, } \text{ two circles of radii }7\text{ cm each are shown. }ABCD
\displaystyle \text{ is a rectangle and } \ AD\text{ and }BC  \ \text{are the radii. The area of the shaded region is }\underline{\hspace{1.5cm}}. \displaystyle \text{Answer:}
\displaystyle AD=BC=7\text{ cm.}
\displaystyle \text{Since the two circles touch externally, }DC=7+7=14\text{ cm.}
\displaystyle \text{Area of rectangle }ABCD=7\times14=98\text{ cm}^2.
\displaystyle \text{The two unshaded regions inside the rectangle are two quadrants of radius }7\text{ cm.}
\displaystyle \text{Area of the two quadrants}=2\times\frac{1}{4}\pi(7)^2=\frac{49\pi}{2}\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=98-\frac{49\pi}{2}.
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 98-\frac{49}{2}\times\frac{22}{7}=98-77=21\text{ cm}^2.
\displaystyle \therefore \text{The blank is }21\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The area of the largest circle that can be drawn inside a rectangle of length }
\displaystyle a\text{ cm and breadth }b  \ \text{cm }(a>b)\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a>b,\text{ the diameter of the largest circle}=b\text{ cm.}
\displaystyle \therefore \text{Radius of the circle}=\frac{b}{2}\text{ cm.}
\displaystyle \text{Area of the circle}=\pi\left(\frac{b}{2}\right)^2=\frac{\pi b^2}{4}\text{ cm}^2.
\displaystyle \therefore \text{The blank is }\frac{\pi b^2}{4}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The number of revolutions made by a circle of radius }r\text{ to cover a distance } \\ s\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Distance covered in one revolution}=2\pi r.
\displaystyle \text{Number of revolutions}=\frac{\text{total distance}}{\text{distance covered in one revolution}}
\displaystyle =\frac{s}{2\pi r}.
\displaystyle \therefore \text{The blank is }\frac{s}{2\pi r}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.