\displaystyle \textbf{Question 1: }\text{Find the circumference and area of a circle of radius }4.2\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius}=4.2\text{ cm.}
\displaystyle \text{Circumference of the circle}=2\pi r
\displaystyle =2\times\frac{22}{7}\times4.2
\displaystyle =26.4\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times(4.2)^2
\displaystyle =55.44\text{ cm}^2.
\displaystyle \therefore \text{The circumference is }26.4\text{ cm and the area is }55.44\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the circumference of a circle whose area is }301.84\text{ cm}^2.
\displaystyle \text{Answer:}
\displaystyle \text{Area}=301.84\text{ cm}^2.
\displaystyle \pi r^2=301.84
\displaystyle r^2=\frac{7}{22}\times301.84=96.04
\displaystyle r=\sqrt{96.04}=9.8\text{ cm.}
\displaystyle \text{Circumference}=2\pi r
\displaystyle =2\times\frac{22}{7}\times9.8
\displaystyle =61.6\text{ cm.}
\displaystyle \therefore \text{The circumference of the circle is }61.6\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the area of a circle whose circumference is }44\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference}=44\text{ cm.}
\displaystyle 2\pi r=44
\displaystyle r=\frac{44}{2\times\frac{22}{7}}=7\text{ cm.}
\displaystyle \text{Area}=\pi r^2
\displaystyle =\frac{22}{7}\times7^2
\displaystyle =154\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }154\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The circumference of a circle exceeds the diameter by }16.8\text{ cm. Find the}
\displaystyle \text{circumference of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: Circumference}-\text{Diameter}=16.8\text{ cm.}
\displaystyle 2\pi r-2r=16.8
\displaystyle r=\frac{16.8}{2\pi-2}
\displaystyle =\frac{16.8\times7}{2\times22-2\times7}=3.92\text{ cm.}
\displaystyle \text{Circumference}=2\pi r
\displaystyle =2\times\frac{22}{7}\times3.92
\displaystyle =24.64\text{ cm.}
\displaystyle \therefore \text{The circumference of the circle is }24.64\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A horse is tied to a pole with a }28\text{ m long string. Find the area where the horse can graze.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius}=28\text{ m.}
\displaystyle \text{Area of the grazing region}=\pi r^2
\displaystyle =\frac{22}{7}\times(28)^2
\displaystyle =2464\text{ m}^2.
\displaystyle \therefore \text{The area where the horse can graze is }2464\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A steel wire when bent in the form of a square encloses an area of }121\text{ cm}^2.
\displaystyle \text{If the same wire is bent in the form of a circle, find the area of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the square}=121\text{ cm}^2.
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle a^2=121
\displaystyle a=11\text{ cm.}
\displaystyle \text{Perimeter of the square}=4\times11=44\text{ cm.}
\displaystyle \text{Since the same wire is bent into a circle, its circumference is }44\text{ cm.}
\displaystyle 2\pi r=44
\displaystyle r=\frac{7\times44}{2\times22}=7\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times7^2
\displaystyle =154\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }154\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The diameters of the front and rear wheels of a tractor are }80\text{ cm and }2\text{ m}
\displaystyle \text{respectively. Find the number of revolutions that the rear wheel will make to cover the distance}
\displaystyle \text{which the front wheel covers in }1400\text{ revolutions.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the front wheel}=80\text{ cm}=0.8\text{ m.}
\displaystyle \text{Radius of the front wheel}=0.4\text{ m.}
\displaystyle \text{Diameter of the rear wheel}=2\text{ m.}
\displaystyle \text{Distance covered by the front wheel}=1400\times2\pi\times0.4
\displaystyle =1400\times2\times\frac{22}{7}\times0.4
\displaystyle =3520\text{ m.}
\displaystyle \text{Let the number of revolutions made by the rear wheel be }n.
\displaystyle n\times2\pi\times1=3520
\displaystyle n\times2\times\frac{22}{7}=3520
\displaystyle n=\frac{7\times3520}{2\times22}=560.
\displaystyle \therefore \text{The rear wheel makes }560\text{ revolutions.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A copper wire when bent in the form of a square encloses an area of }121\text{ cm}^2.
\displaystyle \text{If the same wire is bent into the form of a circle, find the area of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the square}=121\text{ cm}^2.
\displaystyle \text{Let the side of the square be }a\text{ cm.}
\displaystyle a^2=121
\displaystyle a=11\text{ cm.}
\displaystyle \text{Perimeter of the square}=4\times11=44\text{ cm.}
\displaystyle \text{Since the same wire is bent into a circle, its circumference is }44\text{ cm.}
\displaystyle 2\pi r=44
\displaystyle r=\frac{7\times44}{2\times22}=7\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times7^2
\displaystyle =154\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }154\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The circumferences of two circles are in the ratio }2:3.\text{ Find the ratio of their areas.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the two circles be }r_1\text{ and }r_2.
\displaystyle \frac{2\pi r_1}{2\pi r_2}=\frac{2}{3}
\displaystyle \Rightarrow \frac{r_1}{r_2}=\frac{2}{3}.
\displaystyle \text{Ratio of their areas}=\frac{\pi r_1^2}{\pi r_2^2}
\displaystyle =\left(\frac{r_1}{r_2}\right)^2
\displaystyle =\left(\frac{2}{3}\right)^2
\displaystyle =\frac{4}{9}.
\displaystyle \therefore \text{The ratio of their areas is }4:9.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The side of a square is }10\text{ cm. Find the area of its circumscribed and}
\displaystyle \text{inscribed circles.}
\displaystyle \text{Answer:}
\displaystyle \text{For the circumscribed circle, the diameter equals the diagonal of the square.}
\displaystyle \text{Diagonal of the square}=10\sqrt{2}\text{ cm.}
\displaystyle \therefore \text{Radius}=5\sqrt{2}\text{ cm.}
\displaystyle \text{Area of the circumscribed circle}=\pi(5\sqrt{2})^2
\displaystyle =50\pi\text{ cm}^2
\displaystyle \approx157.14\text{ cm}^2.
\displaystyle \text{For the inscribed circle, the diameter equals the side of the square.}
\displaystyle \therefore \text{Diameter}=10\text{ cm and Radius}=5\text{ cm.}
\displaystyle \text{Area of the inscribed circle}=\pi(5)^2
\displaystyle =25\pi\text{ cm}^2
\displaystyle \approx78.57\text{ cm}^2.
\displaystyle \therefore \text{The areas of the circumscribed and inscribed circles are }50\pi\text{ cm}^2
\displaystyle \text{and }25\pi\text{ cm}^2\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The sum of the radii of two circles is }140\text{ cm and the difference of their}
\displaystyle \text{circumferences is }88\text{ cm. Find the diameters of the circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii of the two circles be }r_1\text{ cm and }r_2\text{ cm, where }r_1>r_2.
\displaystyle r_1+r_2=140\qquad\ldots\text{(i)}
\displaystyle 2\pi r_1-2\pi r_2=88
\displaystyle 2\pi(r_1-r_2)=88
\displaystyle r_1-r_2=\frac{88}{2\times\frac{22}{7}}=14.\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2r_1=154
\displaystyle r_1=77\text{ cm.}
\displaystyle r_2=140-77=63\text{ cm.}
\displaystyle \therefore \text{The diameters are }2\times77=154\text{ cm and }2\times63=126\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the area of the circle in which a square of area }64\text{ cm}^2\text{ is inscribed.}
\displaystyle \text{Take }\pi=3.14.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the square}=64\text{ cm}^2.
\displaystyle \therefore \text{Side of the square}=\sqrt{64}=8\text{ cm.}
\displaystyle \text{Diagonal of the square}=8\sqrt{2}\text{ cm.}
\displaystyle \text{Since the square is inscribed in the circle, the diagonal of the square equals the diameter.}
\displaystyle \therefore \text{Diameter of the circle}=8\sqrt{2}\text{ cm.}
\displaystyle \text{Radius}=4\sqrt{2}\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =3.14(4\sqrt{2})^2
\displaystyle =3.14\times32=100.48\text{ cm}^2.
\displaystyle \therefore \text{The area of the circle is }100.48\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A field is in the form of a circle. A fence is to be erected around the field. The cost}
\displaystyle \text{of fencing is Rs. }2640\text{ at the rate of Rs. }12\text{ per metre. The field is then to be thoroughly}
\displaystyle \text{ploughed at the rate of Rs. }0.50\text{ per square metre. Find the amount required to plough the field.}
\displaystyle \text{Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Cost of fencing}=Rs.\ 2640.
\displaystyle \text{Circumference of the field}=\frac{2640}{12}=220\text{ m.}
\displaystyle 2\pi r=220
\displaystyle r=\frac{220\times7}{2\times22}=35\text{ m.}
\displaystyle \text{Area of the field}=\pi r^2
\displaystyle =\frac{22}{7}\times35^2
\displaystyle =3850\text{ m}^2.
\displaystyle \text{Cost of ploughing}=3850\times0.50=Rs.\ 1925.
\displaystyle \therefore \text{The amount required to plough the field is Rs. }1925.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the square}=a.
\displaystyle \text{The diameter of the circle equals the diagonal of the square.}
\displaystyle \therefore \text{Diameter of the circle}=a\sqrt{2}.
\displaystyle \therefore \text{Radius of the circle}=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}.
\displaystyle \text{Ratio of the areas}=\frac{\pi\left(\frac{a}{\sqrt{2}}\right)^2}{a^2}
\displaystyle =\frac{\pi\cdot\frac{a^2}{2}}{a^2}
\displaystyle =\frac{\pi}{2}.
\displaystyle \therefore \text{The ratio of the areas of the circle and the square is }\pi:2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A park is in the form of a rectangle }120\text{ m}\times100\text{ m. At the centre of the park}
\displaystyle \text{there is a circular lawn. The area of the park excluding the lawn is }8700\text{ m}^2.\text{ Find the}
\displaystyle \text{radius of the circular lawn. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circular lawn be }r\text{ m.}
\displaystyle \text{Area of the park}=120\times100=12000\text{ m}^2.
\displaystyle \text{Area of the circular lawn}=\pi r^2.
\displaystyle 12000-\pi r^2=8700
\displaystyle \pi r^2=12000-8700=3300
\displaystyle r^2=\frac{3300\times7}{22}=1050
\displaystyle r=\sqrt{1050}\approx32.40\text{ m.}
\displaystyle \therefore \text{The radius of the circular lawn is }32.40\text{ m (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The radii of two circles are }8\text{ cm and }6\text{ cm respectively. Find the radius of}
\displaystyle \text{the circle having its area equal to the sum of the areas of the two circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }r\text{ cm.}
\displaystyle \pi r^2=\pi(8)^2+\pi(6)^2
\displaystyle r^2=64+36
\displaystyle r^2=100
\displaystyle r=10\text{ cm.}
\displaystyle \therefore \text{The radius of the required circle is }10\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The radii of two circles are }19\text{ cm and }9\text{ cm respectively. Find the radius and area}
\displaystyle \text{of the circle whose circumference is equal to the sum of the circumferences of the two circles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the required circle be }r\text{ cm.}
\displaystyle 2\pi r=2\pi(19)+2\pi(9)
\displaystyle r=19+9=28\text{ cm.}
\displaystyle \text{Area of the required circle}=\pi r^2
\displaystyle =\frac{22}{7}\times28^2
\displaystyle =2464\text{ cm}^2.
\displaystyle \therefore \text{The radius is }28\text{ cm and the area is }2464\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{A car travels }1\text{ kilometre in which each wheel makes }450\text{ complete revolutions.}
\displaystyle \text{Find the radius of its wheels.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the wheel be }r\text{ m.}
\displaystyle \text{Distance travelled}=1\text{ km}=1000\text{ m.}
\displaystyle 450\times2\pi r=1000
\displaystyle r=\frac{1000\times7}{2\times450\times22}
\displaystyle =\frac{35}{99}\text{ m}
\displaystyle \approx0.3535\text{ m}
\displaystyle =35.35\text{ cm.}
\displaystyle \therefore \text{The radius of the wheel is }0.3535\text{ m or }35.35\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The area enclosed between two concentric circles is }770\text{ cm}^2.\text{ If the radius of the}
\displaystyle \text{outer circle is }21\text{ cm, find the radius of the inner circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the inner circle be }r\text{ cm.}
\displaystyle \pi(21)^2-\pi r^2=770
\displaystyle \pi r^2=\frac{22}{7}\times21^2-770
\displaystyle \pi r^2=1386-770=616
\displaystyle r^2=\frac{7}{22}\times616=196
\displaystyle r=14\text{ cm.}
\displaystyle \therefore \text{The radius of the inner circle is }14\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The wheel of a car is making }5\text{ revolutions per second. If the diameter of the}
\displaystyle \text{wheel is }84\text{ cm, find its speed in km/hr. Give your answer correct to the nearest kilometre.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of revolutions per second}=5.
\displaystyle \text{Radius of the wheel}=\frac{84}{2}=42\text{ cm.}
\displaystyle \text{Circumference of the wheel}=2\pi r
\displaystyle =2\times\frac{22}{7}\times42=264\text{ cm.}
\displaystyle \text{Distance covered in }1\text{ second}=5\times264=1320\text{ cm}=13.2\text{ m.}
\displaystyle \text{Distance covered in }1\text{ hour}=13.2\times3600=47520\text{ m}
\displaystyle =47.52\text{ km.}
\displaystyle \therefore \text{Speed of the car}\approx48\text{ km/hr (nearest kilometre).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{A sheet is }11\text{ cm long and }2\text{ cm wide. Circular pieces of diameter }0.5\text{ cm}
\displaystyle \text{are cut from it to prepare discs. Calculate the number of discs that can be prepared.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of discs along the length}=\frac{11}{0.5}=22.
\displaystyle \text{Number of discs along the breadth}=\frac{2}{0.5}=4.
\displaystyle \therefore \text{Total number of discs}=22\times4=88.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{A copper wire when bent in the form of an equilateral triangle has area }121\sqrt{3}\text{ cm}^2.
\displaystyle \text{If the same wire is bent into the form of a circle, find the area enclosed.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle \text{Height}=\sqrt{a^2-\left(\frac{a}{2}\right)^2}=\frac{\sqrt{3}}{2}a.
\displaystyle 121\sqrt{3}=\frac{1}{2}\times a\times\frac{\sqrt{3}}{2}a
\displaystyle 121\sqrt{3}=\frac{\sqrt{3}}{4}a^2
\displaystyle a^2=484
\displaystyle a=22\text{ cm.}
\displaystyle \therefore \text{Length of the wire}=3\times22=66\text{ cm.}
\displaystyle \text{Let the radius of the circle be }r\text{ cm.}
\displaystyle 2\pi r=66
\displaystyle r=\frac{66\times7}{2\times22}=10.5\text{ cm.}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times(10.5)^2
\displaystyle =346.5\text{ cm}^2.
\displaystyle \therefore \text{The area enclosed by the circle is }346.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{A plot is in the form of a rectangle }ABCD\text{ having a semicircle on }BC\text{ as shown}
\displaystyle \text{in the adjoining figure. If }AB=60\text{ m and }BC=28\text{ m, find the area of the plot.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of rectangle }ABCD=60\times28=1680\text{ m}^2.
\displaystyle \text{Diameter of the semicircle}=BC=28\text{ m.}
\displaystyle \therefore \text{Radius of the semicircle}=\frac{28}{2}=14\text{ m.}
\displaystyle \text{Area of the semicircle}=\frac{1}{2}\pi r^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times14^2
\displaystyle =308\text{ m}^2.
\displaystyle \text{Area of the plot}=1680+308=1988\text{ m}^2.
\displaystyle \therefore \text{The area of the plot is }1988\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A playground has the shape of a rectangle with two semicircles on its smaller sides}
\displaystyle \text{as diameters, added to its outside. If the sides of the rectangle are }36\text{ m and }24.5\text{ m,}
\displaystyle \text{find the area of the playground. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the rectangle}=36\times24.5=882\text{ m}^2.
\displaystyle \text{Radius of each semicircle}=\frac{24.5}{2}=12.25\text{ m.}
\displaystyle \text{The two semicircles together form one complete circle.}
\displaystyle \text{Area of the two semicircles}=\pi(12.25)^2
\displaystyle =\frac{22}{7}\times(12.25)^2
\displaystyle =471.625\text{ m}^2.
\displaystyle \text{Total area of the playground}=882+471.625=1353.625\text{ m}^2.
\displaystyle \therefore \text{The area of the playground is }1353.625\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The outer circumference of a circular race-track is }528\text{ m. The track is}
\displaystyle \text{everywhere }14\text{ m wide. Calculate the cost of levelling the track at the rate of }50\text{ paise}
\displaystyle \text{per square metre. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the outer radius of the track be }r\text{ m.}
\displaystyle 2\pi r=528
\displaystyle r=\frac{528\times7}{2\times22}=84\text{ m.}
\displaystyle \text{Inner radius}=84-14=70\text{ m.}
\displaystyle \text{Area of the track}=\pi(84^2-70^2)
\displaystyle =\frac{22}{7}(7056-4900)
\displaystyle =6776\text{ m}^2.
\displaystyle \text{Cost of levelling}=6776\times0.50=Rs.\ 3388.
\displaystyle \therefore \text{The cost of levelling the track is Rs. }3388.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{A rectangular piece is }20\text{ m long and }15\text{ m wide. From its four corners,}
\displaystyle \text{quadrants of radius }3.5\text{ m have been cut. Find the area of the remaining part.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the rectangle}=20\times15=300\text{ m}^2.
\displaystyle \text{The four quadrants together form one complete circle of radius }3.5\text{ m.}
\displaystyle \text{Area of the four quadrants}=\pi(3.5)^2
\displaystyle =\frac{22}{7}\times(3.5)^2=38.5\text{ m}^2.
\displaystyle \text{Area of the remaining part}=300-38.5=261.5\text{ m}^2.
\displaystyle \therefore \text{The area of the remaining part is }261.5\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Four equal circles, each of radius }5\text{ cm, touch each other as shown in the adjoining}
\displaystyle \text{figure. Find the area included between them. Take }\pi=3.14.
\displaystyle \text{Answer:}
\displaystyle \text{The centres of the four circles form a square of side }10\text{ cm.}
\displaystyle \text{Area of the square}=10\times10=100\text{ cm}^2.
\displaystyle \text{The four quadrants together form one complete circle of radius }5\text{ cm.}
\displaystyle \text{Area of the four quadrants}=\pi(5)^2
\displaystyle =3.14\times25=78.5\text{ cm}^2.
\displaystyle \text{Area included between the circles}=100-78.5=21.5\text{ cm}^2.
\displaystyle \therefore \text{The area included between the four circles is }21.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Four cows are tethered at the four corners of a square plot of side }50\text{ m, so that}
\displaystyle \text{they just cannot reach one another. What area will be left ungrazed?}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the square}=50\text{ m.}
\displaystyle \text{Since the cows just cannot reach one another, the length of each rope}=\frac{50}{2}=25\text{ m.}
\displaystyle \text{Area of the square}=50\times50=2500\text{ m}^2.
\displaystyle \text{The four grazed quadrants together form one complete circle of radius }25\text{ m.}
\displaystyle \text{Grazed area}=\pi(25)^2
\displaystyle =\frac{22}{7}\times625\approx1964.29\text{ m}^2.
\displaystyle \text{Ungrazed area}=2500-1964.29=535.71\text{ m}^2.
\displaystyle \therefore \text{The area left ungrazed is }535.71\text{ m}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A road }7\text{ m wide surrounds a circular park whose circumference is }352\text{ m. Find the}
\displaystyle \text{area of the road.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the park be }r\text{ m.}
\displaystyle 2\pi r=352
\displaystyle r=\frac{352\times7}{2\times22}=56\text{ m.}
\displaystyle \text{Outer radius}=56+7=63\text{ m.}
\displaystyle \text{Area of the road}=\pi(63^2-56^2)
\displaystyle =\frac{22}{7}(3969-3136)
\displaystyle =\frac{22}{7}\times833=2618\text{ m}^2.
\displaystyle \therefore \text{The area of the road is }2618\text{ m}^2.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Four equal circles, each of radius }a,\text{ touch each other. Show that the area between}
\displaystyle \text{them is }\frac{6}{7}a^2.\text{ Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{The centres of the four circles form a square of side }2a.
\displaystyle \text{Area of the square}=(2a)(2a)=4a^2.
\displaystyle \text{The four quadrants together form one complete circle of radius }a.
\displaystyle \text{Area of the four quadrants}=\pi a^2=\frac{22}{7}a^2.
\displaystyle \text{Area between the circles}=4a^2-\frac{22}{7}a^2
\displaystyle =\frac{28a^2-22a^2}{7}
\displaystyle =\frac{6}{7}a^2.
\displaystyle \therefore \text{The area between the four circles is }\frac{6}{7}a^2.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Two circular pieces of equal radii and maximum area, touching each other, are cut out}
\displaystyle \text{from a rectangular cardboard of dimensions }14\text{ cm}\times7\text{ cm. Find the area of the remaining}
\displaystyle \text{cardboard. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the cardboard}=14\times7=98\text{ cm}^2.
\displaystyle \text{Diameter of each largest circle}=7\text{ cm.}
\displaystyle \therefore \text{Radius of each circle}=3.5\text{ cm.}
\displaystyle \text{Area of the two circles}=2\pi(3.5)^2
\displaystyle =2\times\frac{22}{7}\times(3.5)^2
\displaystyle =77\text{ cm}^2.
\displaystyle \text{Area of the remaining cardboard}=98-77=21\text{ cm}^2.
\displaystyle \therefore \text{The area of the remaining cardboard is }21\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{In the adjoining figure, a square }OABC\text{ is inscribed in a quadrant }ODBE\text{ of a}
\displaystyle \text{circle. If }OA=21\text{ cm, find the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: }OA=21\text{ cm and }ABCO\text{ is a square.}
\displaystyle \text{Radius of the quadrant}=OB
\displaystyle =\sqrt{21^2+21^2}
\displaystyle =21\sqrt{2}\text{ cm.}
\displaystyle \text{Area of the quadrant}=\frac14\pi(21\sqrt2)^2
\displaystyle =\frac14\times\frac{22}{7}\times2\times21^2
\displaystyle =693\text{ cm}^2.
\displaystyle \text{Area of the square}=21\times21=441\text{ cm}^2.
\displaystyle \therefore \text{Shaded area}=693-441=252\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{In the adjoining figure, }ABC\text{ is a right-angled triangle in which }\angle A=90^\circ,
\displaystyle AB=21\text{ cm and }AC=28\text{ cm. Semicircles are described on }AB,\ BC\text{ and }AC\text{ as}
\displaystyle \text{diameters. Find the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle BC=\sqrt{AB^2+AC^2}
\displaystyle =\sqrt{21^2+28^2}
\displaystyle =\sqrt{441+784}=\sqrt{1225}=35\text{ cm.}
\displaystyle \text{Area of the semicircle on }AB=\frac{1}{2}\pi\left(\frac{21}{2}\right)^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times\left(\frac{21}{2}\right)^2
\displaystyle =173.25\text{ cm}^2.
\displaystyle \text{Area of the semicircle on }AC=\frac{1}{2}\pi\left(\frac{28}{2}\right)^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times14^2
\displaystyle =308\text{ cm}^2.
\displaystyle \text{Area of the semicircle on }BC=\frac{1}{2}\pi\left(\frac{35}{2}\right)^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times\left(\frac{35}{2}\right)^2
\displaystyle =481.25\text{ cm}^2.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times21\times28=294\text{ cm}^2.
\displaystyle \text{Area of the part of the semicircle on }BC\text{ outside }\triangle ABC
\displaystyle =481.25-294=187.25\text{ cm}^2.
\displaystyle \text{Shaded area}=173.25+308-187.25
\displaystyle =294\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }294\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{In the adjoining figure, }AB=36\text{ cm and }O\text{ is the mid-point of }AB.
\displaystyle \text{Semicircles are drawn on }AB,\ AO\text{ and }OB\text{ as diameters. A circle with centre }C\text{ touches}
\displaystyle \text{all the three semicircles. Find the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle AB=36\text{ cm.}
\displaystyle \therefore \text{Radius of the large semicircle}=18\text{ cm.}
\displaystyle \text{Radius of each smaller semicircle}=\frac{18}{2}=9\text{ cm.}
\displaystyle \text{Area of the large semicircle}=\frac{1}{2}\pi(18)^2=162\pi\text{ cm}^2.
\displaystyle \text{Area of the two smaller semicircles}=2\left[\frac{1}{2}\pi(9)^2\right]=81\pi\text{ cm}^2.
\displaystyle \text{Let the radius of the circle with centre }C\text{ be }r\text{ cm.}
\displaystyle \text{Since the figure is symmetrical, }OC=18-r.
\displaystyle \text{The centre of each smaller semicircle is }9\text{ cm from }O.
\displaystyle \text{Since the circle with centre }C\text{ touches a smaller semicircle externally,}
\displaystyle (9+r)^2=9^2+(18-r)^2
\displaystyle 81+18r+r^2=81+324-36r+r^2
\displaystyle 54r=324
\displaystyle r=6\text{ cm.}
\displaystyle \text{Area of the circle with centre }C=\pi(6)^2=36\pi\text{ cm}^2.
\displaystyle \text{Shaded area}=162\pi-81\pi-36\pi
\displaystyle =45\pi\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }45\pi\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{In the adjoining figure, the boundary of the shaded region consists of four semicircular arcs,}
\displaystyle \text{the smallest two being equal. If the diameter of the largest is }14\text{ cm and the diameter of the}
\displaystyle \text{smallest is }3.5\text{ cm, find (i) the length of the boundary and (ii) the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the largest semicircle}=\frac{14}{2}=7\text{ cm.}
\displaystyle \text{Radius of each smallest semicircle}=\frac{3.5}{2}=1.75\text{ cm.}
\displaystyle \text{Diameter of the middle semicircle}=14-3.5-3.5=7\text{ cm.}
\displaystyle \therefore \text{Radius of the middle semicircle}=3.5\text{ cm.}
\displaystyle \text{(i) Length of the boundary}=\pi(7)+\pi(3.5)+2\pi(1.75)
\displaystyle =7\pi+3.5\pi+3.5\pi
\displaystyle =14\pi
\displaystyle =14\times\frac{22}{7}=44\text{ cm.}
\displaystyle \therefore \text{The length of the boundary is }44\text{ cm.}
\displaystyle \text{(ii) Area of the shaded region}=\frac{1}{2}\pi(7)^2+\frac{1}{2}\pi(3.5)^2
\displaystyle -2\left[\frac{1}{2}\pi(1.75)^2\right]
\displaystyle =24.5\pi+6.125\pi-3.0625\pi
\displaystyle =27.5625\pi
\displaystyle =27.5625\times\frac{22}{7}
\displaystyle =86.625\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }86.625\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{In the adjoining figure, }O\text{ is the centre of a circular arc and }AOB\text{ is a straight}
\displaystyle \text{line. Find the perimeter and the area of the shaded region correct to one decimal place. Take }\pi=3.14.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\angle ACB=90^\circ,
\displaystyle AB=\sqrt{AC^2+BC^2}
\displaystyle =\sqrt{12^2+16^2}
\displaystyle =\sqrt{144+256}=\sqrt{400}=20\text{ cm.}
\displaystyle \therefore \text{Radius of the semicircle}=\frac{20}{2}=10\text{ cm.}
\displaystyle \text{Perimeter of the shaded region}=\pi r+AC+BC
\displaystyle =3.14\times10+12+16
\displaystyle =31.4+28=59.4\text{ cm.}
\displaystyle \text{Area of the shaded region}=\text{Area of semicircle}-\text{Area of }\triangle ABC
\displaystyle =\frac{1}{2}\pi(10)^2-\frac{1}{2}\times12\times16
\displaystyle =\frac{1}{2}\times3.14\times100-96
\displaystyle =157-96=61.0\text{ cm}^2.
\displaystyle \therefore \text{The perimeter is }59.4\text{ cm and the area is }61.0\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{In the adjoining figure, there are three semicircles }A,\ B\text{ and }C\text{ having diameter}
\displaystyle 3\text{ cm each, and another semicircle }E\text{ containing a circle }D\text{ of diameter }4.5\text{ cm. Calculate:}
\displaystyle \text{(i) the area of the shaded region, (ii) the cost of painting the shaded region at }25\text{ paise per cm}^2,
\displaystyle \text{to the nearest rupee.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the largest semicircle }E=3+3+3=9\text{ cm.}
\displaystyle \therefore \text{Radius of semicircle }E=4.5\text{ cm.}
\displaystyle \text{Radius of each semicircle }A,\ B\text{ and }C=\frac{3}{2}=1.5\text{ cm.}
\displaystyle \text{Radius of circle }D=\frac{4.5}{2}=2.25\text{ cm.}
\displaystyle \text{(i) Area of the shaded region}=\frac{1}{2}\pi(4.5)^2-2\left[\frac{1}{2}\pi(1.5)^2\right]
\displaystyle +\frac{1}{2}\pi(1.5)^2-\pi(2.25)^2
\displaystyle =10.125\pi-2.25\pi+1.125\pi-5.0625\pi
\displaystyle =3.9375\pi\text{ cm}^2
\displaystyle =3.9375\times\frac{22}{7}=12.375\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }12.375\text{ cm}^2.
\displaystyle \text{(ii) }25\text{ paise}=Rs.\ 0.25.
\displaystyle \text{Cost of painting}=12.375\times0.25=Rs.\ 3.09375.
\displaystyle \therefore \text{The cost of painting the shaded region is Rs. }3\text{ to the nearest rupee.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{In the adjoining figure, }AB\text{ and }CD\text{ are two diameters of a circle perpendicular}
\displaystyle \text{to each other and }OD\text{ is the diameter of the smaller circle. If }OA=7\text{ cm, find the area of the}
\displaystyle \text{shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the larger circle}=OA=7\text{ cm.}
\displaystyle \text{Since }OD\text{ is also a radius of the larger circle, }OD=7\text{ cm.}
\displaystyle \text{Since }OD\text{ is the diameter of the smaller circle,}
\displaystyle \text{Radius of the smaller circle}=\frac{7}{2}=3.5\text{ cm.}
\displaystyle \text{Area of the larger circle}=\pi(7)^2=49\pi\text{ cm}^2.
\displaystyle \text{Area of the smaller circle}=\pi(3.5)^2=12.25\pi\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=49\pi-12.25\pi
\displaystyle =36.75\pi\text{ cm}^2.
\displaystyle =36.75\times\frac{22}{7}=115.5\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }115.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{In the adjoining figure, }OACB\text{ is a quadrant of a circle with centre }O\text{ and}
\displaystyle \text{radius }3.5\text{ cm. If }OD=2\text{ cm, find the area of (i) quadrant }OACB\text{ and (ii) the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Area of quadrant }OACB=\frac{1}{4}\pi(3.5)^2
\displaystyle =\frac{1}{4}\times\frac{22}{7}\times(3.5)^2
\displaystyle =9.625\text{ cm}^2.
\displaystyle \therefore \text{The area of quadrant }OACB\text{ is }9.625\text{ cm}^2.
\displaystyle \text{(ii) Area of the smaller quadrant}=\frac{1}{4}\pi(2)^2
\displaystyle =\frac{1}{4}\times\frac{22}{7}\times4
\displaystyle =\frac{22}{7}\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=9.625-\frac{22}{7}
\displaystyle \approx6.482\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }6.482\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{From each of two opposite corners of a square of side }8\text{ cm, a quadrant of a circle}
\displaystyle \text{of radius }1.4\text{ cm is cut. Another circle of diameter }4.2\text{ cm is also cut from the centre,}
\displaystyle \text{as shown in the figure. Find the area of the remaining shaded portion. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of the square}=8\times8=64\text{ cm}^2.
\displaystyle \text{Area of the two quadrants}=2\left[\frac{1}{4}\pi(1.4)^2\right]
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times(1.4)^2
\displaystyle =3.08\text{ cm}^2.
\displaystyle \text{Radius of the central circle}=\frac{4.2}{2}=2.1\text{ cm.}
\displaystyle \text{Area of the central circle}=\pi(2.1)^2
\displaystyle =\frac{22}{7}\times(2.1)^2=13.86\text{ cm}^2.
\displaystyle \text{Area of the shaded portion}=64-3.08-13.86
\displaystyle =47.06\text{ cm}^2.
\displaystyle \therefore \text{The area of the remaining shaded portion is }47.06\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Find the area of the shaded region in the adjoining figure, if }AC=24\text{ cm, }BC=10\text{ cm}
\displaystyle \text{and }O\text{ is the centre of the circle. Take }\pi=3.14.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\angle ACB=90^\circ,
\displaystyle AB=\sqrt{AC^2+BC^2}
\displaystyle =\sqrt{24^2+10^2}
\displaystyle =\sqrt{576+100}=\sqrt{676}=26\text{ cm.}
\displaystyle \text{Since }AB\text{ is the diameter of the circle,}
\displaystyle \text{Radius}=\frac{26}{2}=13\text{ cm.}
\displaystyle \text{Area of the semicircle}=\frac{1}{2}\pi(13)^2
\displaystyle =\frac{1}{2}\times3.14\times169=265.33\text{ cm}^2.
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times10\times24=120\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=265.33-120
\displaystyle =145.33\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }145.33\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{In the adjoining figure, }OABC\text{ is a square of side }7\text{ cm. If }OAPC\text{ is a}
\displaystyle \text{quadrant of a circle with centre }O,\text{ find the area of the shaded region. Take }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Area of square }OABC=7\times7=49\text{ cm}^2.
\displaystyle \text{Radius of quadrant }OAPC=7\text{ cm.}
\displaystyle \text{Area of quadrant }OAPC=\frac{1}{4}\pi(7)^2
\displaystyle =\frac{1}{4}\times\frac{22}{7}\times49
\displaystyle =38.5\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=49-38.5=10.5\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }10.5\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{In the adjoining figure, }ABCD\text{ is a rectangle, having }AB=20\text{ cm and}
\displaystyle BC=14\text{ cm. Two sectors of }180^\circ\text{ have been cut off. Calculate: (i) the area of the shaded}
\displaystyle \text{region, (ii) the length of the boundary of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each semicircle}=\frac{14}{2}=7\text{ cm.}
\displaystyle \text{(i) Area of the rectangle}=20\times14=280\text{ cm}^2.
\displaystyle \text{Area of the two semicircles}=2\left[\frac{1}{2}\pi(7)^2\right]
\displaystyle =\pi\times49
\displaystyle =\frac{22}{7}\times49=154\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=280-154=126\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }126\text{ cm}^2.
\displaystyle \text{(ii) Length of each semicircular arc}=\pi r
\displaystyle =\frac{22}{7}\times7=22\text{ cm.}
\displaystyle \text{Length of the boundary}=20+20+22+22
\displaystyle =84\text{ cm.}
\displaystyle \therefore \text{The length of the boundary of the shaded region is }84\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{A circle is inscribed in an equilateral triangle }ABC\text{ of side }12\text{ cm, touching its}
\displaystyle \text{sides as shown in the adjoining figure. Find the radius of the inscribed circle and the area of the shaded part.}
\displaystyle \text{Answer:}
\displaystyle \text{Altitude of the equilateral triangle}=\sqrt{12^2-6^2}
\displaystyle =\sqrt{144-36}=\sqrt{108}=6\sqrt{3}\text{ cm.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times12\times6\sqrt{3}
\displaystyle =36\sqrt{3}\text{ cm}^2.
\displaystyle \text{Let the radius of the inscribed circle be }r\text{ cm.}
\displaystyle \text{Semi-perimeter of }\triangle ABC=\frac{12+12+12}{2}=18\text{ cm.}
\displaystyle \text{Area of a triangle}=r\times\text{Semi-perimeter}
\displaystyle 36\sqrt{3}=18r
\displaystyle r=2\sqrt{3}\text{ cm.}
\displaystyle \therefore \text{The radius of the inscribed circle is }2\sqrt{3}\text{ cm.}
\displaystyle \text{Area of the circle}=\pi(2\sqrt{3})^2
\displaystyle =12\pi\text{ cm}^2.
\displaystyle \text{Area of the shaded region}=36\sqrt{3}-12\pi
\displaystyle =36\sqrt{3}-12\times\frac{22}{7}
\displaystyle \approx24.639\text{ cm}^2.
\displaystyle \therefore \text{The area of the shaded region is }24.639\text{ cm}^2\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{The adjoining figure shows the cross-section of a railway tunnel. The radius }OA\text{ of the}
\displaystyle \text{circular part is }2\text{ m. If }\angle AOB=90^\circ,\text{ calculate: (i) the height of the tunnel, (ii) the perimeter}
\displaystyle \text{of the cross-section, and (iii) the area of the cross-section.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{OA^2+OB^2}
\displaystyle =\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\text{ m.}
\displaystyle \text{(i) Let the perpendicular distance from }O\text{ to }AB\text{ be }h\text{ m.}
\displaystyle h=\sqrt{2^2-(\sqrt{2})^2}
\displaystyle =\sqrt{4-2}=\sqrt{2}\text{ m.}
\displaystyle \text{Height of the tunnel}=2+\sqrt{2}\text{ m}
\displaystyle \approx3.414\text{ m.}
\displaystyle \therefore \text{The height of the tunnel is }(2+\sqrt{2})\text{ m}\approx3.414\text{ m.}
\displaystyle \text{(ii) The circular arc subtends an angle of }360^\circ-90^\circ=270^\circ.
\displaystyle \text{Length of the }270^\circ\text{ arc}=\frac{270}{360}\times2\pi(2)=3\pi\text{ m.}
\displaystyle \text{Perimeter}=3\pi+AB
\displaystyle =3\pi+2\sqrt{2}\text{ m.}
\displaystyle \therefore \text{The perimeter of the cross-section is }(3\pi+2\sqrt{2})\text{ m.}
\displaystyle \text{(iii) Area of the }270^\circ\text{ sector}=\frac{270}{360}\pi(2)^2=3\pi\text{ m}^2.
\displaystyle \text{Area of }\triangle AOB=\frac{1}{2}\times2\times2=2\text{ m}^2.
\displaystyle \text{Area of the cross-section}=3\pi+2\text{ m}^2.
\displaystyle \therefore \text{The area of the cross-section is }(3\pi+2)\text{ m}^2.
\displaystyle \\

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