Class 10: Areas Related to Circles

$\displaystyle \textbf{Question 1: }\text{Find the circumference and area of a circle of radius }4.2\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, radius of the circle }r=4.2\text{ cm.}$
$\displaystyle \text{Circumference of the circle}=2\pi r$
$\displaystyle =2\times\frac{22}{7}\times4.2$
$\displaystyle =26.4\text{ cm.}$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle =\frac{22}{7}\times4.2\times4.2$
$\displaystyle =55.44\text{ cm}^2.$
$\displaystyle \therefore \text{The circumference is }26.4\text{ cm and the area is }55.44\text{ cm}^2.$
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$\displaystyle \textbf{Question 2: }\text{Find the circumference of a circle whose area is }301.84\text{ cm}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle \therefore \frac{22}{7}r^2=301.84$
$\displaystyle r^2=301.84\times\frac{7}{22}=96.04$
$\displaystyle \therefore r=\sqrt{96.04}=9.8\text{ cm.}$
$\displaystyle \text{Circumference of the circle}=2\pi r$
$\displaystyle =2\times\frac{22}{7}\times9.8$
$\displaystyle =61.6\text{ cm.}$
$\displaystyle \therefore \text{The circumference of the circle is }61.6\text{ cm.}$
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$\displaystyle \textbf{Question 3: }\text{Find the area of a circle whose circumference is }44\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the circle be }r\text{ cm.}$
$\displaystyle \text{Circumference of the circle}=2\pi r$
$\displaystyle \therefore 2\times\frac{22}{7}\times r=44$
$\displaystyle r=44\times\frac{7}{44}=7\text{ cm.}$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle =\frac{22}{7}\times7\times7$
$\displaystyle =154\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the circle is }154\text{ cm}^2.$
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$\displaystyle \textbf{Question 4: }\text{A road which is }7\text{ m wide surrounds a circular park whose circumference is}$
$\displaystyle 352\text{ m.} \ \text{Find the area of the road.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the circular park be }r\text{ m.}$
$\displaystyle \text{Circumference of the park}=2\pi r$
$\displaystyle \therefore 2\times\frac{22}{7}\times r=352$
$\displaystyle \therefore r=352\times\frac{7}{44}=56\text{ m.}$
$\displaystyle \text{Width of the road}=7\text{ m.}$
$\displaystyle \therefore \text{Outer radius}=R=56+7=63\text{ m.}$
$\displaystyle \text{Area of the road}=\pi(R^2-r^2)$
$\displaystyle =\frac{22}{7}(63^2-56^2)$
$\displaystyle =\frac{22}{7}(63+56)(63-56)$
$\displaystyle =\frac{22}{7}\times119\times7$
$\displaystyle =2618\text{ m}^2.$
$\displaystyle \therefore \text{The area of the road is }2618\text{ m}^2.$
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$\displaystyle \textbf{Question 5: }\text{The circumferences of two circles are in the ratio }2:3.\text{ Find the ratio of their areas.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radii of the two circles be }r_1\text{ and }r_2.$
$\displaystyle \frac{2\pi r_1}{2\pi r_2}=\frac{2}{3}$
$\displaystyle \therefore \frac{r_1}{r_2}=\frac{2}{3}.$
$\displaystyle \text{Ratio of their areas}=\frac{\pi r_1^2}{\pi r_2^2}$
$\displaystyle =\frac{r_1^2}{r_2^2}=\left(\frac{2}{3}\right)^2=\frac{4}{9}.$
$\displaystyle \therefore \text{The ratio of the areas of the two circles is }4:9.$
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$\displaystyle \textbf{Question 6: }\text{The sum of the radii of two circles is }140\text{ cm and the difference of their}$
$\displaystyle \text{circumferences is }88\text{ cm. Find the diameters of the circles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radii of the two circles be }r_1\text{ cm and }r_2\text{ cm, where }r_1>r_2.$
$\displaystyle r_1+r_2=140 \qquad\ldots(1)$
$\displaystyle \text{Difference of their circumferences}=88\text{ cm.}$
$\displaystyle 2\pi r_1-2\pi r_2=88$
$\displaystyle 2\times\frac{22}{7}(r_1-r_2)=88$
$\displaystyle \therefore r_1-r_2=14 \qquad\ldots(2)$
$\displaystyle \text{Adding (1) and (2),}$
$\displaystyle 2r_1=154$
$\displaystyle \therefore r_1=77\text{ cm.}$
$\displaystyle \therefore r_2=140-77=63\text{ cm.}$
$\displaystyle \text{Diameter of the first circle}=2r_1=2\times77=154\text{ cm.}$
$\displaystyle \text{Diameter of the second circle}=2r_2=2\times63=126\text{ cm.}$
$\displaystyle \therefore \text{The diameters of the two circles are }154\text{ cm and }126\text{ cm.}$
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$\displaystyle \textbf{Question 7: }\text{The radii of two circles are }8\text{ cm and }6\text{ cm respectively. Find the radius of the}$
$\displaystyle \text{circle having its area equal to the sum of the areas of the two circles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}$
$\displaystyle \text{Area of the required circle}=\text{sum of the areas of the two circles}$
$\displaystyle \pi R^2=\pi(8)^2+\pi(6)^2$
$\displaystyle R^2=64+36=100$
$\displaystyle \therefore R=\sqrt{100}=10\text{ cm.}$
$\displaystyle \therefore \text{The radius of the required circle is }10\text{ cm.}$
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$\displaystyle \textbf{Question 8: }\text{The radii of two circles are }19\text{ cm and }9\text{ cm respectively. Find the radius and area}$
$\displaystyle \text{of the circle which has its circumference equal to the sum of the circumferences of the two circles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the required circle be }R\text{ cm.}$
$\displaystyle \text{Circumference of the required circle}=\text{sum of the circumferences of the two circles}$
$\displaystyle 2\pi R=2\pi(19)+2\pi(9)$
$\displaystyle 2\pi R=2\pi(19+9)$
$\displaystyle \therefore R=28\text{ cm.}$
$\displaystyle \text{Area of the required circle}=\pi R^2$
$\displaystyle =\frac{22}{7}\times28\times28$
$\displaystyle =2464\text{ cm}^2.$
$\displaystyle \therefore \text{The radius is }28\text{ cm and the area is }2464\text{ cm}^2.$
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$\displaystyle \textbf{Question 9: }\text{The outer circumference of a circular race-track is }528\text{ m. The track is everywhere}$
$\displaystyle 14\text{ m wide. Calculate the cost of levelling the track at the rate of }50\text{ paise per square metre.}$
$\displaystyle \text{Use }\pi=\frac{22}{7}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the outer radius of the circular race-track be }R\text{ m.}$
$\displaystyle \text{Outer circumference}=2\pi R=528$
$\displaystyle 2\times\frac{22}{7}\times R=528$
$\displaystyle \therefore R=528\times\frac{7}{44}=84\text{ m.}$
$\displaystyle \text{Width of the track}=14\text{ m.}$
$\displaystyle \therefore \text{Inner radius }r=84-14=70\text{ m.}$
$\displaystyle \text{Area of the track}=\pi(R^2-r^2)$
$\displaystyle =\frac{22}{7}(84^2-70^2)$
$\displaystyle =\frac{22}{7}(84+70)(84-70)$
$\displaystyle =\frac{22}{7}\times154\times14$
$\displaystyle =6776\text{ m}^2.$
$\displaystyle \text{Rate of levelling}=50\text{ paise per m}^2=\text{Rs. }0.50\text{ per m}^2.$
$\displaystyle \text{Cost of levelling}=6776\times0.50=\text{Rs. }3388.$
$\displaystyle \therefore \text{The cost of levelling the track is Rs. }3388.$
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$\displaystyle \textbf{Question 10: }\text{A circular park is surrounded by a road }21\text{ m wide. If the radius of the park is}$
$\displaystyle 105\text{ m, find the area of the road.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the circular park }r=105\text{ m.}$
$\displaystyle \text{Width of the road}=21\text{ m.}$
$\displaystyle \therefore \text{Outer radius }R=105+21=126\text{ m.}$
$\displaystyle \text{Area of the road}=\pi(R^2-r^2)$
$\displaystyle =\frac{22}{7}(126^2-105^2)$
$\displaystyle =\frac{22}{7}(126+105)(126-105)$
$\displaystyle =\frac{22}{7}\times231\times21$
$\displaystyle =15246\text{ m}^2.$
$\displaystyle \therefore \text{The area of the road is }15246\text{ m}^2.$
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$\displaystyle \textbf{Question 11: }\text{A circle of radius }7.5\text{ cm is inscribed in a square. Find the area outside the}$
$\displaystyle \text{circle and inside the square.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the circle}=7.5\text{ cm.}$
$\displaystyle \therefore \text{Diameter of the circle}=2\times7.5=15\text{ cm.}$
$\displaystyle \text{Since the circle is inscribed in the square, side of the square}=15\text{ cm.}$
$\displaystyle \text{Area of the square}=15^2=225\text{ cm}^2.$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle =\frac{22}{7}\times7.5\times7.5=176.79\text{ cm}^2\text{ (approx.).}$
$\displaystyle \text{Required area}=225-176.79=48.21\text{ cm}^2\text{ (approx.).}$
$\displaystyle \therefore \text{The area outside the circle and inside the square is }48.21\text{ cm}^2\text{ (approx.).}$
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$\displaystyle \textbf{Question 12: }\text{The area of a circle inscribed in an equilateral triangle is }154\text{ cm}^2.$
$\displaystyle \text{ Find the perimeter} \ \text{of the triangle. }\left[\text{Use }\pi=\frac{22}{7}\text{ and }\sqrt3=1.73\right]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the inscribed circle be }r\text{ cm.}$
$\displaystyle \pi r^2=154$
$\displaystyle \frac{22}{7}r^2=154$
$\displaystyle r^2=49$
$\displaystyle \therefore r=7\text{ cm.}$
$\displaystyle \text{In an equilateral triangle, the inradius is one-third of its altitude.}$
$\displaystyle \therefore \text{Altitude}=3r=3\times7=21\text{ cm.}$
$\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}$
$\displaystyle \text{Altitude}=\frac{\sqrt3}{2}a$
$\displaystyle \therefore 21=\frac{1.73}{2}a$
$\displaystyle a=\frac{42}{1.73}=24.28\text{ cm (approx.).}$
$\displaystyle \text{Perimeter of the triangle}=3a$
$\displaystyle =3\times24.28=72.83\text{ cm (approx.).}$
$\displaystyle \therefore \text{The perimeter of the equilateral triangle is }72.83\text{ cm (approx.).}$
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$\displaystyle \textbf{Question 13: }\text{A car travels }1\text{ kilometre distance in which each wheel makes }450$
$\displaystyle \text{ complete revolutions. Find} \ \text{the radius of its wheels.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance travelled}=1\text{ km}=1000\text{ m.}$
$\displaystyle \text{Number of revolutions}=450.$
$\displaystyle \text{Let the radius of each wheel be }r\text{ m.}$
$\displaystyle \text{Distance travelled}=\text{number of revolutions}\times\text{circumference of the wheel}$
$\displaystyle 1000=450\times2\pi r$
$\displaystyle 1000=450\times2\times\frac{22}{7}\times r$
$\displaystyle r=\frac{1000\times7}{450\times44}=\frac{35}{99}\text{ m.}$
$\displaystyle r=\frac{35}{99}\times100=35.35\text{ cm (approx.).}$
$\displaystyle \therefore \text{The radius of each wheel is }35.35\text{ cm (approx.).}$
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$\displaystyle \textbf{Question 14: }\text{A square of diagonal }8\text{ cm is inscribed in a circle. Find the area of}$
$\displaystyle \text{the region lying inside the circle and outside the square.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the square is inscribed in the circle, its diagonal is the diameter of the circle.}$
$\displaystyle \therefore \text{Diameter of the circle}=8\text{ cm.}$
$\displaystyle \therefore r=\frac{8}{2}=4\text{ cm.}$
$\displaystyle \text{Area of the circle}=\pi r^2$
$\displaystyle =\frac{22}{7}\times4^2=\frac{352}{7}\text{ cm}^2.$
$\displaystyle \text{If }d\text{ is the diagonal of a square, its area}=\frac{d^2}{2}.$
$\displaystyle \therefore \text{Area of the square}=\frac{8^2}{2}=32\text{ cm}^2.$
$\displaystyle \text{Required area}=\frac{352}{7}-32$
$\displaystyle =\frac{352-224}{7}=\frac{128}{7}\text{ cm}^2$
$\displaystyle =18.29\text{ cm}^2\text{ (approx.).}$
$\displaystyle \therefore \text{The required area is }18.29\text{ cm}^2\text{ (approx.).}$
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$\displaystyle \textbf{Question 15: }\text{A path of }4\text{ m width runs round a circular grassy plot whose circumference is } \\ 163\frac{3}{7}\text{ m. Find:}$
$\displaystyle \text{(i) the area of the path}$
$\displaystyle \text{(ii) the cost of gravelling the path at the rate of Rs. }1.50\text{ per square metre}$
$\displaystyle \text{(iii) the cost of turfing the plot at the rate of }45\text{ paise per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the grassy plot be }r\text{ m.}$
$\displaystyle \text{Circumference of the grassy plot}=163\frac{3}{7}\text{ m}=\frac{1144}{7}\text{ m.}$
$\displaystyle 2\pi r=\frac{1144}{7}$
$\displaystyle 2\times\frac{22}{7}\times r=\frac{1144}{7}$
$\displaystyle 44r=1144$
$\displaystyle \therefore r=26\text{ m.}$
$\displaystyle \text{Width of the path}=4\text{ m.}$
$\displaystyle \therefore \text{Outer radius }R=26+4=30\text{ m.}$
$\displaystyle \text{(i) Area of the path}=\pi(R^2-r^2)$
$\displaystyle =\frac{22}{7}(30^2-26^2)$
$\displaystyle =\frac{22}{7}(900-676)$
$\displaystyle =\frac{22}{7}\times224=704\text{ m}^2.$
$\displaystyle \therefore \text{The area of the path is }704\text{ m}^2.$

$\displaystyle \text{(ii) Cost of gravelling the path}=704\times\text{Rs. }1.50$
$\displaystyle =\text{Rs. }1056.$
$\displaystyle \therefore \text{The cost of gravelling the path is Rs. }1056.$

$\displaystyle \text{(iii) Area of the grassy plot}=\pi r^2$
$\displaystyle =\frac{22}{7}\times26\times26=\frac{14872}{7}\text{ m}^2.$
$\displaystyle 45\text{ paise}=\text{Rs. }0.45.$
$\displaystyle \text{Cost of turfing}=\frac{14872}{7}\times0.45$
$\displaystyle =\text{Rs. }956.06\text{ (approx.).}$
$\displaystyle \therefore \text{The cost of turfing the plot is Rs. }956.06\text{ (approx.).}$
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$\displaystyle \textbf{Question 16: }\text{A path of width }3.5\text{ m runs around a semi-circular grassy plot whose}$
$\displaystyle \text{perimeter is 72 m. Find the area of the path. }\left(\text{Use }\pi=\frac{22}{7}\right)\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of the semi-circular grassy plot be }r\text{ m.}$
$\displaystyle \text{Perimeter of a semicircle}=\pi r+2r$
$\displaystyle \therefore \pi r+2r=72$
$\displaystyle \frac{22}{7}r+2r=72$
$\displaystyle \frac{36}{7}r=72$
$\displaystyle \therefore r=72\times\frac{7}{36}=14\text{ m.}$
$\displaystyle \text{Width of the path}=3.5\text{ m.}$
$\displaystyle \therefore \text{Outer radius }R=14+3.5=17.5\text{ m.}$
$\displaystyle \text{Area of the path}=\frac{1}{2}\pi(R^2-r^2)$
$\displaystyle =\frac{1}{2}\times\frac{22}{7}\left[(17.5)^2-(14)^2\right]$
$\displaystyle =\frac{11}{7}(306.25-196)$
$\displaystyle =\frac{11}{7}\times110.25$
$\displaystyle =173.25\text{ m}^2.$
$\displaystyle \therefore \text{The area of the path is }173.25\text{ m}^2.$
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$\displaystyle \textbf{Question 17: }\text{Find the area enclosed between two concentric circles of radii }$
$\displaystyle 3.5\text{ cm and } \ 7\text{ cm. A third} \ \text{concentric circle is drawn outside the }7\text{ cm circle, such that}$
$\displaystyle \text{the area enclosed between it and the 7 cm circle is same as that between the two inner}$
$\displaystyle \text{circles. Find the radius of the third circle correct to one decimal place.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area enclosed between the two inner circles}=\pi(7^2-3.5^2)$
$\displaystyle =\frac{22}{7}(49-12.25)$
$\displaystyle =\frac{22}{7}\times36.75=115.5\text{ cm}^2.$
$\displaystyle \text{Let the radius of the third circle be }R\text{ cm.}$
$\displaystyle \text{Area enclosed between the third circle and the }7\text{ cm circle}=115.5\text{ cm}^2.$
$\displaystyle \therefore \pi(R^2-7^2)=115.5$
$\displaystyle \frac{22}{7}(R^2-49)=115.5$
$\displaystyle R^2-49=115.5\times\frac{7}{22}=36.75$
$\displaystyle R^2=85.75$
$\displaystyle \therefore R=\sqrt{85.75}=9.26\text{ cm (approx.).}$
$\displaystyle \therefore \text{The radius of the third circle is }9.3\text{ cm, correct to one decimal place.}$
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$\displaystyle \textbf{Question 18: }\text{An archery target has three regions formed by three concentric circles}$
$\displaystyle \text{as shown in the adjoining figure. If the diameters of the concentric circles are in the ratio }$
$\displaystyle 1:2:3, \ \text{ then find the ratio of the areas} \ \text{of the three regions.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since the diameters are in the ratio }1:2:3,\text{ the radii are also in the ratio }1:2:3.$
$\displaystyle \text{Let the radii of the three concentric circles be }r,\ 2r\text{ and }3r.$
$\displaystyle \text{Area of the innermost region}=\pi r^2.$
$\displaystyle \text{Area of the middle region}=\pi(2r)^2-\pi r^2$
$\displaystyle =4\pi r^2-\pi r^2=3\pi r^2.$
$\displaystyle \text{Area of the outermost region}=\pi(3r)^2-\pi(2r)^2$
$\displaystyle =9\pi r^2-4\pi r^2=5\pi r^2.$
$\displaystyle \therefore \text{Ratio of the areas of the three regions}=1:3:5.$
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