1. Find the circumference of a circle whose radius is 7 cm.
\[ \begin{align*} \text{Given} & \\ \text{Radius of the circle} & = 7 \, \text{cm} \\ \text{Circumference of a circle:} & \quad = 2 \pi r \\ \text{Circumference} & = 2 \times \frac{22}{7} \times 7 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times {\cancelto{1}{7}} \\ & = 2 \times 22 \\ & = 44 \, \text{cm} \\ \text{Answer:} & \quad \text{The circumference of the circle is 44 cm.} \end{align*} \]
2. Find the diameter of a circle whose circumference is 66 m.
\[ \begin{align*} \text{Given} & \\ \text{Circumference of the circle} & = 66 \, \text{m} \\ \ 2 \pi r & = 66 \\ \ 2 \times \frac{22}{7} \times r & = 66 \\ \ r & = \frac{66 \times 7}{2 \times 22} \\ & = \frac{{\cancelto{3}{66}} \times 7}{2 \times {\cancelto{1}{22}}} \\ & = \frac{3 \times 7}{2} \\ \ r & = \frac{21}{2} \text{m}\\ \text{Diameter } d & = 2r \\ & = 2 \times {\frac{21}{2}} \\ & = \cancelto{1}{2} \times {\frac{21}{\cancelto{1}{2}}} \\ d & = 21 \, \text{m} \\ \text{Answer:} & \quad \text{The diameter of the circle is 21 m.} \end{align*} \]
3. If circumference of a circle is 176 cm, find its radius.
\[ \begin{align*} \text{Given} & \\ \text{Circumference of the circle} & = 176 \, \text{cm} \\ \ 2 \pi r & = 176 \\ \ 2 \times \frac{22}{7} \times r & = 176 \\ \ r & = \frac{176 \times 7}{2 \times 22} \\ & = \frac{{\cancelto{8}{176}} \times 7}{2 \times {\cancelto{1}{22}}} \\ & = \frac{{\cancelto{4}{8}} \times 7}{{\cancelto{1}{2}} \times 1} \\ & = 4 \times 7 \\ \ r & = 28 \text{cm}\\ \text{Answer:} & \quad \text{The raduis of the circle is 28 cm.} \end{align*} \]
4. The diameter of a wheel is 1.4 m, find its circumference.
\[ \begin{align*} \text{Given} & \\ \text{Diameter of the wheel } (d) & = 1.4 \, \text{m} \\ \text{Radius } (r) & = \frac{d}{2} \\ & = \frac{1.4}{2} \\ r & = 0.7 \text{m} \\ \text{Circumference of a circle} & \quad = 2 \pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{0.1}{0.7} \\ & = 44 \times 0.1 \\ & = 4.4 \, \text{m} \\ \text{Answer:} & \quad \text{The circumference of the wheel is 4.4 m.} \end{align*} \]
5. The radii of two circles are in the ratio 2: 3. What is the ratio of their circumferences?.
\[ \begin{align*} \text{Let the radii of first circle be } r_1 & = 2x \\ \text{Let the radii of second circle be } r_2 & = 3x \\ \text{The circumference of the first circle } (C_1) & = 2\pi (2x) \\ & = 4\pi x \\ \text{The circumference of the second circle } (C_2) & = 2\pi (3x) \\ & = 6\pi x \\ \text{The ratio of their circumferences} & = C_1 : C_2 \\ & = \frac{C1}{C2} \\ & = \frac{4\pi x}{6\pi x} \\ & = \frac{4}{6} \\ & = \frac{2}{3} \\ \text{Answer:} & \quad \text{The ratio of their circumferences is 2:3.} \end{align*} \]
6. The moon is nearly 385000 km away from the earth. It takes a round of the earth every month. How much distance does it travel in one month?
\[ \begin{align*} \text{Given} & \\ \text{Distance from the Earth to the Moon (radius)} & = 385000 \, \text{km} \\ \text{Circumference}& = 2\pi r \\ \text{Distance} & = 2\pi \times 385000 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{55000}{385000} \quad \\ & = 2 \times 22 \times 55000 \\ & = 44 \times 55000 \\ & = 2420000 \, \text{km} \\ \text{Answer:} & \quad \text{The moon travels approximately 2420000 km in one month.} \end{align*} \]
7. The diameter of the wheel of a car is 35 cm. How much distance will it cover in 1000 revolutions?
\[ \begin{align*} \text{Given} & \\ \text{Diameter of the wheel} & = 35 \, \text{cm} \\ \text{Radius of the wheel} & = \frac{35}{2} \, \text{cm} \\ \text{Number of revolutions} & = 1000 \\ \text{Distance covered in 1 revolutions} & = \text{Circumference} \\ \text{Circumference} & = 2 \pi r \\ & = 2 \times \frac{22}{7} \times \frac{35}{2} \quad \\ & = \cancelto{1}{2} \times \frac{22}{\cancelto{1}{7}} \times \frac{\cancelto{5}{35}}{\cancelto{1}{2}} \quad \\ & = 22 \times 5 \\ & = 110 \, \text{cm} \\ \text{Distance covered in 1000 revolutions} & = 1000 \times \text{Circumference} \\ & = 1000 \times 110 \\ & = 110000 \, \text{cm} \\ \text{Convert cm to km} & = (\text{since } 100000 \, \text{cm} = 1 \, \text{km}) \\ & = \frac{110000}{100000} \\ & = 1.1 \text{km} \\ \text{Answer:} & \quad \text{The wheel will cover 1.1 km in 1000 revolutions.} \end{align*} \]
8. The radius of a wheel of a bus is 0.70 m. How many revolutions wil it make in covering 22 km?
\[ \begin{align*} \text{Given} & \\ \text{Radius of the wheel} & = 0.70 \, \text{m} \\ & = 0.7 \, \text{m} \\ \text{Total distance} & = 22 \, \text{km} \\ & = 22000 \, \text{m} \\ \text{Distance covered in 1 revolution} & = \text{Circumference} \\ \text{Circumference} & = 2\pi r \\ & = 2\pi \times 0.7 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{0.1}{0.7} \\ & = 2 \times 22 \times 0.1 \\ & = 44 \times 0.1 \\ & = 4.4 \, \text{m} \\ \text{Number of revolutions} & = \frac{\text{Total distance}}{\text{Circumference}} \\ \text{Revolutions} & = \frac{22000}{4.4} \\ & = 5000 \\ \text{Answer:} & \quad \text{The wheel will make approximately 5000 revolutions in covering 22 km.} \end{align*} \]
9. A wire is in the form of a circle with radius 42 cm. It is bent into a square. Find the sides of the square.
\[ \begin{align*} \text{Given} & \\ \text{Radius of the Circle} & = 42 \, \text{cm} \\ \text{Perimeter of the square} & = \text{Circumference of the Circle} \\ 4 \times \text{Side} & = 2\pi r \\ 4 \times \text{Side} & = 2 \times \frac{22}{7} \times 42 \\ \text{Side} & = 2 \times \frac{22}{7} \times \frac{42}{4} \\ \text{Side} & = \frac{44}{7} \times \frac{42}{4} \\ \text{Side} & = \frac{\cancelto{11}{44}}{\cancelto{1}{7}} \times \frac{\cancelto{6}{42}}{\cancelto{1}{4}} \\ \text{Side} & = 11 \times 6 \\ \text{Side} & = 66 \, \text{cm} \\ \text{Answer:} & \quad \text{Side of the square is 66 cm} \end{align*} \]
10. The diameter of a circular park is 140 m. Around it on the outside, a path having the width of 7 m is constructed. If the path has to be fenced from inside and outside at the rate of Rs. 7 per metre, find its total cost.
\[ \begin{align*} \text{Given} & \\ \text{Diameter of the park} & = 140 \, \text{m} \\ \text{Radius} & = \frac{d}{2} \\ \text{Radius of the Inner circle } (r_1) & = 70 \, \text{m} \\ \text{Width of the path} & = 7 \, \text{m} \\ \text{Radius of the Outer cicle } (r_2) & = 70 \, \text{m} + 7 \, \text{m} \\ (r_2) & = 77 \, \text{m} \\ \\ \text{Circumference of the inner circle} & = 2\pi r_1 \\ & = 2\pi \times 70 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{10}{70} \\ & = 44 \times 10 \\ & = 440 \, \text{m} \\ \text{Circumference of the outer circle} & = 2\pi r_2 \\ & = 2\pi \times 77 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{11}{77} \\ & = 44 \times 11 \\ & = 484 \, \text{m} \\ \text{Total length to be fenced} & = 440 \, \text{m} + 484 \, \text{m} \\ & = 924 \, \text{m} \\ \text{Cost of fencing per meter} & = Rs. \, 7 \\ \text{Cost} & = 924 \times 7 \\ & = Rs. \, 6468 \\ \text{Answer:} & \quad \text{The total cost of fencing is Rs. 6468.} \end{align*} \]
11. An athlete runs around a circular park 10 times. If the diameter of park is 280 m, find the distance covered by the athlete in kilometres.
\[ \begin{align*} \text{Given} & \\ \text{Diameter} & = 280 \, \text{m} \\ \text{Radius} & = \frac{280}{2} \, \text{m} \\ \text{Radius} & = 140 \, \text{m} \\ \text{Distance covered in 1 revolution} & = \text{Circumference of the circle} \\ \text{Circumference of the circle} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{20}{140} \\ & = 44 \times 20 \\ & = 880 \, \text{m} \\ \text{Distance covered in 10 revolution} & = 10 \times \text{Circumference} \\ & = 10 \times 880 \\ & = 8800 \, \text{m} \\ \text{Converting to kilometers: } & 1000 \, \text{m} = 1 \, \text{km} \\ & = \frac{8800}{1000} \, \text{km} \\ & = 8.8 \, \text{km} \\ \text{Answer:} & \quad \text{The athlete covers a distance of 8.8 km.} \end{align*} \]
12. A circular piece of thin wire is converted into a rhombus of side 11 cm. Find the diameter of the circular piece.
\[ \begin{align*} \text{Let radius of Circle} & = r \, \text{cm} \\ \text{Circumference of circle} & = \text{Perimeter of rhombus} \\ 2 \pi r & = 4 \times \text{side} \\ 2 \times \frac{22}{7} \times r & = 4 \times 11 \\ r & = \frac{4 \times 11 \times 7}{2 \times 22} \\ r & = 7 \, \text{cm} \\ r & = 7 \, \text{cm} \\ \text{Diameter} & = 2 \times r \\ & = 2 \times 7 \\ \text{Diameter} & = 14 \, \text{cm} \\ \text{Answer:} & \quad \text{The diameter of the circular piece is 14 cm.} \end{align*} \]
13. A race track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track.
\[ \begin{align*} \text{Given} & \\ \quad \text{Let inner radius be } 'r_1' \text{and outer radius be } 'r_2' \\ \text{Inner circumference} & = 352 \, \text{m} \\ 2 \pi r_1 = 352 \, \text{m} \\ 2 \times \frac{22}{7} \times r_1 & = 352 \\ r_1 & = \frac{352 \times 7}{2 \times 22} \\ r_1 & = \frac{\cancelto{16}{352} \times 7}{2 \times \cancelto{1}{22}} \\ r_1 & = \frac{\cancelto{8}{16} \times 7}{\cancelto{1}{2} \times {1}} \\ r_1 & = 8 \times 7 \\ r_1 & = 56 \, \text{m} \\ \text{Outer circumference} & = 396 \, \text{m} \\ 2\pi r_2 = 396 \, \text{m} \\ 2 \times \frac{22}{7} \times r_2 & = 396 \\ r_2 & = \frac{396 \times 7}{2 \times 22} \\ r_2 & = \frac{\cancelto{18}{396} \times 7}{2 \times \cancelto{1}{22}} \\ r_2 & = \frac{\cancelto{9}{18} \times 7}{\cancelto{1}{2} \times {1}} \\ r_2 & = 9 \times 7 \\ r_2 & = 63 \, \text{m} \\ \text{Width of the track} & = {\text{outer radius}} - {\text{inner radius}} \\ & = r_2 - r_1 \\ & = 63 \, \text{m} - 56 \, \text{m} \\ & = 7 \, \text{m} \\ \text{Answer:} & \quad \text{The width of the track is 7 m.} \end{align*} \]
14. Radius of a circular region is 63 cm. Find the least length of rope which is sufficient to encircle the circular region.
\[ \begin{align*} \text{Given} & \\ \text{Radius (r)} & = 63 \, \text{cm} \\ \text{Length of rope} & = \text{Circumference} \\ \text{Circumference} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{9}{63} \\ & = 2 \times 22 \times 9 \\ & = 44 \times 9 \\ & = 396 \, \text{cm} \\ \text{Answer:} & \quad \text{The least length of rope required is 396 cm.} \end{align*} \]