1. Find the circumference of a circle whose radius is \( \color{black} 7 \, cm \).
\[ \begin{align*} \text{Radius of the circle} & = 7 \, cm \\ \text{Circumference of a circle} & = 2 \pi r \\ \text{Circumference} & = 2 \times \frac{22}{7} \times 7 \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times {\cancelto{1}{7}} \\ & = 2 \times 22 \\ & = 44 \, \text{cm} \\ \end{align*} \]
Answer The circumference of the circle \( = \color{red} 44 \, cm \)
2. Find the diameter of a circle whose circumference is \( \color{black} 66 \, m \).
\[ \begin{align*} \text{Circumference of the circle} & = 66 \, \text{m} \\ \text{Circumference} & = \pi \times d\\ \pi \times d & = 66 \, m \\ \frac{22}{7} \times d & = 66 \, m \\ d & = \cancelto{3}{66} \, m \times \frac{7}{\cancelto{1}{22}} \\ & = 3 \,m \times 7 \\ d & = 21 \, m \\ \end{align*} \]
Answer The diameter of the circle \( = \color{red} 21 \, m \)
3. If circumference of a circle is \( \color{black} 176 \, cm \), find its radius.
\[ \begin{align*} \text{Circumference of the circle} & = 176 \, \text{cm} \\ 2 \pi r & = 176 \\ 2 \times \frac{22}{7} \times r & = 176 \\ r & = \frac{176 \times 7}{2 \times 22} \\ \\ & = \frac{{\cancelto{8}{176}} \times 7}{2 \times {\cancelto{1}{22}}} \\ \\ & = \frac{{\cancelto{4}{8}} \times 7}{{\cancelto{1}{2}} \times 1} \\ \\ & = 4 \times 7 \\ r & = 28 \, cm\\ \end{align*} \]
Answer The radius of the circle \( = \color{red} 28 \, cm \)
4. The diameter of a wheel is \( \color{black} 1.4 \, m \), find its circumference.
\[ \begin{align*} \text{Diameter of the wheel } (d) & = 1.4 \, \text{m} \\ \text{Circumference} & = \pi d \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{0.2}{1.4} \\ & = 22 \times 0.2 \\ & = 4.4 \, m \\ \end{align*} \]
Answer The circumference of the wheel \( = \color{red} 4.4 \, m \)
5. The radii of two circles are in the ratio \( \color{black} 2:3 \). What is the ratio of their circumferences?
\[ \begin{align*} \text{Let the radii of first circle be } r_1 & = 2x \\ \text{Let the radii of second circle be } r_2 & = 3x \\ \text{The circumference of the first circle } (C_1) & = 2\pi (r_1) \\ & = 2\pi (2x) \\ C_1& = 4\pi x \\ \text{The circumference of the second circle } (C_2) & = 2\pi (r_2) \\ & = 2\pi (3x) \\ C_2& = 6\pi x \\ \text{The ratio of their circumferences} & = C_1 : C_2 \\ & = \frac{C1}{C2} \\ & = \frac{4\pi x}{6\pi x} \\ & = \frac{4}{6} \\ & = \frac{2}{3} \\ &= 2:3 \end{align*} \]
Answer The ratio of their circumferences \( = \color{red} 2:3 \)
6. The moon is nearly \( \color{black} 385000 \, km \) away from the earth. It takes a round of the earth every month. How much distance does it travel in one month?
\[ \begin{align*} \text{Distance from the Earth to the Moon (radius)} & = 3,85,000 \, \text{km} \\ \text{Circumference}& = 2\pi r \\ & = 2 \times \frac{22}{7} \times 385000 \\ \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{55000}{385000} \quad \\ \\ & = 2 \times 22 \times 55000 \\ & = 44 \times 55000 \\ & = 24,20,000 \, \text{km} \\ \end{align*} \]
Answer The moon travels \( \color{red} 24,20,000 \, km \) in one month.
7. The diameter of the wheel of a car is \( \color{black} 35 \, cm \). How much distance will it cover in \( \color{black} 1000 \) revolutions?
\[ \begin{align*} \text{Diameter of the wheel} & = 35 \, \text{cm} \\ \text{Number of revolutions} & = 1000 \\ \text{Distance covered in 1 revolutions} & = \text{Circumference} \\ \text{Circumference} & = \pi d \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{5}{35} \, cm \\ \\ & = 22 \times 5 \, cm \\ \text{Distance covered in 1 revolutions} & = 110 \, \text{cm} \\ \\ \text{Distance covered in 1000 revolutions} & = 1000 \times \text{Circumference} \\ & = 1000 \times 110 \\ & = 110000 \, \text{cm} \\ \\ \text{Convert cm to km} & \implies \color{green} 1 \, cm = \frac{1}{100000}km \\ \\ 110000 \, cm & = \frac{110000}{100000} \, km \\ \\ & = 1.1 \, km \\ \end{align*} \]
Answer Distance covered by the wheel in 1000 revolutions \( = \color{red} 1.1 \, km \)
8. The radius of a wheel of a bus is \( \color{black} 0.70 \, m \). How many revolutions will it make in covering \( \color{black} 22 \, km \)?
\[ \begin{align*} \text{Radius of the wheel} & = 0.70 \, m \\ \text{Distance covered in 1 revolution} & = \text{Circumference} \\ \text{Circumference} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{0.10}{0.70} \, m \\ & = 2 \times 22 \times 0.10 \, m \\ & = 44 \times 0.10 \, m\\ & = 4.40 \, m \\ \text{Circumference} & = 4.4 \, m \\ \\ \text{Total distance} & = 22 \, km \\ \\ \text{Convert km to m} & \implies \color{green} 1 \, km = 1000 \, m \\ & = 22 \times 1000 \\ \text{Total distance} &= 22000 \, m \\ \\\text{Number of revolutions} & = \frac{\text{Total distance}}{\text{Circumference}} \\ \\ \text{Revolutions} & = \frac{22000}{4.4} \\ \\ & = \frac{22000}{4.4} \times \frac{10}{10}\\ \\ & = \frac{\cancelto{20000}{220000}}{\cancelto{4}{44}} \\ \\ &= \frac{20000}{4} \\ \\ & = 5000 \\ \end{align*} \]
Answer Number of revolutions covered in 22 km \( = \color{red} 5000 \)
9. A wire is in the form of a circle with radius \( \color{black}42 \, cm \). It is bent into a square. Find the sides of the square.
\[ \begin{align*} Radius & = 42 \, \text{cm} \\ \text{Perimeter of the square} & = \text{Circumference of the Circle} \\ 4 \times Side & = 2\pi r \\ \\ 4 \times Side & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{6}{42} \\ \\ 4 \times Side & = 44 \times 6 \\ \\ Side & = \frac{\cancelto{11}{44} \times 6}{\cancelto{1}{4}} \\ \\ Side & = 11 \times 6 \\ \\ Side & = 66 \, \text{cm} \\ \end{align*} \]
Answer Sides of the square \( = \color{red} 66 \, cm \)
10. The diameter of a circular park is \( \color{black} 140 \, m \). Around it on the outside, a path having the width of \( \color{black} 7 \, m \) is constructed. If the path has to be fenced from inside and outside at the rate of \( \color{black} \text{₹ }7 \) per metre, find its total cost.
\[ \begin{align*} \text{Diameter of the park} & = 140 \, \text{m} \\ \text{Radius} & = \frac{d}{2} \\ \text{Radius of the Inner circle } (r_1) & = 70 \, \text{m} \\ \text{Width of the path} & = 7 \, \text{m} \\ \text{Radius of the Outer cicle } (r_2) & = 70 \, \text{m} + 7 \, \text{m} \\ (r_2) & = 77 \, m \\ \\ \text{Circumference of the inner circle} & = 2\pi (r_1) \\ \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{10}{70} \, m \\ \\ & = 44 \times 10 \, m\\ & = 440 \, m \\ \text{Circumference of the outer circle} & = 2\pi (r_2) \\ \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{11}{77} \, m \\ \\ & = 44 \times 11 \, m\\ & = 484 \, m \\ \text{Total length to be fenced} & = 440 \, m + 484 \, m \\ & = 924 \, m \\ \text{Cost of fencing per meter} & = \text{₹ } 7 \\ \text{Cost} & = 924 \times 7 \\ & = \text{₹ } 6468 \\ \end{align*} \]
Answer The total cost of fencing \( = \color{red} \text{₹ } 6468 \)
11. An athlete runs around a circular park 10 times. If the diameter of park is \( \color{black} 280 \, m \), find the distance covered by the athlete in kilometres.
\[ \begin{align*} \text{Diameter} & = 280 \, \text{m} \\ \text{Distance covered in 1 revolution} & = \text{Circumference of the circle} \\ \text{Circumference of the circle} & = \pi d \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{40}{280} \\ \\ & = 22 \times 40 \\ & = 880 \, m \\ \text{Distance covered in 10 revolution} & = 10 \times \text{Circumference} \\ & = 10 \times 880 \\ & = 8800 \, \text{m} \\ \text{Convert } & m \text{ to } km \\ \\ 1 \, m & = \frac{1}{1000} \, km \\ \\ 8000 \, m & = \frac{8800}{1000} \, \text{km} \\ \\ & = 8.8 \, \text{km} \\ \\ \end{align*} \]
Answer Distance covered by the athlete \( = \color{red} 8.8 \, km \)
12. A circular piece of thin wire is converted into a rhombus of side \( \color{black} 11 \, cm \). Find the diameter of the circular piece.
\[ \begin{align*} \text{Side} & = 11 \, cm \\ \text{Circumference of circle} & = \text{Perimeter of rhombus} \\ 2 \pi r & = 4 \times \text{side} \\ \\ 2 \times \frac{22}{7} \times r & = 4 \times 11 \, cm \\ \\ \frac{44}{7} \times r & = 44 \, cm \\ \\ r & = \cancelto{1}{44} \, cm \times \frac{7}{\cancelto{1}{44}} \\ \\ r & = 7 \, \text{cm} \\ \text{Diameter} & = 2 \times r \\ & = 2 \times 7 \, cm \\ \text{Diameter} & = 14 \, cm \\ \end{align*} \]
Answer The diameter of the circular piece \( = \color{red} 14 \, cm \)
13. A race track is in the form of a ring whose inner circumference is \( \color{black} 352 \, m \), and the outer circumference is \( \color{black} 396 \, m \). Find the width of the track.
\[ \begin{align} \text{Let inner radius } & = r_1 \\ \text{ outer radius } & = r_2 \\ \\ \text{Inner circumference } (C_1) & = 352 \, m \\ 2 \pi r_1 & = 352 \, m \\ 2 \times \frac{22}{7} \times r_1 & = 352 \\ r_1 & = \frac{352 \times 7}{2 \times 22} \\ r_1 & = \frac{\cancelto{16}{352} \times 7}{2 \times \cancelto{1}{22}} \\ r_1 & = \frac{\cancelto{8}{16} \times 7}{\cancelto{1}{2} \times {1}} \\ r_1 & = 8 \times 7 \\ r_1 & = 56 \, m \\ \\ \text{Outer circumference} (C_2) & = 396 \, m \\ 2\pi r_2 &= 396 \, m \\ 2 \times \frac{22}{7} \times r_2 & = 396 \\ r_2 & = \frac{396 \times 7}{2 \times 22} \\ r_2 & = \frac{\cancelto{18}{396} \times 7}{2 \times \cancelto{1}{22}} \\ r_2 & = \frac{\cancelto{9}{18} \times 7}{\cancelto{1}{2} \times {1}} \\ r_2 & = 9 \times 7 \\ r_2 & = 63 \, m \\ \text{Width of the track} & = {\text{outer radius}} - {\text{inner radius}} \\ & = r_2 - r_1 \\ & = 63 \, m - 56 \, m \\ & = 7 \, m \\ \end{align} \]
Answer Width of the track \( = \color{red} 7 \, m \)
14. Radius of a circular region is \( \color{black} 63 \, cm \). Find the least length of rope which is sufficient to encircle the circular region.
\[ \begin{align*} \text{Radius (r)} & = 63 \, \text{cm} \\ \text{Length of rope} & = \text{Circumference} \\ \text{Circumference} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{9}{63} \, cm \\ & = 2 \times 22 \times 9 \, cm \\ & = 44 \times 9 \, cm \\ & = 396 \, \text{cm} \\ \end{align*} \]
Answer Least length of rope required \( = \color{red} 396 \, cm \)