1. Find the area of a circle of diameter \( \color{black} 8.4 \, m \).
\[ \begin{align*} \text{Diameter } (d) & = 8.4 \, \text{m} \\ \text{Radius } (r) & = \frac{d}{2} \\ & = \frac{8.4}{2} \, m \\ \text{Radius of the circle } (r) & = 4.2 \, m \\ \text{Area} & = \pi r^2 \\ \\ & = \frac{22}{7} \times (4.2)^2 \\ \\ & = \frac{22}{7} \times 4.2 \times 4.2 \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{0.6}{4.2} \times 4.2 \\ \\ & = 22 \times 0.6 \times 4.2 \\ & = 55.44 \, m^2 \\ \end{align*} \]
Answer Area of the circle \( = \color{red} 55.44 \, m^2 \)
2. The circumference of a circle is \( \color{black} 17.6 \, cm \). Find the area of the circle.
\[ \begin{align*} \text{Circumference of the circle} & = 17.6 \, \text{cm} \\ 2\pi r & = 17.6 \\ \\ 2 \times \frac{22}{7} \times r & = 17.6 \\ \\ r & = \frac{\cancelto{8.8}{17.6} \times 7}{\cancelto{1}{2} \times 22} \\ \\ r & = \frac{\cancelto{0.4}{8.8} \times 7}{{1} \times \cancelto{1}{22}} \\ \\ r & = 0.4 \times 7 \\ r & = 2.8 \, cm \\ \\ \text{Area of a circle} & = \pi r^2 \\ \\ & = \frac{22}{7} \times (2.8)^2 \\ \\ & = \frac{22}{7} \times 2.8 \times 2.8 \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{0.4}{2.8} \times 2.8 \\ \\ & = 22 \times 0.4 \times 2.8 \\ & = 24.64 \, cm^2 \\ \end{align*} \]
Answer Area of the circle \( = \color{red} 26.64 \, cm^2 \)
3. The area of a circular lawn is \( \color{black} 154 \, m^2 \), find its circumference.
\[ \begin{align*} \text{Area} & = 154 \, \text{m}^2 \\ \text{Area of a circle} & = \pi r^2 \\ \pi r^2 & = 154 \\ \frac{22}{7} \times r^2 & = 154 \\ r^2 & = \cancelto{7}{154} \times \frac{7}{\cancelto{1}{22}} \\ r^2 & = 7 \times 7 \\ r^2 & = 7^2 \\ r & = 7 \, m \\ \\ \text{Circumference} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{1}{7} \\ & = 44 \, m \\ \end{align*} \]
Answer Circumference of the lawn \( = \color{red} 44 \, m \)
4. A circular plot is of radius \( \color{black} 77 \, m \). Find its value at \( \color{black} \text{₹ }175 \, \) per square metre.
\[ \begin{align*} \text{Radius } (r)& = 77 \, m \\ \text{Area} &= \pi r^2 \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{11}{77} \times 77 \\ \\ & = 22 \times 11 \times 77 \\ & = 22 \times 847 \\ & = 18634 \, m^2 \\ \\ \text{Cost} & = \text{₹ } 175 \\ \text{Value of the plot} & = \text{Area} \times \text{Cost} \\ & = 18634 \times 175 \\ & = \text{₹ } 32,60,950 \\ \end{align*} \]
Answer Value of the circular plot \( = \color{red} \text{₹ } 32,60,950\)
5. The diameters of two silver discs are in the ratio \( \color{black} 2:3 \). What will be the ratio of their areas?
\[ \begin{align*} \text{Let diameters of two dics be } & (d_1) \text{ and } (d_2) \\ d_1 : d_2 & = 2:3 \\ \\ \text{Let raduis of two dics be } & (r_1) \text{ and } (r_2) \\ \\ r_1 : r_2 & = \frac{d_1}{2}: \frac{d_2}{2} \\ \\ r_1 : r_2 & = \frac{2}{2}: \frac{3}{2} \\ \\ r_1 : r_2 & = 1: \frac{3}{2} \\ \\ \text{Let Area of two dics be } & (A_1) \text{ and } (A_2) \\ \\ A_1 : A_2 & = \pi (r_1)^2 : \pi (r_1)^2 \\ \\ \frac{A_1}{A_2} & = \frac{\pi (r_1)^2}{\pi (r_2)^2} \\ \\ \frac{A_1}{A_2} & = \frac{\cancel\pi (r_1)^2}{\cancel\pi (r_2)^2} \\ \\ & = \frac{1^2}{\left(\frac{3}{2}\right)^2} \\ \\ & = \frac{1}{\left(\frac{9}{4}\right)} \\ \\ & = 1 \times \frac{4}{9} \\ \\ \frac{A_1}{A_2} & = \frac{4}{9} \\ \\ A_1 : A_2 &= 4:9 \end{align*} \]
Answer Ratio of their areas \( = \color{red} 4:9 \)
6. If the perimeter of a semi-circular park is \( \color{black} 72 \, m \), find its area and radius.
\[ \begin{align*} \text{Perimeter } & = 72 \, \text{m} \\ \text{Perimeter of semi circular park } & = \pi r + 2r \\ \pi r + 2r & = 72 \\ (\pi + 2)r & = 72 \\ \\ \left({\frac{22}{7} + 2}\right)r & = 72 \\ \\ \left({\frac{22 +14}{7}} \right)r & = 72 \\ \\ \left({\frac{36}{7}} \right)r & = 72 \\ \\ r & = \cancelto{2}{72} \times \frac{7}{\cancelto{1}{36}} \\ \\ r & = 2 \times 7 \\ r & = 14 \, m \\ \\ \text{Area of semi circular park }(A) & = \frac{1}{2} \pi r^2 \\ \\ & = \frac{1}{2} \times \frac{22}{7} \times 14^2 \\ \\ & = \frac{1}{\cancelto{1}{2}} \times \frac{\cancelto{11}{22}}{\cancelto{1}{7}} \times \cancelto{2}{14} \times 14 \\ \\ & = 11 \times 2 \times 14 \\ & = 11 \times 28 \\ Area & = 308 \, \text{m}^2 \\ \end{align*} \]
Answer Radius \( = \color{red} 14 \,m \) , Area \( = \color{red} 308 \,m^2 \)
7. If a wire of \( \color{black} 440 \, m \) length is moulded in the form of a circle and a square turn-by-turn, find the ratio of the area of the circle to that of square.
\[ \begin{align*} \text{Length of the wire} & = 440 \, \text{m} \\ \text{Circumference of circle} & = \text{Length of the wire} \\ 2\pi r & = 440 \\ 2 \times \frac{22}{7} \times r & = 440 \\ \\ \frac{44}{7} \times r & = 440 \\ \\ r & = \cancelto{10}{440} \times \frac{7}{\cancelto{1}{44}} \\ \\ r & = 10 \times 7 \\ r & = 70 \, m \\ \\\text{Perimeter of the square} & = \text{Length of the wire} \\ 4 \times \text{Side} & = 440 \, m \\ \\ \text{Side} & = \frac{440}{4} \, m \\ \\ \text{Side} & = 110 \, m \\ \\= \text{Area of the Circle} : & \text{ Area of the Square} \\ \\ & = \frac{\text{Area of the Circle}}{\text{Area of the Square}} \\ \\ & = \frac{\pi r^2}{(Side)^2} \\ \\ & = \frac{22 \times \cancelto{10}{70} \times 70}{\cancelto{1}{7} \times 110 \times 110} \\ \\ & = \frac{22 \times 1\cancel{0} \times 7\cancel{0}}{11\cancel{0} \times 11\cancel{0}} \\ \\ & = \frac{\cancelto{2}{22} \times 7}{\cancelto{1}{11} \times 11} \\ \\ & = \frac{14}{11} \\ \\ & = 14 : 11 \\ \end{align*} \]
Answer Ratio of the area of the circle to that of square \( = \color{red} 14 : 11 \)
8. Find the area of the square that can be inscribed in a circle of radius \( \color{black} 8 \, cm \).
\[ \begin{align*} \text{Radius} & = 8 \, cm \\ \text{Diagonal of the Square (Diameter)} & = 2 \times \text{Radius} \\ \text{Let Diagonal (AC) be } d_1 & = 16 \, cm \\ \text{Let Diagonal (BD) be } d_2 & = 16 \, cm \\ \text{Area of the Square} & = \text{Area of the Rhombus} \\ \\ & = \frac{1}{2} \times d_1 \times d_2 \\ \\ & = \frac{1}{\cancelto{1}{2}} \times \cancelto{8}{16} \times 16 \\ & = 8 \times 16 \\ & = 128 \, cm^2 \\ \end{align*} \]
Answer Area of the square \( = \color{red} 128 \, cm^2 \)
9. The diameter of a circular park is \( \color{black} 24 \, m \). A \( \color{black} 3 \, m \) wide path runs around it from inside. Find the cost of paving the path at \( \color{black} \text{₹ } 25 \) per square meter.
\[ \begin{align*} \text{Outer Diameter} & = 24 \, \text{m} \\ \text{Outer Radius } (r_2) & = \frac{24}{2} \\ r_2 & = 12 \, m \\ \text{Width of the path} & = 3 \, m \\ \text{Inner Radius } (r_1) & = 12 \, m - 3 \, m \\ (r_1) & = 9 \, m \\ \text{Area of the Path} & = \text{Area of outer circle} - \text{Area of inner circle} \\ & = \pi (r_2)^2 - \pi (r_1)^2 \\ \\ & = \pi [(r_2)^2 - (r_1)^2] \\ \\ & = \frac{22}{7} [(12)^2 - (9)^2] \\ \\ & = \frac{22}{7} (144 - 81) \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{9}{63} \\ & = 22 \times 9 \\ \text{Area of the Path} & = 198 \, m^2 \\ \\ \text{Cost per square meter} & = \text{₹} 25 \\ \text{Total Cost} & = Area \times Cost \\ & = 198 \times 25 \\ & = \text{₹} 4950 \\ \end{align*} \]
Answer The cost of paving the path \( = \color{red} \text{₹} 4950 \)
10. A paper is in the form of a rectangle PQRS in which \( \color{black} PQ = 10 \, cm \) and \( \color{black} QR = 7 \, cm \). Two semi-circular portions with QR and PS as diameters are cut off (As shown in the diagram). Find the area of the remaining part.
\[ \begin{align*} \text{Length of the rectangle (PQ)} & = 10 \, cm \\ \text{Width of the rectangle (QR)} & = 7 \, cm \\ \text{Diameter} & = 7 \, cm \\ \text{Radius }(r) & = \frac{7}{2} \, cm \\ \text{Area of the remaining part} & = \text{Area of the rectangle} - \text{Area of the circle} \\ & = (L \times B) - (\pi r^2) \\ \\ & = (10 \times 7) - \left(\frac{\cancelto{11}{22}}{\cancelto{1}{7}} \times \frac{\cancelto{1}{7}}{\cancelto{1}{2}} \times \frac{7}{2} \right) \\ \\ & = 70 - \left(\frac{11 \times 7}{2} \right) \\ \\ & = 70 - \left(\frac{77}{2} \right) \\ \\ & = 70 - 38.5 \\ & = 31.5 \, cm^2 \\ \end{align*} \]
Answer The area of the remaining part \( = \color{red} 31.5 \, cm^2 \)
11. A horse is tied to a pole with \( \color{black} 21 \, m \) long string. Find the area where the horse can graze.
\[ \begin{align*} \text{Radius } (r) & = 21 \, \text{m} \\ \text{Area of a circle } (A) & = \pi r^2 \\ \\ A & = \frac{22}{7} \times 21^2 \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{3}{21} \times 21 \\ & = 22 \times 63 \\ A & = 1386 \, \text{m}^2 \end{align*} \]
Answer The area where the horse can graze \( = \color{red} 1386 \, m^2 \)
12. Two concentric circles have radii of \( \color{black} 14 \, cm \) and \( \color{black} 7 \, cm \) respectively. Find the area of space between them.
\[ \begin{align*} \text{Radius of the larger circle } (R) & = 14 \, \text{cm} \\ \text{Radius of the smaller circle } (r) & = 7 \, \text{cm} \\ \text{Area of the space between the circles} & = {\text{Area of the larger circle}} - {\text{Area of the smaller circle}} \\ & = \pi R^2 - \pi r^2 \\ & = \pi (R^2 - r^2) \\ \\ & = \frac{22}{7} (14^2 - 7^2) \\ \\ & = \frac{22}{7} (196 - 49) \\ \\ & = \frac{22}{7} \times (147) \\ \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{21}{147} \\ \\ & = 22 \times 21 \\ & = 462 \, \text{cm}^2 \\ \end{align*} \]
Answer The area of the space between the circles \( = \color{red} 462 \, cm^2 \)
13. The sum of circumferences of four small circles of equal radius is equal to the circumference of a bigger circle. Find the ratio of the area of the bigger circle to that of the smaller circle.
\[ \begin{align*} \text{Smaller circle radius} = r \\ \text{Bigger circle radius} = R \\ \text{Circumference of bigger circle} & = \text{Circumference of the 4 smaller circles} \\ 2\pi R & = 4 \times 2\pi r \\ 2\pi R & = 8\pi r \\ \\ R & = \frac{\cancelto{4}{8} \cancel{\pi} r}{\cancelto{1}{2} \cancel{\pi}} \\ \\ R & = 4r \\ \\ \text{Area of the bigger circle} &: \text{Area of the smaller circle} \\ & = \pi R^2 : \pi r^2 \\ & = \pi (4r)^2 : \pi r^2 \\ & = \pi 16r^2 : \pi r^2 \\ \\ & = \frac{\cancel{\pi} 16 \cancel{r^2}}{\cancel{\pi r^2}} \\ \\ & = \frac{16}{1} \\ \\ & = 16:1 \end{align*} \]
Answer The ratio \( = \color{red} 16:1 \)
14. The area of a circle is equal to the area of a rectangle with sides \( \color{black} 112 \, m \) and \( \color{black} 88 \, m \) respectively. Find the circumference of the circle.
\[ \begin{align*} \text{Length of the rectangle} & = 112 \, m \\ \text{Breadth of the rectangle} & = 88 \, m \\ \text{Area of the Circle} & = \text{Area of the rectangle} \\ \pi r^2 & = \text{L} \times \text{B} \\ \frac{22}{7} \times r^2 & = 112 \times 88 \\ r^2 & = 112 \times \cancelto{4}{88} \times \frac{7}{\cancelto{1}{22}} \\ r^2 & = 112 \times 4 \times 7 \\ r^2 & = (2 \times 2 \times 2 \times 2 \times 7) \times (2 \times 2) \times 7 \\ r^2 & = (2 \times 2 \times 2 \times 7) \times (2 \times 2 \times 2 \times 7) \\ r^2 & = (56)(56) \\ r^2 & = (56)^2 \\ r & = 56 \, \text{m} \\ \\\text{Circumference of the circle} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{8}{56} \\ & = 2 \times 22 \times 8 \\ & = 352 \, \text{m} \\ \end{align*} \]
Answer Circumference of the circle \( = \color{red} 352 \, m \)