1. Find the area of a circle of diameter 8.4 m.
\[ \begin{align*} \text{Given:} & \\ \text{Diameter} & = 8.4 \, \text{m} \\ \text{Radius of the circle} & = \frac{8.4}{2} \, \text{m} \\ & r = 4.2 \, \text{m} \\ \text{Area} & = \pi r^2 \\ & = \frac{22}{7} \times 4.2 \times 4.2 \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{0.6}{4.2} \times 4.2 \\ & = 22 \times 0.6 \times 4.2 \\ & = 55.44 \, \text{m}^2 \\ \text{Answer:} & \quad \text{The area of the circle is approximately 55.44 m}^2. \end{align*} \]
2. The circumference of a circle is 17.6 cm. Find the area of the circle.
\[ \begin{align*} \text{let radius of circle } & = r \, \text{cm} \\ \text{Circumference of the circle} & = 17.6 \, \text{cm} \\ 2\pi r & = 17.6 \\ 2 \times \frac{22}{7} \times r & = 17.6 \\ r & = \frac{\cancelto{8.8}{17.6} \times 7}{\cancelto{1}{2} \times 22} \\ r & = \frac{\cancelto{0.4}{8.8} \times 7}{{1} \times \cancelto{1}{22}} \\ r & = 0.4 \times 7 \\ r & = 2.8 \, \text{cm} \\ \text{Area of a circle} & = \pi r^2 \\ \text{Area} & = \frac{22}{7} \times 2.8^2 \\ \text{Area} & = \frac{22}{7} \times 2.8 \times 2.8 \\ \text{Area} & = \frac{22}{\cancelto{1}{7}} \times \cancelto{0.4}{2.8} \times 2.8 \\ & = 22 \times 0.4 \times 2.8 \\ & = 24.64 \, \text{cm}^2 \\ \text{Answer:}& \quad \text{The area of the circle is 24.64 cm}^2. \end{align*} \]
3. The area of a circular lawn is 154 m2, find its circumference.
\[ \begin{align*} \text{Given} & \\ \text{Area} & = 154 \, \text{m}^2 \\ \text{Area of a circle} & = \pi r^2 \\ \pi r^2 & = 154 \\ \frac{22}{7} \times r^2 & = 154 \\ r^2 & = \frac{\cancelto{7}{154} \times 7}{\cancelto{1}{22}} \\ r^2 & = 7 times 7 \\ r^2 & = 7^2 \\ r & = \sqrt{7^2} \\ r & = 7 \, \text{m} \\ \text{Circumference of a circle} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{1}{7} \\ & = 44 \, \text{m} \\ \text{Answer:} & \quad \text{The circumference of the lawn is 44 m.} \end{align*} \]
4. A circular plot is of radius 77 m. Find its value at Rs. 175 per square metre.
\[ \begin{align*} \text{Given} & \\ \text{Radius} & = 77 \, \text{m} \\ \text{Area of a circle} = \pi r^2 \\ \text{Area} & = \frac{22}{\cancelto{1}{7}} \times \cancelto{11}{77} \times 77 \\ & = 22 \times 11 \times 77 \\ & = 22 \times 847 \\ & = 18634 \, \text{m}^2 \\ \text{Rate per square meter} & = Rs.\, 175 \\ \text{Value of the Circular plot} & = \text{Area} \times \text{Rate} \\ & = 18634 \times 175 \\ & = Rs.\, 32,60,950 \\ \text{Answer:}& \quad \text{The value is Rs. 32,60,950.} \end{align*} \]
5.The diameters of two silver discs are in the ratio 2: 3. What will be the ratio of their areas?
\[ \begin{align*} \text{Given:} & \\ \text{Let diameters of two dics be } & (d_1) \text{ and } (d_2) \\ \text{Let } d_1 & = 2x \\ \text{Let } d_2 & = 3x \\ \text{Let raduis of two dics be } & (r_1) \text{ and } (r_2) \\ r_1 & = \frac{d_1}{2} \\ r_1 & = \frac{2x}{2} \\ r_1 & = x \\ r_2 & = \frac{d_2}{2} \\ r_2 & = \frac{3x}{2} \\ \text{Let Area of two dics be } & (A_1) \text{ and } (A_2) \\ A_1 : A_2 & = \frac{A_1}{A_2} \\ \frac{A_1}{A_2} & = \frac{\pi (r_1)^2}{\pi (r_2)^2} \\ & = \frac{\cancelto{1}{\pi} (x)^2}{\cancelto{1}{\pi} (\frac{3x}{2})^2} \\ & = \frac{x^2}{(\frac{9}{4})x^2} \\ & = \frac{4}{9} \\ \text{Answer:} & \quad \text{The ratio of their areas is 4:9.} \end{align*} \]
6. If the perimeter of a semi-circular park is 72 m, find its area and radius.
\[ \begin{align*} \text{Let radius} & = r \, \text{m} \\ \text{Perimeter of the semi-circular park} & = 72 \, \text{m} \\ \text{Perimeter } (P) & = \pi r + 2r \\ \pi r + 2r & = 72 \\ (\pi + 2)r & = 72 \\ ({\frac{22}{7} + 2})r & = 72 \\ ({\frac{22}{7} + \frac{14}{7}})r & = 72 \\ ({\frac{36}{7}})r & = 72 \\ r & = \cancelto{2}{72} \times \frac{7}{\cancelto{1}{36}} \\ r & = 2 \times 7 \\ r & = 14 \, \text{m} \\ \text{Area of semi circular park }(A) & = \frac{1}{2} \pi r^2 \\ & = \frac{1}{2} \times \frac{22}{7} \times 14^2 \\ & = \frac{1}{\cancelto{1}{2}} \times \frac{\cancelto{11}{22}}{\cancelto{1}{7}} \times \cancelto{2}{14} \times 14 \\ & = 22 \times 14 \\ A & = 308 \, \text{m}^2 \\ \text{Answer:} & \quad \text{The radius is 14 m and its area is 308 m}^2. \end{align*} \]
7. If a wire of 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.
\[ \begin{align*} \text{Given} & \\ \text{Length of the wire} & = 440 \, \text{m} \\ \text{Circumference of circle} & = \text{Length of the wire} \\ 2\pi r & = 440 \\ 2 \times \frac{22}{7} \times r & = 440 \\ r & = \frac{440 \times 7}{2 \times 22} \\ r & = \frac{\cancelto{220}{440} \times 7}{\cancelto{1}{2} \times 22} \\ r & = \frac{\cancelto{10}{220} \times 7}{1 \times \cancelto{1}{22}} \\ r & = 10 \times 7 \\ r & = 70 \, \text{m} \\ \\ \text{Perimeter of the square} & = \text{Wire length}\\ 4 \times \text{Side} & = 440 \\ \text{Side} & = \frac{440}{4} \\ \text{Side} & = 110 \, \text{m} \\ \\ \text{Area of the Circle : Area of the Square} & = \frac{\text{Area of the Circle}}{\text{Area of the Square}} \\ \frac{\text{Area of the Circle}}{\text{Area of the Square}} & = \frac{\pi r^2}{\text{Side}^2} \\ & = \frac{22 \times \cancelto{10}{70} \times 7\cancelto{}0}{\cancelto{1}{7} \times 110 \times 11\cancelto{}0} \\ & = \frac{\cancelto{2}{22} \times 1\cancelto{}0 \times 7}{1 \times 11\cancelto{}0 \times \cancelto{1}{11}} \\ & = \frac{2 \times 7}{11} \\ \frac{\text{Area of the Circle}}{\text{Area of the Square}} & = \frac{14}{11} \\ \\ \text{Answer:} & \quad \text{Area of the Circle : Area of the Square is }\\ & \quad 14 : 11 \\ \end{align*} \]
8. Find the area of the square that can be inscribed in a circle of radius 8 cm.
\[ \begin{align*} \text{Given:} & \\ \text{Radius of the circle} & = 8 \, \text{cm} \\ \text{Diameter (Diagonal of the square)} & = 2 \times \text{Radius} \\ \text{Let Diagonal (AC) be } d_1 & = 2 \times 8 \\ d_1 & = 16 \, \text{cm} \\ \text{Let Diagonal (BD) be } d_2 & = 2 \times 8 \\ d_2 & = 16 \, \text{cm} \\ \text{Area of the Square} & = \text{Area of the Rhombus} \\ & = \frac{1}{2} \times d_1 \times d_2 \\ & = \frac{1}{2} \times 16 \times 16 \\ & = 8 \times 16 \\ & = 128 \, \text{cm}^2 \\ \text{Answer: } & \text{The area of the inscribed square is 128 cm}^2. \end{align*} \]
9. The diameter of a circular park is 24 m. A 3 m wide path runs around it from inside. Find the cost of paving the path at Rs. 25 per square meter.
\[ \begin{align*} \text{Given} & \\ \text{Outer Diameter} & = 24 \, \text{m} \\ \text{Outer Radius } (r_2) & = \frac{24}{2} = 12 \, \text{m} \\ \text{Width of the path} & = 3 \, \text{m} \\ \text{Inner Radius } (r_1) & = 12 \, \text{m} - 3 \, \text{m} \\ (r_1) & = 9 \, \text{m} \\ \text{Area of the Path} & = \text{Area of outer circle} - \text{Area of inner circle} \\ & = \pi (r_2)^2 - \pi (r_1)^2 \\ & = \frac{22}{7} [(r_2)^2 - (r_1)^2] \\ & = \frac{22}{7} [(12)^2 - (9)^2] \\ & = \frac{22}{7} (144 - 81) \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{9}{63} \\ & = 22 \times 9 \\ \text{Area of the Path} & = 198 \, \text{m}^2 \\ \text{Cost per square meter} & = Rs.\, 25 \\ \text{Total Cost} & = \text{Area of the Path} \times \text{Cost per square meter} \\ & = 198 \, \text{m}^2 \times Rs.\, 25 \\ & = Rs.\, 4950 \\ \text{Answer:} & \quad \text{The cost of paving the path is Rs. 4950.} \end{align*} \]
10. A paper is in the form of a rectangle PQRS in which PQ = 10 cm and 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.
\[ \begin{align*} \text{Given:} \\ \text{Length of the rectangle (PQ)} & = 10 \, \text{cm} \\ \text{Width of the rectangle (QR)} & = 7 \, \text{cm} \\ \text{Diameter} & = 7 \, \text{cm} \\ \text{Radius }(r) & = \frac{7}{2} \, \text{cm} \\ \text{Area of the remaining part} & = \text{Area of the rectangle} - \text{Area of the circle} \\ & = (L \times B) - (\pi r^2) \\ & = (10 \times 7) - (\frac{\cancelto{11}{22}}{\cancelto{1}{7}} \times \frac{\cancelto{1}{7}}{\cancelto{1}{2}} \times \frac{7}{2}) \\ & = 70 - (\frac{11 \times 7}{2}) \\ & = 70 - (\frac{77}{2}) \\ & = 70 - 38.5 \\ & = 31.5 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the remaining part is 31.5 cm}^2. \end{align*} \]
11. A horse is tied to a pole with 21 m long string. Find the area where the horse can graze.
\[ \begin{align*} \text{Given:} \\ \text{Radius of the circle (r)} & = 21 \, \text{m} \\ \text{Area of a circle } (A) & = \pi r^2 \\ A & = \frac{22}{7} \times 21^2 \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{3}{21} \times 21 \\ & = 22 \times 63 \\ & = 1386 \, \text{m}^2 \\ \text{Answer:} & \quad \text{The area where the horse can graze is 1386 m}^2. \end{align*} \]
12. Two concentric circles have radii of 14 cm and 7cm respectively. Find the area of space between them.
\[ \begin{align*} \text{Given:} & \\ \text{Radius of the larger circle (R)} & = 14 \, \text{cm} \\ \text{Radius of the smaller circle (r)} & = 7 \, \text{cm} \\ \text{Area of the larger circle} & = \pi R^2 \\ & = \frac{22}{7} \times 14^2 \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{2}{14} \times 14 \\ & = 22 \times 28 \\ & = 616 \, \text{cm}^2 \\ \text{Area of the smaller circle} & = \pi r^2 \\ & = \frac{22}{7} \times 7^2 \\ & = \frac{22}{\cancelto{1}{7}} \times \cancelto{1}{7} \times 7 \\ & = 22 \times 7 \\ & = 154 \, \text{cm}^2 \\ \text{Area of the space between the circles:} & = {\text{Area of the larger circle}} - {\text{Area of the smaller circle}} \\ & = 616 \, \text{cm}^2 - 154 \, \text{cm}^2 \\ & = 462 \, \text{cm}^2 \\ \text{Answer:} & \quad \text{The area of the space between the circles is 462 cm}^2. \end{align*} \]
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 ot that of the smaller circle.
\[ \begin{align*} \text{Given:} & \\ \text{Circumferences of 4 small circles} & = \text{Circumference of the bigger circle} \\ 4 \times 2\pi r_1 & = 2\pi r_2 \\ 8\pi r_1 & = 2\pi r_2 \\ r_1 & = \frac{2\pi r_2}{8\pi} \\ r_1 & = \frac{\cancelto{1}{2} \cancelto{}{\pi} r_2}{\cancelto{4}{8} \cancelto{}{\pi}} \\ r_1 & = \frac{r_2}{4} \\ \\ \text{Area of the bigger circle} &: \text{Area of the smaller circle} \\ & = \pi (r_2)^2 : \pi (r_1)^2 \\ & = \frac{\pi (r_2)^2}{\pi (r_1)^2} \\ & = \frac{\cancelto{}{\pi} (r_2)^2}{\cancelto{}{\pi} (r_1)^2} \\ & = \frac{(r_2)^2}{(r_1)^2} \\ & = \frac{(r_2)^2}{(\frac{r_2}{4})^2} \\ & = \frac{16(r_2)^2}{(r_2)^2} \\ & = \frac{16\cancelto{}{(r_2)^2}}{\cancelto{}{(r_2)^2}} \\ & = \frac{16}{1} \\ & = 16:1 \\ \text{Answer:}& \quad \text{The ratio is 16:1} \end{align*} \]
14. The area of a circle is equal to the area of a rectangle with sides 112 mand 88 m respectively. Find the circumference of the circle.
\[ \begin{align*} \text{Given:} & \\ \text{Length of the rectangle} & = 112 \, \text{m} \\ \text{Width of the rectangle} & = 88 \, \text{m} \\ \text{Area of the Circle} & = \text{Area of the rectangle} \\ \pi r^2 & = \text{L} \times \text{B} \\ \frac{22}{7} \times r^2 & = 112 \times 88 \\ r^2 & = \frac{112 \times 88 \times 7}{22} \\ r^2 & = \frac{112 \times \cancelto{4}{88} \times 7}{\cancelto{1}{22}} \\ r^2 & = 112 \times 4 \times 7 \\ r^2 & = 2 \times 2 \times 2 \times 2 \times 7 \times 2 \times 2 \times 7 \\ r^2 & = (2 \times 2 \times 2 \times 7) (\times 2 \times 2 \times 2 \times 7) \\ r^2 & = (56)(56) \\ r^2 & = (56)^2 \\ r & = \sqrt{(56)^2} \\ r & = 56 \, \text{m} \\ \\ \text{Circumference of the circle} & = 2\pi r \\ & = 2 \times \frac{22}{\cancelto{1}{7}} \times \cancelto{8}{56} \\ & = 2 \times 22 \times 8 \\ & = 352 \, \text{m} \\ \text{Answer:} & \quad \text{The circumference of the circle is 352 m.} \end{align*} \]