Detailed solutions and full explanations to grade 8 problems and questions on circles are presented.
Diameter of \( C_1 \) is \( 20 \) ⇒ \( r_1 = 10 \). Area of \( C_1 \): \( A = \pi (10)^2 \) Diameter of \( C_2 = 10 \) ⇒ \( r_2 = 5 \) Diameter of \( C_3 = 5 \) ⇒ \( r_3 = 2.5 \) Area of \( C_3 \): \( B = \pi (2.5)^2 \)
\[ \frac{A}{B} = \frac{\pi (10)^2}{\pi (2.5)^2} = \left( \frac{10}{2.5} \right)^2 = 4^2 = 16 \]Ratio: \( 16 : 1 \)
Total square area: \( 4 \times 4 = 16\ \text{m}^2 \) Side length: \( 2 \ \text{m} \) ⇒ semicircle radius \( r = 2 \ \text{m} \) Two semicircles form one full circle: area \( = \pi (2)^2 = 4\pi \)
\[ \text{Total area} = 16 + 4\pi \approx 28.56\ \text{m}^2 \]Inner radius: \( 10 \ \text{m} \), Outer radius: \( 11 \ \text{m} \)
\[ A = \pi (11)^2 - \pi (10)^2 = 121\pi - 100\pi = 21\pi \ \text{m}^2 \]Radius: \( 18\ \text{cm} \) Area: \( \pi (18)^2 = 324\pi \approx 1017 \ \text{cm}^2 \)
\[ \text{Cost/cm}^2 \approx \frac{19.99}{1017} \approx 0.02\ \text{USD} \ (\text{2 cents}) \]\(\pi r^2 = 5 \ \Rightarrow \ r \approx 1.26 \ \text{m}\)
\[ C = 2\pi r \approx 8\ \text{m} \]Old area: \( \pi r^2 \) New radius: \( 1.2r \), New area: \( 1.44\pi r^2 \)
\[ \% \text{ change} = \frac{1.44\pi r^2 - \pi r^2}{\pi r^2} \times 100\% = 44\% \]Diameter: \( 130\ \text{in} \) ⇒ radius \( 65\ \text{in} \)
\[ A = \pi (65)^2 = 4225\pi \approx 13267\ \text{in}^2 \]