Answer :
To complete the table, we need to find the diameter, radius, and area of four different circles. Below is the step-by-step solution:
[tex]\[ \pi = \frac{22}{7} \][/tex]
### Circle A
- Given: Radius [tex]\( r = 49 \text{ cm} \)[/tex]
- Diameter: [tex]\( d = 2r = 2 \times 24.5 = 49.0 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (24.5)^2 = 1886.5 \text{ cm}^2 \)[/tex]
### Circle B
- Given: Diameter [tex]\( d = 7 \text{ cm} \)[/tex]
- Radius: [tex]\( r = \frac{d}{2} = \frac{7}{2} = 3.5 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (3.5)^2 = 38.5 \text{ cm}^2 \)[/tex]
### Circle C
- Given: Diameter [tex]\( d = 21 \text{ cm} \)[/tex]
- Radius: [tex]\( r = \frac{d}{2} = \frac{21}{2} = 10.5 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (10.5)^2 = 346.5 \text{ cm}^2 \)[/tex]
### Circle D
- Given: Radius [tex]\( r = 35 \text{ cm} \)[/tex]
- Diameter: [tex]\( d = 2r = 2 \times 17.5 = 35.0 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (17.5)^2 = 962.5 \text{ cm}^2 \)[/tex]
Now we can fill in the table:
[tex]\[ \begin{tabular}{|l|l|l|l|} \hline \text{Circle} & \text{Diameter (cm)} & \text{Radius (cm)} & \text{Area (cm}^2\text{)} \\ \hline \text{A} & 49.0 & 24.5 & 1886.5 \\ \hline \text{B} & 7 & 3.5 & 38.5 \\ \hline \text{C} & 21 & 10.5 & 346.5 \\ \hline \text{D} & 35.0 & 17.5 & 962.5 \\ \hline \end{tabular} \][/tex]
[tex]\[ \pi = \frac{22}{7} \][/tex]
### Circle A
- Given: Radius [tex]\( r = 49 \text{ cm} \)[/tex]
- Diameter: [tex]\( d = 2r = 2 \times 24.5 = 49.0 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (24.5)^2 = 1886.5 \text{ cm}^2 \)[/tex]
### Circle B
- Given: Diameter [tex]\( d = 7 \text{ cm} \)[/tex]
- Radius: [tex]\( r = \frac{d}{2} = \frac{7}{2} = 3.5 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (3.5)^2 = 38.5 \text{ cm}^2 \)[/tex]
### Circle C
- Given: Diameter [tex]\( d = 21 \text{ cm} \)[/tex]
- Radius: [tex]\( r = \frac{d}{2} = \frac{21}{2} = 10.5 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (10.5)^2 = 346.5 \text{ cm}^2 \)[/tex]
### Circle D
- Given: Radius [tex]\( r = 35 \text{ cm} \)[/tex]
- Diameter: [tex]\( d = 2r = 2 \times 17.5 = 35.0 \text{ cm} \)[/tex]
- Area: [tex]\( \text{Area} = \pi r^2 = \left(\frac{22}{7}\right) \times (17.5)^2 = 962.5 \text{ cm}^2 \)[/tex]
Now we can fill in the table:
[tex]\[ \begin{tabular}{|l|l|l|l|} \hline \text{Circle} & \text{Diameter (cm)} & \text{Radius (cm)} & \text{Area (cm}^2\text{)} \\ \hline \text{A} & 49.0 & 24.5 & 1886.5 \\ \hline \text{B} & 7 & 3.5 & 38.5 \\ \hline \text{C} & 21 & 10.5 & 346.5 \\ \hline \text{D} & 35.0 & 17.5 & 962.5 \\ \hline \end{tabular} \][/tex]