The formula for the surface area of a sphere is 4πr2 . Sphere A has a radius of 3 cm and sphere B has a radius of 6 cm. Calculate the surface area of each sphere and compare their surface areas.

The formula for the surface area of a sphere is:

\[
\text{Surface Area} = 4\pi r^2
\]

### Surface area of Sphere A (radius = 3 cm):

\[
\text{Surface Area of A} = 4\pi (3)^2 = 4\pi \times 9 = 36\pi \ \text{cm}^2
\]
Approximating \(\pi \approx 3.1416\):
\[
\text{Surface Area of A} \approx 36 \times 3.1416 = 113.0976 \ \text{cm}^2
\]

### Surface area of Sphere B (radius = 6 cm):

\[
\text{Surface Area of B} = 4\pi (6)^2 = 4\pi \times 36 = 144\pi \ \text{cm}^2
\]
Approximating \(\pi \approx 3.1416\):
\[
\text{Surface Area of B} \approx 144 \times 3.1416 = 452.3904 \ \text{cm}^2
\]

### Comparison:
- Surface area of Sphere A: \(113.1 \ \text{cm}^2\)
- Surface area of Sphere B: \(452.4 \ \text{cm}^2\)

Sphere B has a surface area four times that of Sphere A, which is consistent with the ratio of their radii squared, since \((6/3)^2 = 4\).

Would you like more details or further clarification?

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Here are 5 related questions to expand on this:
1. How does the volume of the spheres compare?
2. What is the effect on surface area if the radius is doubled again?
3. Can you derive the surface area formula for a sphere from its geometry?
4. What is the surface area of a hemisphere with radius 6 cm?
5. If Sphere B's radius were reduced by 50%, what would its new surface area be?

**Tip:** The surface area of a sphere is proportional to the square of its radius, so doubling the radius results in four times the surface area.

The formula for the surface area of a sphere is 4πr². Sphere A has a radius of 3 cm and sphere B has a radius of 6 cm. Calculate the surface area of each sphere and compare their surface areas.

Calculate and compare the surface areas of Sphere A (radius 3 cm) and Sphere B (radius 6 cm) using the surface area formula 4πr². Sphere B has four times the surface area of Sphere A. Learn the key geometric concepts behind this problem, ideal for Grades 6-8.

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