Answer:
Difference = Surface Area_small - Surface Area_large ≈ 63.72 cm^2 - 85.39 cm^2 ≈ -21.67 cm^2
Explanation:
Radius of the larger hollow ball (r_outer) = 3 cm
For the smaller hollow balls, since they are identical, we assume that the outer and inner radii are the same:
Radius of the smaller hollow balls (r_inner) = r_outer/2 = 3 cm / 2 = 1.5 cm
Now, let's calculate the surface area for both cases:
Surface area of the larger hollow ball:
Surface Area_large = 4π(r_outer^2 - r_inner^2)
Surface Area_large = 4π(3^2 - 1.5^2)
Surface Area_large = 4π(9 - 2.25)
Surface Area_large = 4π(6.75)
Surface Area_large ≈ 85.39 cm^2
Surface area of the smaller hollow balls (each):
Surface Area_small = 4π(r_outer^2 - r_inner^2)
Surface Area_small = 4π(1.5^2 - 0.75^2)
Surface Area_small = 4π(2.25 - 0.5625)
Surface Area_small = 4π(1.6875)
Surface Area_small ≈ 21.24 cm^2
Now, let's compare the surface areas:
The total surface area of the three smaller hollow balls = 3 * Surface Area_small ≈ 3 * 21.24 cm^2 ≈ 63.72 cm^2
Therefore, it takes more paint to paint the three smaller hollow balls compared to the larger hollow ball. The difference in surface area is given by:
Difference = Surface Area_small - Surface Area_large ≈ 63.72 cm^2 - 85.39 cm^2 ≈ -21.67 cm^2
The difference is negative because the surface area of the larger hollow ball is greater than the combined surface area of the three smaller hollow balls.