B. What is the radius of a ball that uses one-half of the amount of rubber coating used to cover the 16-inch ball? Write your answer in simplest form. The radius is
inches.
Question 2
A playground ball with a 16-inch diameter has a rubber coating on its surface.

a. Does a ball with a diameter that is $\frac{1}{4}$
times the diameter of the given ball need $\frac{1}{4}$
times the amount of rubber coating? Explain.

The radius, r₂, of the ball that uses one-half the amount of rubber coating used to cover the 16-inch ball is approximately 4.66 inches

Step-by-step explanation:

The dimension of the ball with known radius = 16-inch

The surface area of the ball with 16-inch radius = 4×π×r² = π·D² = π×16² = 804.24772 in.²

Given that the ball uses one-half the rubber material coating used to cover the 16-inch ball, we have the surface area of the ball = 804.24772 in.²/2 = 402.12386 in.²

The radius, r₂ of the new ball is found as follows;

402.12386 in.² = 4×π×r₂²

r₂² = 402.12386 in.² /(4×π) ≈ 32

r₂ = √32 = 4·√2 ≈ 4.66 inches

The radius, r₂, of the ball that uses one-half the amount of rubber coating used to cover the 16-inch ball ≈ 4.66 inches.