Answer:
The correct answer is;
The replacement cone holds 4·π cubic inches more than the volume of the original cone
Explanation:
The given measurements of the cones are;
The radius of the original cone, r₁ = 3 inches
The height of the original cone, h₁ = 4 inches
The radius of the replacement cone, r₂ = 4 inches
The height of the replacement cone, h₂ = 3 inches
The volume of a cone, V = (1/3)·π·r²·h
The volume of the original cone, V₁ = (1/3)·π·r₁²·h₁
∴ V₁ = (1/3) × π × (3 in.)² × 4 in. ≈ 37.7 in.³
The volume of the replacement cone, V₂ = (1/3)·π·r₂²·h₂
∴ V₂ = (1/3) × π × (4 in.)² × 3 in. ≈ 50.27 in.³
The ratio of the volume of the two cones is given as follows;
Volume ratio = V₁/V₂
∴ Volume ratio = (1/3) × π × (3 in.)² × 4 in. /((1/3) × π × (4 in.)² × 3 in.) = 3/4
The difference in volume, ΔV = V₂ - V₁ = (1/3) × π × ((4 in.)² × 3 in. - (3 in.)² × 4 in. = 4·π in.³
Therefore, the volume of the replacement cone is 4·π in.³ more than the volume of the original cone.