Answer :
The answer is :
The volume of the replacement cone holds 4·π cubic inches more than the volume of the original cone.
The given measurements of the cones are;
- The original cone's radius is, r₁ = 3 inches
- The original cone's height is, h₁ = 4 inches
- The replacement cone's radius is, r₂ = 4 inches
- The replacement cone's height is, h₂ = 3 inches
The volume of cone, V = ([tex]\frac{1}{3}[/tex])·π·r²·h
The volume of an original cone, V₁ = ([tex]\frac{1}{3}[/tex])·π·r₁²·h₁
∴ V₁ = ([tex]\frac{1}{3}[/tex]) × π × (3)² × 4 ≈ 37.7 in.³
The replacement cone's volume is, V₂ = ([tex]\frac{1}{3}[/tex])·π·r₂²·h₂
∴ V₂ = ([tex]\frac{1}{3}[/tex]) × π × (4)² × 3 ≈ 50.27 in.³
The volume ratio of the cone = V₁/V₂
∴ Volume ratio = ([tex]\frac{1}{3}[/tex]) × π × (3)² × 4 in. /(([tex]\frac{1}{3}[/tex]) × π × (4 )² × 3) = 3/4
The difference in volume of two cone, ΔV = V₂ - V₁ = ([tex]\frac{1}{3}[/tex]) × π × ((4)² × 3- (3 )² × 4 = 4·π in.³
Therefore, the volume of the replacement cone is 4·π cubic inches more than the volume of the original cone.
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