A 10-by-6 inch piece of paper is used to form the lateral surface of a cylinder. If the entire piece of paper is used to make the lateral surface, which of the following must be true of the two possible cylinders that can be formed?

A. The volume of the cylinder with height 10 is 60π60π cubic inches greater than the volume of the cylinder with height 6.

B. The volume of the cylinder with height 6 is 60π60π cubic inches greater than the volume of the cylinder with height 10.

C. The volume of the cylinder with height 10 is 60π60π cubic inches greater than the volume of the cylinder with height 6.

D. The volume of the cylinder with height 6 is 60π60π cubic inches greater than the volume of the cylinder with height 10.

E. The volume of the cylinder with height 6 is 240π240π cubic inches greater than the volume of the cylinder with height 10.

Question

Mathematics

Kamila Cordova

24 August, 20:31

A 10-by-6 inch piece of paper is used to form the lateral surface of a cylinder. If the entire piece of paper is used to make the lateral surface, which of the following must be true of the two possible cylinders that can be formed?

A. The volume of the cylinder with height 10 is 60π60π cubic inches greater than the volume of the cylinder with height 6.

B. The volume of the cylinder with height 6 is 60π60π cubic inches greater than the volume of the cylinder with height 10.

C. The volume of the cylinder with height 10 is 60π60π cubic inches greater than the volume of the cylinder with height 6.

D. The volume of the cylinder with height 6 is 60π60π cubic inches greater than the volume of the cylinder with height 10.

E. The volume of the cylinder with height 6 is 240π240π cubic inches greater than the volume of the cylinder with height 10.

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Answers (1)

Know the Answer?

Cason Robertson · Answer 1

the cylinder with height 6 has a volume of 60/π in³ greater than the volume of the cylinder with height 10 (option B, if 60π is changed for 60/π)

Step-by-step explanation:

The volume of a cylinder is

V = π*R²*H (H=height)

since the length L of the piece of paper is L=2*π*R →R=L / (2*π) (since is rolled to form the cylinder), then:

V = π*R²*H = π*L² / (2*π) ²*H = L²*H / (4*π)

with L=10 in and H = 6 in we have

V₂ = L²*H / (4*π)

the other way around is changing H for L

V₁ = H²*L / (4*π)

the difference between the volumes will be

V₂ - V₁ = L²*H / (4*π) - H²*L / (4*π) = L*H * (L-H) / (4*π)

replacing values

V₂ - V₁ = L*H * (L-H) / (4*π) = 10*6 * (10-6) / (4*π) = 60/π in³

then the cylinder with height 6 has a volume of 60/π in³ greater than the volume of the cylinder with height 10