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# A rectangular label with an area of 18 square inches. is wrapped aro

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Re: A rectangular label with an area of 18 square inches. is wrapped aro [#permalink]
Bunuel wrote:
A rectangular label with an area of 18 square inches. is wrapped around a can that is 3 inches tall, such that the label exactly covers the outside of the can excluding the top and the bottom. What is the volume of the can, in cubic inches?

A. 27π

B. 27

C. 18

D. 27/π

E. 18/π

Area of the rectangular label = 18 inch²
Since, Height of the label will be same as height of the can, therefore (h) = 3 inch (given in question stem)
And Surface Area of Can/cylinder (2πrh)= 18
So …..r=3/π (after solving)

Volume of the Cylinder= πr²h= π x (3/π)² x 3= 27/π
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Re: A rectangular label with an area of 18 square inches. is wrapped aro [#permalink]
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Bunuel wrote:
A rectangular label with an area of 18 square inches. is wrapped around a can that is 3 inches tall, such that the label exactly covers the outside of the can excluding the top and the bottom. What is the volume of the can, in cubic inches?

A. 27π

B. 27

C. 18

D. 27/π

E. 18/π

We are given that a rectangular label with an area of 18 square inches is wrapped around a can that is 3 inches tall.

This information actually tells us that the height of the cylindrical can is 3 inches and the lateral surface area (i.e., the surface area without the top or the bottom) of the cylindrical can is 18 square inches. Recall that the lateral surface area, A, of a cylinder is given by the formula A = 2πrh, in which r is the radius of the cylinder and h is the height of the cylinder. Thus, we have:

18 = 2πr(3)

18 = 6πr

r = 18/(6π) = 3/π

Finally, we can determine the volume of the cylinder using this formula: volume = πr^2h

volume = π x (3/π)^2 x 3

volume = π x 9/π^2 x 3

volume = 27/π

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Re: A rectangular label with an area of 18 square inches. is wrapped aro [#permalink]
Circumference = 18 = 6πr

r = 3/π
Volume = (π∗r2∗hπ∗r2∗h) = 27/π