McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
2. Surface Areas of Prisms and Cylinders
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Exercise 42 Page 820

Notice that the base areas are different. What must happen if you want the cylinders to have bases with different areas?

See solution.

Practice makes perfect

Let's consider two right cylinders C_1 and C_2. Below, we write the surface area of each cylinder.

Surface area of C_1 Surface area of C_2
S_1 = π r_1^2 + 2π r_1h_1 S_2 = π r_2^2 + 2π r_2h_2
We are told that the lateral areas are equal, and so, if we want the cylinders to have different surface areas, the radius of the bases must be different. That is, r_1≠ r_2. r_1 ≠ r_2 ⇒ h_1≠ h_2

However, we must consider numbers so that r_1h_1 = r_2h_2. Below, we pick an example combination of numbers that works. r_1 = 3, h_1 = 4 and r_2 = 4, h_2 = 3

Finding the Lateral Areas

Let's find the lateral area of each cylinder, starting with C_1.
L_(C_1) = 2π r_1h_1
L_(C_1) = 2π* 3* 4
L_(C_1) = 24π
Next, let's find the lateral area of C_2.
L_(C_2) = 2π r_2h_2
L_(C_2) = 2π* 4* 3
L_(C_2) = 24π
As we can see, both cylinders have the same lateral area.

Finding the Surface Areas

Let's finding the surface area C_1.
S_1 = π r_1^2 + L_(C_1)
S_1 = π* 3^2 + 24π
S_1 = 9π + 24π
S_1 = 33π
Finally, let's find the surface area of C_2.
S_2 = π r_2^2 + L_(C_2)
S_2 = π* 4^2 + 24π
S_2 = 16π + 24π
S_2 = 40π
As we wanted, both cylinders have different surface areas.

Drawing the Cylinders

Below, we show the two cylinders with the dimensions used before.
Keep in mind that this is just an example, so your answer may vary.