Question

a cylindrical cardboard tube with a diameter of 8 centimeters and a height of 20 centimeters is used to package a gift.

what is the approximate volume of the tube? round to the nearest whole cubic centimeter.

\\(1,005\text{ cm}^3\\)
\\(1,340\text{ cm}^3\\)
\\(3,351\text{ cm}^3\\)
\\(4,021\text{ cm}^3\\)

Explanation:

Identify the given dimensions

We are given a cylindrical cardboard tube with:

• Diameter \(d = 8\text{ cm}\)
• Height \(h = 20\text{ cm}\)

Calculate the radius of the cylinder

The radius \(r\) is half of the diameter:

$$r = \frac{d}{2} = \frac{8}{2} = 4\text{ cm}$$

Apply the cylinder volume formula

The volume \(V\) of a cylinder is calculated using the formula:

$$V = \pi r^2 h$$

Substituting our values:

$$V = \pi \cdot (4)^2 \cdot 20$$

$$V = \pi \cdot 16 \cdot 20$$

$$V = 320\pi$$

Calculate the approximate numerical value

Using the approximation \(\pi \approx 3.14159\):

$$V \approx 320 \cdot 3.14159 \approx 1005.31\text{ cm}^3$$

Rounding to the nearest whole cubic centimeter gives:

$$V \approx 1005\text{ cm}^3$$

Answer:

• (A) 1,005 cm³ (Correct answer)
• (B) 1,340 cm³
• (C) 3,351 cm³
• (D) 4,021 cm³