Question

consider the cube shaped container below with a centered, cone shaped hole drilled out. (not drawn to scale.) use 3.1416 for the approximate value of π. round your answer to two decimal places. determine the side length and volume of the cube. s = \boxed{ } cm v_{cube} = \boxed{ } cm³ determine the radius, height, and volume of the cone. r = \boxed{ } cm h = \boxed{ } cm v_{cone} = \boxed{ } cm³ determine the volume of the container after the cone has been removed. v_{total} = \boxed{ } cm³

Explanation:

Part 1: Side length and volume of the cube

Step1: Determine side length of cube

The cube's side length is given by the diameter of the cone's base (since the cone is centered). The radius of the cone is 3 cm, so diameter (side length \( s \)) is \( 2 \times 3 = 6 \)? Wait, no, wait the diagram has 24 cm? Wait, no, maybe I misread. Wait, the diagram: the arrow is 24 cm? Wait, no, maybe the cube's side length is 24 cm? Wait, the cone has radius 3 cm. Wait, let's re-examine.

Wait, the problem says "cube shaped container with a centered, cone shaped hole drilled out". The diagram shows a cube with a cone inside. The label "24 cm" – maybe that's the side length? Wait, no, the dashed line is 3 cm (radius of cone). Wait, maybe the side length of the cube is 24 cm? Wait, no, that can't be, because the cone's radius is 3 cm, so diameter 6 cm. Wait, maybe I misread the diagram. Wait, the user's diagram: the arrow is 24 cm, and the dashed line is 3 cm (radius of cone). Wait, perhaps the cube's side length \( s = 24 \) cm? Wait, that makes more sense. Let's check:

Step1: Side length of cube

From the diagram, the side length \( s \) of the cube is 24 cm (as per the label). So \( s = 24 \) cm.

Step2: Volume of cube

Volume of a cube is \( V_{\text{cube}} = s^3 \). Substituting \( s = 24 \):

\( V_{\text{cube}} = 24^3 = 24 \times 24 \times 24 = 13824 \) \( \text{cm}^3 \).

Part 2: Radius, height, and volume of the cone

Step1: Radius of cone

From the diagram, the radius \( r \) of the cone is 3 cm (dashed line). So \( r = 3 \) cm.

Step2: Height of cone

The height \( h \) of the cone is equal to the side length of the cube (since it's drilled through the cube), so \( h = 24 \) cm.

Step3: Volume of cone

Volume of a cone is \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \). Substituting \( r = 3 \), \( h = 24 \), \( \pi = 3.1416 \):

\( V_{\text{cone}} = \frac{1}{3} \times 3.1416 \times 3^2 \times 24 \)

First, calculate \( 3^2 = 9 \). Then:

\( V_{\text{cone}} = \frac{1}{3} \times 3.1416 \times 9 \times 24 \)

Simplify \( \frac{1}{3} \times 9 = 3 \), so:

\( V_{\text{cone}} = 3.1416 \times 3 \times 24 = 3.1416 \times 72 = 226.1952 \) \( \text{cm}^3 \).

Part 3: Volume of the container after cone removal

Step1: Subtract cone volume from cube volume

\( V_{\text{total}} = V_{\text{cube}} - V_{\text{cone}} \)

Substituting \( V_{\text{cube}} = 13824 \) and \( V_{\text{cone}} = 226.1952 \):

\( V_{\text{total}} = 13824 - 226.1952 = 13597.8048 \approx 13597.80 \) (rounded to two decimal places).

Final Answers:

Cube:

• \( s = \boldsymbol{24} \) cm
• \( V_{\text{cube}} = \boldsymbol{13824.00} \) \( \text{cm}^3 \)

Cone:

• \( r = \boldsymbol{3} \) cm
• \( h = \boldsymbol{24} \) cm
• \( V_{\text{cone}} = \boldsymbol{226.20} \) \( \text{cm}^3 \) (rounded to two decimals)

Total Volume:

• \( V_{\text{total}} = \boldsymbol{13597.80} \) \( \text{cm}^3 \)

Answer:

Step1: Subtract cone volume from cube volume

\( V_{\text{total}} = V_{\text{cube}} - V_{\text{cone}} \)

Substituting \( V_{\text{cube}} = 13824 \) and \( V_{\text{cone}} = 226.1952 \):

\( V_{\text{total}} = 13824 - 226.1952 = 13597.8048 \approx 13597.80 \) (rounded to two decimal places).

Final Answers:

Cube:

• \( s = \boldsymbol{24} \) cm
• \( V_{\text{cube}} = \boldsymbol{13824.00} \) \( \text{cm}^3 \)

Cone:

• \( r = \boldsymbol{3} \) cm
• \( h = \boldsymbol{24} \) cm
• \( V_{\text{cone}} = \boldsymbol{226.20} \) \( \text{cm}^3 \) (rounded to two decimals)

Total Volume:

• \( V_{\text{total}} = \boldsymbol{13597.80} \) \( \text{cm}^3 \)