Suppose a sphere, cylinder, and cone all share the same radius. The cylinder and cone also share the same height, which is twice the radius. Which of the following is true about the relationship between the volume formulas for the sphere, cylinder, and cone?(1 point)

cylinder = sphere – cone
cylinder = sphere – cone

sphere = cylinder + cone
sphere = cylinder + cone

cone = sphere – cylinder
cone = sphere – cylinder

cone = cylinder – sphere

9 answers

To find the correct relationship between the volumes of the sphere, cylinder, and cone when they share the same radius and where the height of the cylinder and cone is twice the radius, let's calculate each volume using the following formulas:

  1. Volume of the Sphere (V_sphere): \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \]

  2. Volume of the Cylinder (V_cylinder): The height of the cylinder (\(h\)) is twice the radius (\(h = 2r\)): \[ V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (2r) = 2 \pi r^3 \]

  3. Volume of the Cone (V_cone): The height of the cone is also \(h = 2r\): \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]

Now we can summarize our volumes:

  • \(V_{\text{sphere}} = \frac{4}{3} \pi r^3\)
  • \(V_{\text{cylinder}} = 2 \pi r^3\)
  • \(V_{\text{cone}} = \frac{2}{3} \pi r^3\)

Next, we check the relationships given in the options:

  1. Option: cylinder = sphere – cone \[ V_{\text{cylinder}} = V_{\text{sphere}} - V_{\text{cone}} \] Substituting the volumes gives: \[ 2 \pi r^3 = \frac{4}{3} \pi r^3 - \frac{2}{3} \pi r^3 \] \[ 2 \pi r^3 = \frac{4}{3} \pi r^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi r^3 \] This does not hold true.

  2. Option: sphere = cylinder + cone \[ V_{\text{sphere}} = V_{\text{cylinder}} + V_{\text{cone}} \] Substituting gives: \[ \frac{4}{3} \pi r^3 = 2 \pi r^3 + \frac{2}{3} \pi r^3 \] \[ \frac{4}{3} \pi r^3 \neq 2 \pi r^3 + \frac{2}{3} \pi r^3 \] This is also not true.

  3. Option: cone = sphere – cylinder \[ V_{\text{cone}} = V_{\text{sphere}} - V_{\text{cylinder}} \] Substituting: \[ \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3 - 2 \pi r^3 \] \[ \frac{2}{3} \pi r^3 \neq \frac{4}{3} \pi r^3 - \frac{6}{3} \pi r^3 \] Not true.

  4. Option: cone = cylinder – sphere \[ V_{\text{cone}} = V_{\text{cylinder}} - V_{\text{sphere}} \] Substituting: \[ \frac{2}{3} \pi r^3 = 2 \pi r^3 - \frac{4}{3} \pi r^3 \] \[ \frac{2}{3} \pi r^3 = \frac{6}{3} \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \] This holds true!

Thus, the correct relationship is:

cone = cylinder – sphere

Use the relationship among the formulas for the volumes of cones, cylinders, and spheres to solve the following problem. The volume of a cylinder is 36 cm3
. What is the volume of a sphere if its radius is the same as the cylinder’s and the height of the cylinder is equal to the sphere’s diameter?(1 point)
Responses

12 cm3
12 cm cubed

54 cm3
54 cm cubed

18 cm3
18 cm cubed

24 cm3

Let's break the problem down step-by-step.

  1. Understanding the Problem:

    • The volume of the cylinder is \(36 , \text{cm}^3\).
    • The radius (\(r\)) of the cylinder is the same as the radius of the sphere.
    • The height (\(h\)) of the cylinder is equal to the diameter of the sphere. Since the diameter of the sphere is \(2r\), we have \(h = 2r\).
  2. Volume of a Cylinder: The formula for the volume of a cylinder is: \[ V_{\text{cylinder}} = \pi r^2 h \] Plugging in \(h = 2r\), we get: \[ V_{\text{cylinder}} = \pi r^2 (2r) = 2\pi r^3 \] Since we know that the volume of the cylinder is \(36 , \text{cm}^3\), we can set up the equation: \[ 2\pi r^3 = 36 \] Solving for \(r^3\): \[ r^3 = \frac{36}{2\pi} = \frac{18}{\pi} \]

  3. Volume of a Sphere: The formula for the volume of a sphere is: \[ V_{\text{sphere}} = \frac{4}{3}\pi r^3 \] Now we can plug \(r^3 = \frac{18}{\pi}\) into the sphere's volume formula: \[ V_{\text{sphere}} = \frac{4}{3} \pi \left(\frac{18}{\pi}\right) \] Simplifying this: \[ V_{\text{sphere}} = \frac{4}{3} \cdot 18 = \frac{72}{3} = 24 , \text{cm}^3 \]

Therefore, the volume of the sphere is 24 cm³.

The volume of a cone is 253 π cm3
. What is the volume of a sphere if its radius is the same as the cone’s and the height of the cone is equal to the sphere’s diameter?(1 point)
Responses

503 π cm3
Start Fraction 50 over 3 End Fraction pi cm cubed

256 π cm3
Start Fraction 25 over 6 End Fraction pi cm cubed

25 π cm3
25 pi cm cubed

252 π cm3

To find the volume of the sphere given the volume of the cone, let's break down the problem into steps.

  1. Understand the Relationship:

    • The volume of the cone is given as \( 253\pi , \text{cm}^3 \).
    • The radius of the sphere is the same as the radius of the cone.
    • The height of the cone is equal to the diameter of the sphere. Since the diameter of the sphere is \( 2r \), the height of the cone will also be \( 2r \).
  2. Volume of the Cone Formula: The volume of a cone is given by: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] Plugging in the height \( h = 2r \): \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]

  3. Setting the Volume Equal: We know the volume of the cone is \( 253\pi \): \[ \frac{2}{3} \pi r^3 = 253\pi \] Dividing both sides by \(\pi\): \[ \frac{2}{3} r^3 = 253 \] Multiplying both sides by \(\frac{3}{2} \): \[ r^3 = 253 \cdot \frac{3}{2} = \frac{759}{2} \]

  4. Volume of the Sphere: The volume of a sphere is given by: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \] Plugging in \( r^3 = \frac{759}{2} \): \[ V_{\text{sphere}} = \frac{4}{3} \pi \left(\frac{759}{2}\right) \] Simplifying this: \[ V_{\text{sphere}} = \frac{4 \times 759}{3 \times 2} \pi = \frac{3036}{6} \pi = 506 \pi \]

So, the volume of the sphere is \( 506\pi , \text{cm}^3 \).

As none of the provided options appear to match \( 506\pi , \text{cm}^3 \), it’s essential to review the list for clarity. Please check again if the values have been provided correctly.

Based on the calculations:

  • The correct answer would be 503π cm³ as the closest feasible value given options. Thus: 503π cm³.
Use the image to answer the question.

A cone is placed inside a cylinder. The apex of the cone touching the center of the top circle of the cylinder is highlighted with a dot. The cone with its base is drawn in dashed lines. The base of the cone is common with the base of the cylinder.

A sculptor created a design by carving a cone out of a cylinder. The cone and cylinder share the same radius and height. If the volume of the cylinder before removing the cone is 54 in.3
, what is the volume of the amount remaining?(1 point)
Responses

27 in.3
27 in. cubed

40 in.3
40 in. cubed

18 in.3
18 in. cubed

36 in.3
36 in. cubed
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To determine the volume of the remaining material after the cone has been carved out of the cylinder, we'll first need to calculate the volume of the cone and then subtract it from the volume of the cylinder.

  1. Volume of the Cylinder: We are given that the volume of the cylinder is \( 54 , \text{in}^3 \).

  2. Volume of the Cone: The formula for the volume of a cone is: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] Since we're told the cone and cylinder have the same radius and height, we can also express the volume of the cylinder in terms of radius and height: \[ V_{\text{cylinder}} = \pi r^2 h \]

    From the relationship between the volumes: \[ V_{\text{cylinder}} = 3 V_{\text{cone}} \] Because the volume of the cone is exactly one-third of the volume of the cylinder (since they share the same radius and height).

    Therefore: \[ V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} = \frac{1}{3} \cdot 54 = 18 , \text{in}^3 \]

  3. Volume of the Remaining Material: To find the volume of the remaining material after the cone is carved out: \[ V_{\text{remaining}} = V_{\text{cylinder}} - V_{\text{cone}} = 54 - 18 = 36 , \text{in}^3 \]

Thus, the volume of the amount remaining after carving out the cone from the cylinder is 36 in³.

Use the image to answer the question.

A hemisphere is placed on top of an inverted cone. The back of the circumference, the radius, and the perpendicular height of the cone are marked with dashed lines. The radius and the height make a right angle and are perpendicular.

An engineer wants to know the volume of the following object composed of a hemisphere and a cone. She knows the height of the cone is equal to the diameter of the hemisphere, and the volume of the cone is 12 m3
. What is the volume of the whole object?(1 point)
Responses

There is not enough information to say.
There is not enough information to say.

24 m3
24 m cubed

36 m3
36 m cubed

18 m3
18 m cubed
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To find the volume of the whole object, which consists of a hemisphere on top of an inverted cone, we need to consider both the volume of the cone and the volume of the hemisphere. We already know the volume of the cone is \(12 , \text{m}^3\), but we need to calculate the volume of the hemisphere.

  1. Understanding the Geometry:

    • Let \(r\) be the radius of the hemisphere.
    • The height of the cone is given as equal to the diameter of the hemisphere. Therefore, the height of the cone \(h\) is: \[ h = 2r \]
  2. Volume of the Cone: The volume of the cone has been provided: \[ V_{\text{cone}} = 12 , \text{m}^3 \]

  3. Finding Radius from Cone Volume: The formula for the volume of a cone is: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] Substituting \(h = 2r\), we find: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \] Setting this equal to the given volume of the cone: \[ \frac{2}{3} \pi r^3 = 12 \] Solving for \(r^3\): \[ r^3 = \frac{12 \cdot 3}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \]

  4. Volume of the Hemisphere: The formula for the volume of a hemisphere is: \[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \] Now, substituting the expression for \(r^3\): \[ V_{\text{hemisphere}} = \frac{2}{3} \pi \left(\frac{18}{\pi}\right) = \frac{2 \cdot 18}{3} = \frac{36}{3} = 12 , \text{m}^3 \]

  5. Total Volume of the Whole Object: Add the volumes of the cone and the hemisphere: \[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 12 + 12 = 24 , \text{m}^3 \]

Thus, the volume of the whole object is 24 m³.