Pete runs an ice cream stand that also sells snow cones served in paper cones. The paper cones he usually uses have a diameter of 6 inches and a height of 2 inches, but his supplier is out of them. As a replacement, he purchases paper cones with a diameter of 2 inches and a height of 6 inches. How do the volumes of the original and replacement cones compare?
a. The original and replacement cones have the same volume.
b. The original cone holds 2 times the amount as the replacement cone.
c. The original cone has a greater volume than the replacement cone.
Selected:d. The replacement cone has a greater volume than the original cone.
d. The replacement cone has a greater volume than the original cone.
Why? Because the volume of a cone is determined by the formula V = (1/3)πr^2h, where r is the radius and h is the height.
For the original cone with a diameter of 6 inches and a height of 2 inches, the radius is 3 inches. Plugging these values into the formula, we get V = (1/3)π(3^2)(2) = 6π cubic inches.
For the replacement cone with a diameter of 2 inches and a height of 6 inches, the radius is 1 inch. Plugging these values into the formula, we get V = (1/3)π(1^2)(6) = 2π cubic inches.
Since 6π > 2π, the replacement cone has a greater volume than the original cone. It seems like Pete made a good choice in buying the replacement cones, he's getting more volume for his ice cream and snow cones!
To compare the volumes of the original and replacement cones, we can use the formula for the volume of a cone: V = (1/3) * π * r^2 * h, where V is the volume, π is a mathematical constant approximately equal to 3.14159, r is the radius of the base of the cone, and h is the height of the cone.
For the original cone with a diameter of 6 inches, the radius (r) is half of the diameter, so r = 6 / 2 = 3 inches. The height (h) is 2 inches. Plugging these values into the formula, we get:
V_original = (1/3) * π * (3 inches)^2 * 2 inches.
For the replacement cone with a diameter of 2 inches, the radius (r) is half of the diameter, so r = 2 / 2 = 1 inch. The height (h) is 6 inches. Plugging these values into the formula, we get:
V_replacement = (1/3) * π * (1 inch)^2 * 6 inches.
Now, let's calculate the volumes:
V_original = (1/3) * 3.14159 * (3 inches)^2 * 2 inches
≈ 3.14159 * 9 inches^2 * 2 inches / 3
≈ 3.14159 * 9 inches * 2 inches / 3
≈ 3.14159 * 18 inches^2 / 3
≈ 56.54867 cubic inches.
V_replacement = (1/3) * 3.14159 * (1 inch)^2 * 6 inches
≈ 3.14159 * 1 inch^2 * 6 inches / 3
≈ 3.14159 * 6 inches / 3
≈ 3.14159 * 2 inches
≈ 6.28318 cubic inches.
Comparing the volumes, we find that V_replacement ≈ 6.28318 cubic inches is greater than V_original ≈ 56.54867 cubic inches.
Therefore, the correct answer is:
d. The replacement cone has a greater volume than the original cone.
To compare the volumes of the original and replacement cones, we can use the formula for the volume of a cone: V = (1/3) * π * r^2 * h, where V is the volume, π is a mathematical constant approximately equal to 3.14, r is the radius, and h is the height of the cone.
Let's calculate the volumes step-by-step:
For the original cone:
- Diameter = 6 inches, so radius = diameter/2 = 6/2 = 3 inches
- Height = 2 inches
Using the formula, the volume of the original cone is:
V_original = (1/3) * π * (3 inches)^2 * 2 inches
V_original = (1/3) * 3.14 * 9 square inches * 2 inches
V_original = (1/3) * 3.14 * 18 cubic inches
V_original ≈ 18.84 cubic inches
For the replacement cone:
- Diameter = 2 inches, so radius = diameter/2 = 2/2 = 1 inch
- Height = 6 inches
Using the formula, the volume of the replacement cone is:
V_replacement = (1/3) * π * (1 inch)^2 * 6 inches
V_replacement = (1/3) * 3.14 * 1 square inch * 6 inches
V_replacement = (1/3) * 3.14 * 6 cubic inches
V_replacement ≈ 6.28 cubic inches
Comparing the volumes, we can see that:
V_replacement ≈ 6.28 cubic inches < V_original ≈ 18.84 cubic inches
Therefore, the replacement cone has a greater volume than the original cone.
So, the correct answer is d. The replacement cone has a greater volume than the original cone.