Question 1

A)Describe an informal argument for the volume of a cone. If you know the volume of a cylinder, what must you do to find the volume of a cone that has the same height and base radius?(1 point)
Responses

You need to multiply the volume of the cylinder by 13.
You need to multiply the volume of the cylinder by Start Fraction 1 over 3 End Fraction .

You need to add 3 cubic units to the volume of the cylinder.
You need to add 3 cubic units to the volume of the cylinder.

You need to divide the volume of the cylinder by 13.
You need to divide the volume of the cylinder by Start Fraction 1 over 3 End Fraction .

You need to multiply the volume of the cylinder by 3.
You need to multiply the volume of the cylinder by 3.
Question 2
A)Eduardo makes a cone out of clay with a height that’s twice the length of the diameter. The diameter of the cone is 30 centimeters. Find the volume of the cone.(1 point)
Responses

2,250π cm3
2,250 pi cm cubed

72,000π cm3
72,000 pi cm cubed

18,000π cm3
18,000 pi cm cubed

4,500π cm3
4,500 pi cm cubed
Question 3
A)An ice cream shop currently sells soft-serve ice cream in cones that have a radius of 3 inches and a height of 6 inches. The shop wants to offer a wider cone for their hard-serve ice cream that has the same volume as their existing cone but a wider radius of 4 inches. What will be the height, to the nearest inch, of the new cone?(1 point)
Responses

14 inches
14 inches

3 inches
3 inches

4 inches
4 inches

18 inches
18 inches
Question 4
A)A cone has a height of 9 inches and a diameter of 4 inches. Calculate the volume of the cone. (1 point)
Responses

36π in.3
36 pi in. cubed

12π in.3
12 pi in. cubed

48π in.3
48 pi in. cubed

27π in.3
27 pi in. cubed
Question 5
A)
Use the image to answer the question.

A cone of height 11 and radius 6 is shown.

Tarik finds the volume of the cone in the image. What is the volume of the cone in cubic units?

(1 point)
Responses

66π cubic units
66 pi cubic units

396π cubic units
396 pi cubic units

132π cubic units

1 answer

Let's go through each question one by one:

Question 1

A) For the volume of a cone, an informal argument is that a cone can be thought of as a pyramid with a circular base. The formula for the volume of a cylinder is V = πr²h, where r is the radius and h is the height. Since a cone takes up one third of the space in the cylinder with the same base and height, the volume of the cone is given by V = (1/3)πr²h.

The correct response is: You need to multiply the volume of the cylinder by \( \frac{1}{3} \).


Question 2

A) The diameter of the cone is 30 cm, so the radius \( r \) is half of that: \( r = 15 \) cm. The height \( h \) is twice the diameter, which gives us a height of \( h = 30 \) cm.

Using the volume formula for a cone \( V = \frac{1}{3}πr^2h \):

  • \( V = \frac{1}{3}π(15)^2(30) = \frac{1}{3}π(225)(30) = \frac{1}{3}π(6750) = 2250π , \text{cm}^3 \).

The correct response is: 2,250π cm³.


Question 3

A) The volume of the original cone is calculated with \( r = 3 \) inches and \( h = 6 \) inches:

  • \( V = \frac{1}{3}π(3)^2(6) = \frac{1}{3}π(9)(6) = \frac{1}{3}π(54) = 18π \) cubic inches.

In the new cone, the radius \( r = 4 \) inches and we need to find the new height \( h' \) such that the volume remains the same:

  • \( V = \frac{1}{3}π(4)^2h' \Rightarrow 18π = \frac{1}{3}π(16)h' \).

Cancelling \( π \) and multiplying both sides by 3:

  • \( 54 = 16h' \Rightarrow h' = \frac{54}{16} = 3.375 \) inches.

Rounding to the nearest inch:

  • This is approximately \( 3 \) inches.

The correct response is: 3 inches.


Question 4

A) The cone has a height of 9 inches and a diameter of 4 inches, so the radius \( r = 2 \) inches. Using the volume formula:

  • \( V = \frac{1}{3}π(2)^2(9) = \frac{1}{3}π(4)(9) = \frac{1}{3}π(36) = 12π \) cubic inches.

The correct response is: 12π in³.


Question 5

A) The cone has a height of 11 and a radius of 6: Using the volume formula:

  • \( V = \frac{1}{3}π(6)^2(11) = \frac{1}{3}π(36)(11) = \frac{1}{3}π(396) = 132π \) cubic units.

The correct response is: 132π cubic units.