Question

question 3 (multiple choice worth 2 points) (06.01 mc) a cone has a volume of 50π in³ and a diameter of 10 in. wilson states that a cylinder with the same height and diameter has the same volume. which statement explains whether or not wilson is correct? a cylinder in which h = 2 and d = 10 has a volume of 50π in³, therefore, wilson is correct a cylinder in which h = 6 and d = 10 has a volume of 50π in³, therefore, wilson is correct a cylinder in which h = 2 and d = 10 has a volume of 150π in³, therefore, wilson is incorrect a cylinder in which h = 6 and d = 10 has a volume of 150π in³, therefore, wilson is incorrect

Explanation:

First, recall the volume formulas for a cone and a cylinder. The volume of a cone is \( V_{cone} = \frac{1}{3}\pi r^2 h \), and the volume of a cylinder is \( V_{cylinder} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. The diameter \( d = 10 \), so the radius \( r = \frac{d}{2} = 5 \).

The cone has a volume of \( 50\pi \, \text{in}^3 \). Let's find the height of the cone using its volume formula.

Step 1: Find the height of the cone

Given \( V_{cone} = 50\pi = \frac{1}{3}\pi (5)^2 h \)
Simplify: \( 50\pi = \frac{1}{3}\pi \times 25 \times h \)
Divide both sides by \( \pi \): \( 50 = \frac{25}{3}h \)
Multiply both sides by \( \frac{3}{25} \): \( h = 50 \times \frac{3}{25} = 6 \). Wait, no, wait. Wait, the problem says the cone has volume \( 50\pi \) and diameter 10. Wait, maybe I misread. Wait, the cone's volume is \( 50\pi \), diameter \( d = 10 \), so radius \( r = 5 \). So \( V_{cone} = \frac{1}{3}\pi r^2 h = 50\pi \). So \( \frac{1}{3}\pi (5)^2 h = 50\pi \). Divide both sides by \( \pi \): \( \frac{25}{3}h = 50 \). Then \( h = 50 \times \frac{3}{25} = 6 \). Wait, but the options have h=2 or h=6. Wait, maybe the cone's height is 6? Wait, no, let's check the cylinder volume.

The volume of a cylinder with the same radius (since diameter is 10, radius 5) and height \( h \) is \( V_{cylinder} = \pi r^2 h = \pi (5)^2 h = 25\pi h \).

The volume of the cone is \( \frac{1}{3}\pi r^2 h_{cone} = 50\pi \). So \( \frac{1}{3} \times 25\pi \times h_{cone} = 50\pi \). So \( \frac{25}{3} h_{cone} = 50 \), so \( h_{cone} = 6 \). So the cone has height 6 and radius 5.

Now, a cylinder with the same diameter (so same radius 5) and height: let's check the options.

Option 1: Cylinder with h=2, d=10 (r=5). Volume: \( \pi (5)^2 (2) = 50\pi \). But the cone's volume is 50π, and the cylinder with h=2 would have volume 50π? Wait, no, the cone's volume is \( \frac{1}{3}\pi r^2 h \), so if the cylinder has the same r and h, its volume is 3 times the cone's. Wait, maybe I made a mistake. Wait, the cone's volume is 50π. If the cylinder has the same radius and height as the cone, then cylinder volume is 3*50π=150π. But let's check the options.

Wait, the options are about cylinders with h=2 or h=6, d=10 (so r=5).

Let's calculate the volume of a cylinder with d=10 (r=5) and h=2: \( V = \pi r^2 h = \pi (5)^2 (2) = 50\pi \). But the cone's volume is 50π, and the formula for cone is \( \frac{1}{3}\pi r^2 h \), so if the cylinder has h=2 and r=5, its volume is 50π, but the cone with the same r and h would have volume \( \frac{1}{3}\pi (5)^2 (2) = \frac{50}{3}\pi \), which is not 50π. Wait, maybe the cone's height is 6? Let's recalculate the cone's height.

If the cone's volume is 50π, and r=5, then \( 50\pi = \frac{1}{3}\pi (5)^2 h \) => \( 50 = \frac{25}{3}h \) => \( h = 6 \). So the cone has h=6, r=5. Then a cylinder with the same r=5 and h=6 would have volume \( \pi (5)^2 (6) = 150\pi \), which is 3 times the cone's volume (50π). So if Wilson says a cylinder with the same height and diameter (so same r) has the same volume as the cone, he is wrong. Now let's check the options:

Option 1: Cylinder with h=2, d=10 (r=5). Volume: \( \pi (5)^2 (2) = 50\pi \). But the cone's volume is 50π, but the cone with h=2 and r=5 would have volume \( \frac{1}{3}\pi (5)^2 (2) = \frac{50}{3}\pi eq 50\pi \). So this is inconsistent.

Option 2: Cylinder with h=6, d=10 (r=5). Volume: \( \pi (5)^2 (6) = 150\pi \). The cone's volume is 50π, so 150π is 3 times 50π. So if Wilson says the cylinder has the same volume as the cone, he is wrong. Now check the options:

Wait, the options are:

A. A cylinder in which h = 2 and d = 10 has a volume of 50π in³, therefore, Wilson is correct

B. A cylinder in which h = 6 and d = 10 has a volume of 50π in³, therefore, Wilson is correct

C. A cylinder in which h = 2 and d = 10 has a volume of 150π in³, therefore, Wilson is incorrect

D. A cylinder in which h = 6 and d = 10 has a volume of 150π in³, therefore, Wilson is incorrect

Wait, maybe I misread the options. Let's re-express:

Wait, the correct calculation:

Volume of cylinder: \( V = \pi r^2 h \), r = d/2 = 5.

For h=2, d=10 (r=5): \( V = \pi (5)^2 (2) = 50\pi \).

For h=6, d=10 (r=5): \( V = \pi (5)^2 (6) = 150\pi \).

The cone's volume is 50π. The relationship between cone and cylinder with same base (same r) and same height: cylinder volume is 3 times cone volume. So if the cone has volume 50π, a cylinder with same r and h would have 150π. So Wilson says the cylinder has the same volume as the cone (50π), which is wrong. So we need to find the option that shows the cylinder's volume is 150π (when h=6, d=10) and concludes Wilson is incorrect.

So the correct option is D: A cylinder in which h = 6 and d = 10 has a volume of 150π in³, therefore, Wilson is incorrect.

Answer:

D. A cylinder in which h = 6 and d = 10 has a volume of 150π in³, therefore, Wilson is incorrect