Question

You have three shapes

A cone with a height of 3r with an unknown radius
A cylinder with a height of 3r and an unknown radius
And a spear with an unknown high and an unknown radius
The volume of the cone is 25 π

Answer each question

How many times larger is the volume of the cylinder than the volume of the cone if the cone and the cylinder had the same dimensions.

The volume of the spear if the radius is the same value as the cone.

How many times larger is the volume of the spear than the volume of the cone if the cone in the sphere had the same dimensions.

In order for the volume of the cone plus the volume of the spear equals the volume of the cylinder, the height would have to be blank not 3r for the cone and cylinder.

The volume of the cylinder.

Answer 1

The pool of answers you have to pick from is,

Three, 25 π, 125 π, 3r, 2r, 1/3, 50 / 3 π, 75 / 2 π, 4r, 100 over 3 π, 75 π, 2

Answer 2

To solve these questions, we will use the formulas for the volumes of a cone, cylinder, and sphere.

1. **Volume of a Cone:**
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]
Given that the volume of the cone is \( 25\pi \) and the height \( h = 3r \), we can substitute:
\[
25\pi = \frac{1}{3} \pi r^2 (3r)
\]
Simplifying gives:
\[
25 = r^3
\]
Therefore, the radius \( r = 5 \).

2. **Volume of the Cylinder with the same dimensions as the cone:**
\[
V_{\text{cylinder}} = \pi r^2 h
\]
Using \( r = 5 \) and \( h = 3r = 15 \):
\[
V_{\text{cylinder}} = \pi (5)^2 (15) = \pi (25)(15) = 375\pi
\]

3. **How many times larger is the volume of the cylinder than the volume of the cone?**
\[
\text{Ratio} = \frac{V_{\text{cylinder}}}{V_{\text{cone}}} = \frac{375\pi}{25\pi} = \frac{375}{25} = 15
\]
So, the volume of the cylinder is **15 times larger** than the volume of the cone.

4. **Volume of the Sphere with the same radius as the cone:**
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3
\]
Using \( r = 5 \):
\[
V_{\text{sphere}} = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi (125) = \frac{500}{3}\pi
\]

5. **How many times larger is the volume of the sphere than the volume of the cone?**
\[
\text{Ratio} = \frac{V_{\text{sphere}}}{V_{\text{cone}}} = \frac{\frac{500}{3}\pi}{25\pi} = \frac{\frac{500}{3}}{25} = \frac{500}{75} = \frac{20}{3} \approx 6.67
\]
So, the volume of the sphere is approximately **\( \frac{20}{3} \) or 6.67 times larger** than the volume of the cone.

6. **For the volume of the cone plus the volume of the spear to equal the volume of the cylinder:**
\[
V_{\text{cone}} + V_{\text{sphere}} = V_{\text{cylinder}}
\]
Plugging in the values:
\[
25\pi + \frac{500}{3}\pi = 375\pi
\]
To find a height \( h \) for the cone and cylinder, we set:
\[
25 + \frac{500}{3} = \frac{375}{h}
\]
Solving for \( h \) gives:
\[
\frac{375}{h} = \frac{75}{3} + \frac{500}{3} = \frac{575}{3}, \quad h = \frac{375 \times 3}{575} = \frac{1125}{575}
\]
Simplifying gives:
\[
h \approx 1.96 \quad (\text{less than } 3r)
\]
Thus, to find \( h \) we need it to be approximately **less than \( 3r \)**.

7. **The volume of the cylinder:**
From earlier calculation, the volume of the cylinder is \( 375\pi \).

### Summary Results:
1. The volume of the cylinder is **15 times** larger than the cone.
2. The volume of the sphere is **\( \frac{500}{3}\pi \)**.
3. The sphere is approximately **\( \frac{20}{3} \) times** larger than the cone.
4. The height needed for the equation \( V_{\text{cone}} + V_{\text{sphere}} = V_{\text{cylinder}} \) would be less than \( 3r \).
5. The volume of the cylinder is **\( 375\pi \)**.