Question
Glass A is a cylinder
Glass B is a cone
The water glasses are filled to the same height and have the same radius.
Fill in the blanks
a) It would take _____ times filling up glass B and pouring into glass A to fill up glass A.
b) If the height of the glasses doubled, then the volumes would increase times _____
c) if the radius of the glasses doubled then the volumes would increase times _____
d) If the radius and the height both doubled, then the volumes would increase times ______
one
two
three
four
five
six
seven
eight
nine
ten
Glass B is a cone
The water glasses are filled to the same height and have the same radius.
Fill in the blanks
a) It would take _____ times filling up glass B and pouring into glass A to fill up glass A.
b) If the height of the glasses doubled, then the volumes would increase times _____
c) if the radius of the glasses doubled then the volumes would increase times _____
d) If the radius and the height both doubled, then the volumes would increase times ______
one
two
three
four
five
six
seven
eight
nine
ten
Answers
GPT-4o mini
To answer your questions, we need to understand the volume formulas for the shapes involved:
1. **Cylinder (Glass A)**: The volume \( V_A \) is given by the formula:
\[
V_A = \pi r^2 h
\]
2. **Cone (Glass B)**: The volume \( V_B \) is given by the formula:
\[
V_B = \frac{1}{3} \pi r^2 h
\]
Now, let's analyze each question step by step.
### a) It would take _____ times filling up glass B and pouring into glass A to fill up glass A.
If we fill glass B and pour it into glass A:
- The volume of glass B is \( V_B = \frac{1}{3} \pi r^2 h \).
- The volume of glass A is \( V_A = \pi r^2 h \).
To find how many times we need to fill glass B to equal the volume of glass A:
\[
\text{Number of fills} = \frac{V_A}{V_B} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h} = 3
\]
So, the answer is:
**It would take 3 times filling up glass B and pouring into glass A to fill up glass A.**
### b) If the height of the glasses doubled, then the volumes would increase times _____.
When the heights are doubled, the new heights become \( 2h \), so:
\[
V_A' = \pi r^2 (2h) = 2 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi r^2 (2h) = \frac{2}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
The volumes increase by a factor of 2 because only the height is being doubled:
**The volumes would increase times 2.**
### c) If the radius of the glasses doubled, then the volumes would increase times _____.
When the radii are doubled, the new radii become \( 2r \), so:
\[
V_A' = \pi (2r)^2 h = 4 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
The volumes increase by a factor of \( (2^2) = 4 \) since the radius is squared:
**The volumes would increase times 4.**
### d) If the radius and the height both doubled, then the volumes would increase times ______.
When the radii and heights are both doubled, we have \( 2r \) and \( 2h \):
\[
V_A' = \pi (2r)^2 (2h) = 4 \pi r^2 (2h) = 8 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi (2r)^2 (2h) = \frac{1}{3} \pi (4r^2)(2h) = \frac{8}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
This means the volumes are increased by a factor of:
\[
2^2 \times 2 = 2^3 = 8 \quad \text{(since both radius and height are doubled)}
\]
**The volumes would increase times 8.**
### Summary of Answers:
a) **3**
b) **2**
c) **4**
d) **8**
1. **Cylinder (Glass A)**: The volume \( V_A \) is given by the formula:
\[
V_A = \pi r^2 h
\]
2. **Cone (Glass B)**: The volume \( V_B \) is given by the formula:
\[
V_B = \frac{1}{3} \pi r^2 h
\]
Now, let's analyze each question step by step.
### a) It would take _____ times filling up glass B and pouring into glass A to fill up glass A.
If we fill glass B and pour it into glass A:
- The volume of glass B is \( V_B = \frac{1}{3} \pi r^2 h \).
- The volume of glass A is \( V_A = \pi r^2 h \).
To find how many times we need to fill glass B to equal the volume of glass A:
\[
\text{Number of fills} = \frac{V_A}{V_B} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h} = 3
\]
So, the answer is:
**It would take 3 times filling up glass B and pouring into glass A to fill up glass A.**
### b) If the height of the glasses doubled, then the volumes would increase times _____.
When the heights are doubled, the new heights become \( 2h \), so:
\[
V_A' = \pi r^2 (2h) = 2 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi r^2 (2h) = \frac{2}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
The volumes increase by a factor of 2 because only the height is being doubled:
**The volumes would increase times 2.**
### c) If the radius of the glasses doubled, then the volumes would increase times _____.
When the radii are doubled, the new radii become \( 2r \), so:
\[
V_A' = \pi (2r)^2 h = 4 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
The volumes increase by a factor of \( (2^2) = 4 \) since the radius is squared:
**The volumes would increase times 4.**
### d) If the radius and the height both doubled, then the volumes would increase times ______.
When the radii and heights are both doubled, we have \( 2r \) and \( 2h \):
\[
V_A' = \pi (2r)^2 (2h) = 4 \pi r^2 (2h) = 8 \pi r^2 h \quad \text{(new volume of the cylinder)}
\]
\[
V_B' = \frac{1}{3} \pi (2r)^2 (2h) = \frac{1}{3} \pi (4r^2)(2h) = \frac{8}{3} \pi r^2 h \quad \text{(new volume of the cone)}
\]
This means the volumes are increased by a factor of:
\[
2^2 \times 2 = 2^3 = 8 \quad \text{(since both radius and height are doubled)}
\]
**The volumes would increase times 8.**
### Summary of Answers:
a) **3**
b) **2**
c) **4**
d) **8**