To solve the problems related to the silo made up of a cylinder and a cone, we will address each part of your question step by step.
8.1 The perpendicular height, \( AB \), of the cone.
To find the perpendicular height of the cone, we can use the information about the triangle formed by points A, B, and C.
Given:
- \( AC = 0.55 , \text{m} \) (the radius of the base of the cone, which is equal to the radius of the cylinder)
- \( BC = 0.32 , \text{m} \) (the height \( AB \))
Since point B is a right angle (90 degrees), we can use the Pythagorean theorem to find \( AB \):
\[ AB^2 + BC^2 = AC^2 \]
Substituting the values we have:
\[ AB^2 + (0.32)^2 = (0.55)^2 \]
Calculating \( (0.32)^2 \):
\[ (0.32)^2 = 0.1024 \]
Calculating \( (0.55)^2 \):
\[ (0.55)^2 = 0.3025 \]
Putting these values in the equation:
\[ AB^2 + 0.1024 = 0.3025 \]
Subtracting \( 0.1024 \) from both sides gives:
\[ AB^2 = 0.3025 - 0.1024 = 0.2001 \]
Now taking the square root to find \( AB \):
\[ AB = \sqrt{0.2001} \approx 0.447 , \text{m} \]
Thus, the perpendicular height \( AB \) of the cone is approximately:
\( AB \approx 0.447 , \text{m} \)
8.2 The height of the cylinder if the volume of the cylinder is \( 0.55 , \text{m}^3 \).
The volume of the cylinder is given by the formula:
\[ V = \pi r^2 h \]
We are given \( V = 0.55 , \text{m}^3 \) and need to solve for the height \( h \).
First, we need to find \( r \):
From the right triangle ABC, we know that \( AC = 0.55 , \text{m} \), which is also the radius \( r \).
Now using the formula for the volume of the cylinder:
\[ 0.55 = \pi (0.55^2) h \]
Calculating \( 0.55^2 \):
\[ 0.55^2 = 0.3025 \]
So the equation becomes:
\[ 0.55 = \pi (0.3025) h \]
Dividing both sides by \( \pi \cdot 0.3025 \):
\[ h = \frac{0.55}{\pi \cdot 0.3025} \]
Calculating \( \pi \cdot 0.3025 \):
\[ \pi \cdot 0.3025 \approx 0.9506 \]
So,
\[ h \approx \frac{0.55}{0.9506} \approx 0.578 , \text{m} \]
Thus, the height of the cylinder is approximately:
\( h \approx 0.578 , \text{m} \)
8.3 The total volume of the silo.
The total volume of the silo consists of the volume of the cylinder and the volume of the cone.
- Volume of the Cone:
The volume \( V_c \) of the cone is given by:
\[ V_c = \frac{1}{3}\pi r^2 h \]
Where we already have \( r = 0.55 , \text{m} \) and \( h = 0.447 , \text{m} \):
\[ V_c = \frac{1}{3} \pi (0.55^2)(0.447) = \frac{1}{3} \pi (0.3025)(0.447) \]
Calculating \( 0.3025 \times 0.447 \):
\[ 0.3025 \times 0.447 \approx 0.1354 \]
Now substituting back:
\[ V_c \approx \frac{1}{3} \pi (0.1354) \approx \frac{0.425 \approx 0.425}{3} \approx 0.14167 , \text{m}^3 \]
- Volume of the Cylinder:
From the previous calculation, we have \( V_{cylinder} = 0.55 , \text{m}^3 \).
- Total Volume:
Now we sum the volumes:
\[ V_{total} = V_{cylinder} + V_{cone} \approx 0.55 + 0.14167 \approx 0.69167 , \text{m}^3 \]
Thus, the total volume of the silo is approximately:
\( V_{total} \approx 0.692 , \text{m}^3 \)
In summary, the answers are:
- 8.1: \( AB \approx 0.447 , \text{m} \)
- 8.2: \( h \approx 0.578 , \text{m} \)
- 8.3: \( V_{total} \approx 0.692 , \text{m}^3 \)