To solve the given problems, let's break them down step by step.
1.1.1 The perpendicular height, \( AB \), of the cone using Pythagoras.
Given:
- From \( A \) to \( C \) is \( 6.5 , \text{m} \) (slant height of the cone).
- From \( B \) to \( C \) is \( 2.5 , \text{m} \) (radius of the cone which is also the radius of the cylinder since they have equal radii).
Using Pythagoras' theorem in triangle \( ABC \):
\[ AC^2 = AB^2 + BC^2 \]
Substituting the known values:
\[ (6.5)^2 = AB^2 + (2.5)^2 \]
Calculating the squares:
\[ 42.25 = AB^2 + 6.25 \]
Now, subtract \( 6.25 \) from both sides:
\[ AB^2 = 42.25 - 6.25 \]
\[ AB^2 = 36 \]
Taking the square root to find \( AB \):
\[ AB = 6 , \text{m} \]
1.1.2 The height of the cylinder, if the volume of the cylinder is \( 157.08 , \text{m}^3 \).
The volume of the cylinder is given by the formula:
\[ V = \pi r^2 h \]
We need to find the height \( h \) of the cylinder.
First, let's find the radius \( r \). As given, \( BC = 2.5 , \text{m} \), thus:
\[ r = 2.5 , \text{m} \]
Now we plug in the volume and the radius into the cylinder volume formula:
\[ 157.08 = \pi (2.5)^2 h \]
Calculating \( (2.5)^2 \):
\[ 157.08 = \pi (6.25) h \]
Now solving for \( h \):
\[ h = \frac{157.08}{\pi \times 6.25} \]
Using \( \pi \approx 3.14159 \):
\[ h \approx \frac{157.08}{3.14159 \times 6.25} \]
Calculating the denominator:
\[ 3.14159 \times 6.25 \approx 19.6349375 \]
Thus,
\[ h \approx \frac{157.08}{19.6349375} \approx 8.0 , \text{m} \]
1.1.3 The total volume of the silo.
The total volume of the silo is the sum of the volume of the cone and the volume of the cylinder. We already have the volume of the cylinder, and we can find the volume of the cone using its formula.
The volume of the cone is given by:
\[ V_{cone} = \frac{1}{3} \pi r^2 h \]
Where:
- \( r = 2.5 , \text{m} \)
- \( h = AB = 6 , \text{m} \)
Calculating the volume of the cone:
\[ V_{cone} = \frac{1}{3} \pi (2.5)^2 (6) \]
Calculating \( (2.5)^2 \):
\[ V_{cone} = \frac{1}{3} \pi (6.25) (6) = \frac{1}{3} \pi (37.5) \]
Now calculating:
\[ V_{cone} = \frac{37.5 \pi}{3} \approx 12.5 \pi \approx 39.27 , \text{m}^3 \]
Now we sum the volumes:
\[ V_{total} = V_{cylinder} + V_{cone} \]
Given \( V_{cylinder} = 157.08 , \text{m}^3 \):
\[ V_{total} \approx 157.08 + 39.27 \approx 196.35 , \text{m}^3 \]
Final Answers
1.1.1 The perpendicular height, \( AB \), of the cone is \( 6 , \text{m} \). 1.1.2 The height of the cylinder is approximately \( 8 , \text{m} \). 1.1.3 The total volume of the silo is approximately \( 196.35 , \text{m}^3 \).