Calculate: 1.1.1 The perpendicular height, 𝐴𝐵, of the cone using Pythagoras. 1.1.2 The height of the cylinder, if the volume of the cylinder is 157, 08 𝑚3 . 1.1.3 The total volume of the silo.

Let's go through each part of the problem step by step.

### **1.1.1 The Perpendicular Height \(AB\) of the Cone using Pythagoras**

Given:
- Radius of the base \( r = 5 \) cm.
- Slant height \( l = 10 \) cm (as interpreted from the image).

We need to find the perpendicular height \( h \) of the cone, denoted as \( AB \).

Using the Pythagorean theorem in the right triangle formed by the radius \( r \), the perpendicular height \( h \), and the slant height \( l \):

\[
l^2 = r^2 + h^2
\]

Substituting the given values:

\[
10^2 = 5^2 + h^2
\]

\[
100 = 25 + h^2
\]

\[
h^2 = 100 - 25 = 75
\]

\[
h = \sqrt{75} \approx 8.66 \, \text{cm}
\]

So, the perpendicular height \( AB \) of the cone is approximately **8.66 cm**.

---

### **1.1.2 The Height of the Cylinder, Given the Volume**

Given:
- Volume of the cylinder \( V_{\text{cylinder}} = 157.08 \, \text{m}^3 \).
- Radius of the base \( r = 5 \) m (assuming the unit is meters as given).

The formula for the volume of a cylinder is:

\[
V_{\text{cylinder}} = \pi r^2 h
\]

We need to find the height \( h \):

\[
h = \frac{V_{\text{cylinder}}}{\pi r^2}
\]

Substituting the given values:

\[
h = \frac{157.08}{\pi \times 5^2} = \frac{157.08}{\pi \times 25} = \frac{157.08}{78.54} \approx 2 \, \text{m}
\]

So, the height of the cylinder is approximately **2 meters**.

---

### **1.1.3 The Total Volume of the Silo**

The silo consists of both the cylinder and the cone. We will calculate the volume of each and then sum them.

**Volume of the Cylinder:**
Already calculated as:

\[
V_{\text{cylinder}} = 157.08 \, \text{m}^3
\]

**Volume of the Cone:**
The formula for the volume of a cone is:

\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}}
\]

We already found the perpendicular height of the cone \( AB = 8.66 \) m.

\[
V_{\text{cone}} = \frac{1}{3} \pi (5^2) (8.66) = \frac{1}{3} \pi (25)(8.66) = \frac{1}{3} \times 681.15 = 227.05 \, \text{m}^3
\]

**Total Volume of the Silo:**

\[
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{cone}} = 157.08 + 227.05 = 384.13 \, \text{m}^3
\]

So, the total volume of the silo is **384.13 cubic meters**.

---

Would you like further clarification on any of these calculations or any additional information?

Here are some related questions:
1. How does the volume of the cone change if the radius is doubled but the height remains the same?
2. What would be the new height of the cylinder if its volume was increased to 200 cubic meters?
3. How would the surface area of the silo change if the radius was increased by 50%?
4. What is the relationship between the slant height and the perpendicular height of a cone?
5. How can we calculate the lateral surface area of the cone?

**Tip:** Always keep track of units when dealing with volumes and areas, especially when converting between different measurement systems.

Learn how to calculate the perpendicular height of a cone using Pythagoras, find the height of a cylinder given its volume, and determine the total volume of a silo. This problem covers key geometric concepts, including volumes of solids and the Pythagorean theorem. Suitable for advanced high school students.

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