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We are asked to calculate the slant height (\( l \)) of a cone, where the total surface area is given as \( 80\pi \, \text{cm}^2 \), and the diameter is \( 10 \, \text{cm} \).

### Given Information:
- Total surface area of the cone: \( 80\pi \, \text{cm}^2 \)
- Diameter: \( 10 \, \text{cm} \), so the radius \( r = \frac{10}{2} = 5 \, \text{cm} \)
- Formula for the total surface area \( A_{\text{total}} \) of a cone is:
  \[
  A_{\text{total}} = \pi r l + \pi r^2
  \]
  Where:
  - \( \pi r l \) is the lateral surface area
  - \( \pi r^2 \) is the area of the base

### Step-by-step Calculation:

1. **Set up the equation using the surface area formula:**
   \[
   80\pi = \pi r l + \pi r^2
   \]
2. **Substitute the radius \( r = 5 \, \text{cm} \):**
   \[
   80\pi = \pi (5) l + \pi (5)^2
   \]
   Simplify:
   \[
   80\pi = 5\pi l + 25\pi
   \]
3. **Divide the entire equation by \( \pi \):**
   \[
   80 = 5l + 25
   \]
4. **Solve for \( l \):**
   \[
   80 - 25 = 5l
   \]
   \[
   55 = 5l
   \]
   \[
   l = \frac{55}{5} = 11 \, \text{cm}
   \]

Thus, the slant height of the cone is \( \mathbf{11 \, \text{cm}} \).

Would you like any further details or clarifications? Here are some related questions:

1. How is the formula for the surface area of a cone derived?
2. What is the difference between lateral surface area and total surface area?
3. How would the answer change if the diameter was 8 cm instead of 10 cm?
4. Can you calculate the height of the cone using the Pythagorean theorem?
5. What are the units for surface area and why are they squared?

**Tip:** When solving for one variable in terms of another, always make sure all other terms are fully simplified before performing algebraic operations.

Work out the slant height of the cone. The cone below has a total surface area of 80π cm² and a diameter of 10 cm.

Discover how to calculate the slant height of a cone when its total surface area and diameter are known. This detailed guide uses fundamental geometry formulas and the Pythagorean theorem, suitable for students in Grades 8-10.

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