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To solve for the **curved surface area** of the cone, we need to use the formula for the lateral surface area of a cone:

\[
\text{Curved Surface Area} = \pi r l
\]

where:
- \( r \) is the radius of the base of the cone,
- \( l \) is the slant height of the cone.

### Step 1: Calculate the Slant Height \( l \)
We can use the Pythagorean theorem to calculate the slant height \( l \), as the cone forms a right-angled triangle with the height \( h \) and the radius \( r \).

\[
l = \sqrt{h^2 + r^2}
\]

Given:
- \( h = 12 \, \text{cm} \) (height of the cone),
- \( r = 4 \, \text{cm} \) (radius of the base of the cone),

We can now calculate \( l \):

\[
l = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} \approx 12.65 \, \text{cm}
\]

### Step 2: Calculate the Curved Surface Area
Now, substitute \( r = 4 \, \text{cm} \) and \( l = 12.65 \, \text{cm} \) into the formula for the curved surface area:

\[
\text{Curved Surface Area} = \pi \times 4 \times 12.65 \approx 159 \, \text{cm}^2
\]

The curved surface area of the cone is approximately **159 cm²**, correct to three significant figures.

Would you like more details or have any questions?

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Here are some related questions:
1. How do you calculate the total surface area of a cone, including the base?
2. Can the slant height of a cone be shorter than its vertical height?
3. What is the relationship between a cylinder and a cone that share the same base?
4. How does the curved surface area of a cone compare with that of a cylinder?
5. Why do we use the Pythagorean theorem to find the slant height of a cone?

**Tip:** Always make sure to use the correct units for each measurement when solving geometry problems.

The height of the cone is 12 cm and the radius of the base of the cone is 4 cm. Work out the curved surface area of the cone. Give your answer correct to 3 significant figures.

Learn how to calculate the curved surface area of a cone with a radius of 4 cm and height of 12 cm using the formula A = πrl. This problem involves key geometric concepts, including the Pythagorean theorem and surface area calculations, suitable for students in Grades 7-9.

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