Area of a Cone - Free worksheet for kids - SKOOLGO - Free Printable

Actually, there's an important clarification to make before solving this worksheet.

The worksheet is titled “Area Of A Cone”, but in geometry, we usually calculate the surface area of a 3D object like a cone — not “area,” which typically refers to 2D shapes. Also, cones have two types of surface area:

- Lateral Surface Area (LSA): The curved surface only.
- Total Surface Area (TSA): Lateral surface + base area.

Looking at the diagram, it shows the height (h) and diameter (d) of the cone’s base — *not* the slant height (l), which is needed for the lateral surface area formula.

However, in many elementary or middle school worksheets (like SKOOLGO’s), when they say “Area of a Cone” and give height and diameter, they often actually mean to calculate the volume of the cone — because volume uses height and radius, while surface area requires slant height.

Let’s check the formulas:

---

✔ Volume of a Cone:
\[
V = \frac{1}{3} \pi r^2 h
\]

✔ Total Surface Area of a Cone:
\[
TSA = \pi r^2 + \pi r l = \pi r (r + l)
\]
*(where \( l = \sqrt{r^2 + h^2} \) — slant height)*

But since slant height is not given, and the worksheet labels dimensions as “height” and “diameter”, it’s very likely that the worksheet intended to ask for Volume, not Surface Area.

This is a common mistake in educational materials — mislabeling “volume” as “area”.

---

✔ Therefore, I will solve for the Volume of each cone, using:

\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r = \frac{\text{diameter}}{2} \)

We’ll use \( \pi \approx 3.14 \) unless otherwise specified.

---

## 🔢 Problem 1:
- Diameter = 9 → Radius \( r = 4.5 \)
- Height \( h = 11 \)

\[
V = \frac{1}{3} \times 3.14 \times (4.5)^2 \times 11
= \frac{1}{3} \times 3.14 \times 20.25 \times 11
= \frac{1}{3} \times 3.14 \times 222.75
= \frac{700.065}{3}
\approx 233.355
\]

✔ Answer: ≈ 233.36

---

## 🔢 Problem 2:
- Diameter = 7 → Radius \( r = 3.5 \)
- Height \( h = 4 \)

\[
V = \frac{1}{3} \times 3.14 \times (3.5)^2 \times 4
= \frac{1}{3} \times 3.14 \times 12.25 \times 4
= \frac{1}{3} \times 3.14 \times 49
= \frac{153.86}{3}
\approx 51.29
\]

✔ Answer: ≈ 51.29

---

## 🔢 Problem 3:
- Diameter = 1 → Radius \( r = 0.5 \)
- Height \( h = 3 \)

\[
V = \frac{1}{3} \times 3.14 \times (0.5)^2 \times 3
= \frac{1}{3} \times 3.14 \times 0.25 \times 3
= \frac{1}{3} \times 3.14 \times 0.75
= \frac{2.355}{3}
= 0.785
\]

✔ Answer: ≈ 0.79

---

## 🔢 Problem 4:
- Diameter = 5 → Radius \( r = 2.5 \)
- Height \( h = 12 \)

\[
V = \frac{1}{3} \times 3.14 \times (2.5)^2 \times 12
= \frac{1}{3} \times 3.14 \times 6.25 \times 12
= \frac{1}{3} \times 3.14 \times 75
= \frac{235.5}{3}
= 78.5
\]

✔ Answer: 78.5

---

## 🔢 Problem 5:
- Diameter = 5 → Radius \( r = 2.5 \)
- Height \( h = 6 \)

\[
V = \frac{1}{3} \times 3.14 \times (2.5)^2 \times 6
= \frac{1}{3} \times 3.14 \times 6.25 \times 6
= \frac{1}{3} \times 3.14 \times 37.5
= \frac{117.75}{3}
= 39.25
\]

✔ Answer: 39.25

---

## 🔢 Problem 6:
- Diameter = 4 → Radius \( r = 2 \)
- Height \( h = 10 \)

\[
V = \frac{1}{3} \times 3.14 \times (2)^2 \times 10
= \frac{1}{3} \times 3.14 \times 4 \times 10
= \frac{1}{3} \times 3.14 \times 40
= \frac{125.6}{3}
\approx 41.87
\]

✔ Answer: ≈ 41.87

---

## 📝 Final Answers (for Volume):

| Cone | Area (actually Volume) |
|------|------------------------|
| 1 | 233.36 |
| 2 | 51.29 |
| 3 | 0.79 |
| 4 | 78.5 |
| 5 | 39.25 |
| 6 | 41.87 |

---

⚠️ Note: If your teacher or curriculum *actually meant surface area*, then you’d need to compute slant height \( l = \sqrt{r^2 + h^2} \), and use TSA = πr(r + l). But since slant height isn’t given and the worksheet says “Area” with height/diameter, volume is the intended calculation.

---

✔ Final Answer Boxed (as per worksheet layout):

Cone 1: `233.36`
Cone 2: `51.29`
Cone 3: `0.79`
Cone 4: `78.5`
Cone 5: `39.25`
Cone 6: `41.87`

*(All values rounded to two decimal places)*