IGCSE Maths 0580 Practice Questions 2022 2024 Revised Syllabus ... - Free Printable

It looks like you've uploaded a handwritten math problem involving a composite solid made up of a cone and a hemisphere, both sharing the same radius. The goal is to calculate the volume and surface area of this combined shape, and possibly its mass if density is provided.

Let’s break down the problem step by step based on what's visible in the image:

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🔹 Given:

- Radius \( r = 3 \) cm
- Height of the cone \( h = 4 \) cm (from diagram)
- The figure consists of:
- A cone with height \( h = 4 \) cm and base radius \( r = 3 \) cm
- A hemisphere attached to the base of the cone (same radius)

We are asked to find:
1. Volume of the composite shape
2. Surface area of the composite shape
3. Possibly mass (if density is given)

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## ✔ Step 1: Volume Calculation

Volume of Cone:

\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi \cdot 9 \cdot 4 = 12\pi \, \text{cm}^3
\]

Volume of Hemisphere:

\[
V_{\text{hemi}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi \cdot 27 = 18\pi \, \text{cm}^3
\]

Total Volume:

\[
V_{\text{total}} = V_{\text{cone}} + V_{\text{hemi}} = 12\pi + 18\pi = 30\pi \, \text{cm}^3
\]

Now, using \( \pi \approx 3.14 \):
\[
30\pi \approx 30 \times 3.14 = 94.2 \, \text{cm}^3
\]

✔ So, Volume ≈ 94.2 cm³

This matches the red box in your image:
> \( V_{\text{total}} = 30\pi \approx 94.2 \, \text{cm}^3 \)

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## ✔ Step 2: Surface Area

Note: The base of the cone is joined to the hemisphere, so we do not include the circular base of the cone or the flat face of the hemisphere (they are internal and not exposed).

So, only consider:
- Curved surface area of the cone
- Curved surface area of the hemisphere

Curved Surface Area of Cone:

\[
A_{\text{cone}} = \pi r l
\]
where \( l \) is the slant height.

We are told:
- \( r = 3 \), \( h = 4 \)
- So, \( l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) cm

Thus:
\[
A_{\text{cone}} = \pi \cdot 3 \cdot 5 = 15\pi \, \text{cm}^2
\]

Curved Surface Area of Hemisphere:

\[
A_{\text{hemi}} = 2\pi r^2 = 2\pi (3)^2 = 2\pi \cdot 9 = 18\pi \, \text{cm}^2
\]

Total Surface Area:

\[
A_{\text{total}} = 15\pi + 18\pi = 33\pi \, \text{cm}^2
\]

Using \( \pi \approx 3.14 \):
\[
33\pi \approx 33 \times 3.14 = 103.62 \, \text{cm}^2
\]

✔ So, Surface Area ≈ 103.6 cm²

This also matches the calculation in your image:
> \( S_{\text{pw}} = 33\pi \approx 103.6 \, \text{cm}^2 \)

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## ✔ Step 3: Mass (if needed)

From the image, there's a note:
> "1 cm³ → 2.85 g"
> So, mass = volume × density

We found:
- Volume = 94.2 cm³
- Density = 2.85 g/cm³

\[
\text{Mass} = 94.2 \times 2.85 = ?
\]

Let’s compute:
\[
94.2 \times 2.85 = ?
\]

Break it down:
\[
94.2 \times 2.85 = 94.2 \times (3 - 0.15) = 94.2 \times 3 - 94.2 \times 0.15
\]
\[
= 282.6 - 14.13 = 268.47 \, \text{g}
\]

But in the image, it says:
> \( 94.2 \times 2.85 = 268.47 \), which rounds to 268 g (as shown)

✔ So, Mass ≈ 268 g

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## ✔ Final Answers:

| Quantity | Value |
|--------|-------|
| Volume | \( 30\pi \, \text{cm}^3 \approx 94.2 \, \text{cm}^3 \) |
| Surface Area | \( 33\pi \, \text{cm}^2 \approx 103.6 \, \text{cm}^2 \) |
| Mass | \( 268 \, \text{g} \) |

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📌 Summary of Key Formulas Used:

| Shape | Formula |
|------|--------|
| Cone Volume | \( \frac{1}{3} \pi r^2 h \) |
| Hemisphere Volume | \( \frac{2}{3} \pi r^3 \) |
| Cone CSA | \( \pi r l \) |
| Hemisphere CSA | \( 2\pi r^2 \) |
| Slant height \( l \) | \( \sqrt{r^2 + h^2} \) |

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✔ Your working in the image is correct — all steps match the standard method for such composite solids.

Let me know if you want a diagram explanation or help with similar problems!