Volume and surface area of 3D shapes worksheet for Grade 6 geometry practice.

Let's solve each problem step by step from the Grade 6 Geometry Worksheet on Volume & Surface Area of 3D Shapes.

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Problem 1: Cube

Shape: Cube
Side length = 3 in

#### ✔ Volume of a cube:
$$
V = s^3 = 3^3 = 27 \text{ in}^3
$$

#### ✔ Surface Area of a cube:
$$
SA = 6s^2 = 6 \times (3)^2 = 6 \times 9 = 54 \text{ in}^2
$$

> Answer:
> - Volume = 27 in³
> - Surface Area = 54 in²

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Problem 2: Triangular Prism

Given:
- Base triangle: height = 2.5 in, base = 2.5 in
- Length (depth) of prism = 8 in

#### Step 1: Volume of triangular prism
$$
V = \text{Base Area} \times \text{Length}
$$

Area of triangle:
$$
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.5 \times 2.5 = \frac{6.25}{2} = 3.125 \text{ in}^2
$$

Now volume:
$$
V = 3.125 \times 8 = 25 \text{ in}^3
$$

#### Step 2: Surface Area
Surface area = sum of areas of all faces

The prism has:
- 2 triangular bases
- 3 rectangular sides

Area of two triangular bases:
$$
2 \times 3.125 = 6.25 \text{ in}^2
$$

Rectangular faces:
We need to know the side lengths of the triangle. Given:
- Base = 2.5 in
- Height = 2.5 in (but this is perpendicular height, not side)

But we are told that the triangle has sides:
- One side is 2.5 in (base)
- The other two sides: one is 2.5 in (height), but actually the triangle appears to be right-angled, with legs 2.5 in and 2.5 in?

Wait — looking at the diagram: it shows a right triangle with:
- Base = 2.5 in
- Height = 2.5 in
- Hypotenuse = ? Let's calculate:

$$
\text{Hypotenuse} = \sqrt{2.5^2 + 2.5^2} = \sqrt{6.25 + 6.25} = \sqrt{12.5} \approx 3.54 \text{ in}
$$

So the three rectangular faces have dimensions:
1. $ 2.5 \times 8 $ → area = $ 20 \text{ in}^2 $
2. $ 2.5 \times 8 $ → area = $ 20 \text{ in}^2 $
3. $ 3.54 \times 8 $ → area ≈ $ 28.32 \text{ in}^2 $

Total lateral surface area:
$$
20 + 20 + 28.32 = 68.32 \text{ in}^2
$$

Add two triangular bases:
$$
6.25 + 68.32 = 74.57 \text{ in}^2
$$

But let’s use exact values instead of approximations.

Use:
- Hypotenuse = $ \sqrt{2.5^2 + 2.5^2} = \sqrt{12.5} = \frac{5\sqrt{2}}{2} \approx 3.5355 $

But better to keep exact.

Alternatively, maybe the triangle is isosceles right triangle, so legs = 2.5 in each.

So:
- Rectangular face 1: $ 2.5 \times 8 = 20 $
- Rectangular face 2: $ 2.5 \times 8 = 20 $
- Rectangular face 3: $ \sqrt{2.5^2 + 2.5^2} \times 8 = \sqrt{12.5} \times 8 = 8\sqrt{12.5} $

But let’s compute numerically for simplicity.

$$
\sqrt{12.5} \approx 3.5355 \Rightarrow 3.5355 \times 8 \approx 28.284
$$

Total SA:
- Triangles: $ 2 \times \left( \frac{1}{2} \times 2.5 \times 2.5 \right) = 2 \times 3.125 = 6.25 $
- Rectangles: $ 20 + 20 + 28.284 = 68.284 $
- Total SA ≈ $ 6.25 + 68.284 = 74.534 \text{ in}^2 $

Rounded to nearest hundredth: 74.53 in²

But perhaps the diagram intends for the hypotenuse to be known or simplified.

Wait — actually, recheck the diagram: it shows a right triangle with base 2.5 in, height 2.5 in, and slant edge labeled as 3.5 in? No, it says “2.5 in” for both legs and the hypotenuse isn’t labeled.

But in the image, the side opposite the right angle is drawn, but no label. So assume it's just a right triangle with legs 2.5 in.

So:

> Volume = 25 in³
> Surface Area ≈ 74.53 in²

But let’s write it more cleanly.

Alternatively, maybe the triangle is not isosceles. Wait — the diagram shows:
- Base = 2.5 in
- Height = 2.5 in (from top vertex to base)
- But the horizontal leg is 2.5 in, and vertical leg is 2.5 in → yes, it's a right triangle with legs 2.5 in.

So calculations are correct.

✔ Final Answer:
- Volume = 25 in³
- Surface Area ≈ 74.53 in² (or leave as exact if needed)

But let’s keep it as:
- SA = $ 2 \times \left(\frac{1}{2} \times 2.5 \times 2.5\right) + (2.5 \times 8) + (2.5 \times 8) + (\sqrt{2.5^2 + 2.5^2} \times 8) $
- $ = 6.25 + 20 + 20 + (8 \times \sqrt{12.5}) $
- $ = 46.25 + 8 \times \sqrt{12.5} $

But for grade 6, likely expect decimal.

So:
- $ \sqrt{12.5} \approx 3.535 $
- $ 8 \times 3.535 = 28.28 $
- Total SA = $ 6.25 + 20 + 20 + 28.28 = 74.53 \text{ in}^2 $

> ✔ Answer:
> - Volume = 25 in³
> - Surface Area ≈ 74.53 in²

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Problem 3: Cylinder

Given:
- Radius = 7 in
- Height = 8 in

Wait — look at the diagram:
- Diameter = 7 in → so radius $ r = 3.5 $ in
- Height = 8 in

#### ✔ Volume of cylinder:
$$
V = \pi r^2 h = \pi \times (3.5)^2 \times 8 = \pi \times 12.25 \times 8 = \pi \times 98
$$

Using $ \pi \approx 3.14 $:
$$
V \approx 3.14 \times 98 = 307.72 \text{ in}^3
$$

#### ✔ Surface Area:
$$
SA = 2\pi r^2 + 2\pi r h
$$

First term: $ 2\pi r^2 = 2\pi \times 12.25 = 24.5\pi $

Second term: $ 2\pi r h = 2\pi \times 3.5 \times 8 = 56\pi $

Total SA = $ (24.5 + 56)\pi = 80.5\pi $

With $ \pi \approx 3.14 $:
$$
80.5 \times 3.14 = 252.67 \text{ in}^2
$$

> ✔ Answer:
> - Volume ≈ 307.72 in³
> - Surface Area ≈ 252.67 in²

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Problem 4: Triangular Prism (different orientation)

This is a triangular prism with a right triangle base.

Given:
- Base triangle: legs = 7 in and 7 in? Wait — check:
- One leg = 7 in (horizontal)
- Other leg = 7 in (vertical)? No — wait.
- The triangle has:
- One leg = 7 in (horizontal)
- One leg = 7 in (vertical)? Wait — the diagram shows:
- A right triangle with:
- Horizontal leg = 7 in
- Vertical leg = 7 in?
- But also labeled: hypotenuse = 9.9 in

Wait — the hypotenuse is labeled as 9.9 in, and the legs are 7 in and 7 in?

Check: $ \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.899 $ → which rounds to 9.9 in

Yes! So it’s a right triangle with legs 7 in and 7 in, hypotenuse ≈ 9.9 in.

And the length of the prism = 8 in (the depth)

So same as Problem 2, but different dimensions.

#### ✔ Volume:
$$
\text{Base Area} = \frac{1}{2} \times 7 \times 7 = \frac{49}{2} = 24.5 \text{ in}^2
$$
$$
V = 24.5 \times 8 = 196 \text{ in}^3
$$

#### ✔ Surface Area:
- Two triangular bases: $ 2 \times 24.5 = 49 \text{ in}^2 $
- Three rectangular faces:
1. $ 7 \times 8 = 56 $
2. $ 7 \times 8 = 56 $
3. $ 9.9 \times 8 = 79.2 $

Total lateral area: $ 56 + 56 + 79.2 = 191.2 $

Total SA: $ 49 + 191.2 = 240.2 \text{ in}^2 $

> ✔ Answer:
> - Volume = 196 in³
> - Surface Area = 240.2 in²

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✔ Final Answers Summary:

| Problem | Shape | Volume | Surface Area |
|--------|---------------|--------------|------------------|
| 1 | Cube | 27 in³ | 54 in² |
| 2 | Triangular Prism | 25 in³ | ≈74.53 in² |
| 3 | Cylinder | ≈307.72 in³ | ≈252.67 in² |
| 4 | Triangular Prism | 196 in³ | 240.2 in² |

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🔍 Explanation Recap:

- Cube: All sides equal → use formulas $ V = s^3 $, $ SA = 6s^2 $
- Triangular Prism: Use $ V = \text{Base Area} \times \text{Length} $, and SA = sum of all face areas
- Cylinder: Use $ V = \pi r^2 h $, $ SA = 2\pi r^2 + 2\pi r h $
- Always identify base shape and dimensions carefully.

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