Volume and surface area worksheet for 3D shapes.

Let's solve each of the 3D shapes step by step, calculating both volume and surface area.

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1. Rectangular Prism (Cuboid)

Dimensions:
- Length = 10 mm
- Width = 6 mm
- Height = 3 mm

#### Volume:
$$
V = l \times w \times h = 10 \times 6 \times 3 = 180\ \text{mm}^3
$$

#### Surface Area:
$$
SA = 2(lw + lh + wh) = 2(10 \times 6 + 10 \times 3 + 6 \times 3) = 2(60 + 30 + 18) = 2 \times 108 = 216\ \text{mm}^2
$$

✔ Volume = 180 mm³, Surface Area = 216 mm²

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2. Cube

Side length = 4 mm

#### Volume:
$$
V = s^3 = 4^3 = 64\ \text{mm}^3
$$

#### Surface Area:
$$
SA = 6s^2 = 6 \times 4^2 = 6 \times 16 = 96\ \text{mm}^2
$$

✔ Volume = 64 mm³, Surface Area = 96 mm²

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3. Cylinder

Radius (r) = 7 cm, Height (h) = 10 cm

> Note: The image says "7 cm r" — so radius is 7 cm.

#### Volume:
$$
V = \pi r^2 h = \pi \times 7^2 \times 10 = \pi \times 49 \times 10 = 490\pi \approx 1539.38\ \text{cm}^3
$$

We'll keep it as $ 490\pi $ unless asked for decimal.

#### Surface Area:
$$
SA = 2\pi r^2 + 2\pi r h = 2\pi(7)^2 + 2\pi(7)(10) = 2\pi(49) + 2\pi(70) = 98\pi + 140\pi = 238\pi \approx 747.66\ \text{cm}^2
$$

✔ Volume = $ 490\pi $ cm³ ≈ 1539.38 cm³, Surface Area = $ 238\pi $ cm² ≈ 747.66 cm²

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4. Triangular Prism

Base triangle:
- Base = 15 cm
- Height = 8 cm
- Side = 17 cm (hypotenuse of right triangle?)
Length of prism = 10 cm

First, confirm if it’s a right triangular prism:
Check: $ 8^2 + 15^2 = 64 + 225 = 289 = 17^2 $ → Yes, right triangle.

So base area:
$$
A_{\text{base}} = \frac{1}{2} \times 15 \times 8 = 60\ \text{cm}^2
$$

#### Volume:
$$
V = \text{Base Area} \times \text{Length} = 60 \times 10 = 600\ \text{cm}^3
$$

#### Surface Area:
- Two triangular bases: $ 2 \times 60 = 120 $
- Three rectangular faces:
- 15 cm × 10 cm = 150
- 8 cm × 10 cm = 80
- 17 cm × 10 cm = 170
- Total lateral area = 150 + 80 + 170 = 400
- Total SA = 120 + 400 = 520 cm²

✔ Volume = 600 cm³, Surface Area = 520 cm²

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5. Square Pyramid

Base side = 4 yd, Slant height = 10 yd

#### Volume:
We need height of pyramid, but not given directly. Wait — slant height is 10 yd.

But we can find height using Pythagoras:

Half of base = 2 yd
Slant height = 10 yd
Then:
$$
h = \sqrt{10^2 - 2^2} = \sqrt{100 - 4} = \sqrt{96} = 4\sqrt{6} \approx 9.8\ \text{yd}
$$

Now volume:
$$
V = \frac{1}{3} \times \text{Base Area} \times h = \frac{1}{3} \times (4 \times 4) \times 4\sqrt{6} = \frac{1}{3} \times 16 \times 4\sqrt{6} = \frac{64\sqrt{6}}{3} \approx \frac{64 \times 2.45}{3} \approx \frac{156.8}{3} \approx 52.27\ \text{yd}^3
$$

But wait — maybe the slant height is given, and we're to use that?

Actually, in the diagram, the slant height is 10 yd, and base is 4 yd.

But let's recheck: Is the height of the pyramid needed? Yes, for volume.

But since only slant height is marked, and no vertical height, perhaps assume the slant height is from apex to midpoint of base edge?

Yes — typically in pyramids, slant height is from apex to base edge.

So, from apex to midpoint of a side: distance = 10 yd.

The horizontal distance from center to side midpoint = 2 yd (half of 4 yd).

So:
$$
h = \sqrt{10^2 - 2^2} = \sqrt{96} = 4\sqrt{6} \approx 9.8\ \text{yd}
$$

So Volume:
$$
V = \frac{1}{3} \times 4 \times 4 \times 4\sqrt{6} = \frac{64\sqrt{6}}{3} \approx 52.27\ \text{yd}^3
$$

But let's keep exact: $ \frac{64\sqrt{6}}{3} $

#### Surface Area:
- Base: $ 4 \times 4 = 16\ \text{yd}^2 $
- Lateral faces: 4 triangles, each with base 4 yd, slant height 10 yd
- Area of one triangle: $ \frac{1}{2} \times 4 \times 10 = 20 $
- Total lateral area = $ 4 \times 20 = 80 $
- Total SA = $ 16 + 80 = 96\ \text{yd}^2 $

✔ Volume ≈ 52.27 yd³, Surface Area = 96 yd²

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6. Triangular Pyramid (Tetrahedron)

Base triangle: base = 8 cm, height = 5 cm
Vertical height of pyramid = 6 cm (from apex to base)

Wait — the diagram shows a triangle with base 8 cm, height 5 cm, and then a vertical line from apex to base labeled 6 cm? That seems inconsistent.

Actually, looking at the diagram:
It shows a triangular pyramid with base triangle (8 cm base, 5 cm height), and the height of the pyramid is 6 cm (perpendicular from apex to base).

So:

#### Volume:
$$
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
$$

Base area:
$$
A = \frac{1}{2} \times 8 \times 5 = 20\ \text{cm}^2
$$

Volume:
$$
V = \frac{1}{3} \times 20 \times 6 = 40\ \text{cm}^3
$$

#### Surface Area:
Need areas of all 4 triangular faces.

- Base: already known = 20 cm²
- Other three faces: not fully specified.

Wait — we don’t have enough info about the other faces. We know the height of the pyramid is 6 cm, but not the slant heights or side lengths.

Unless the pyramid is regular? But base is 8 cm, and height 5 cm — not equilateral.

So we cannot compute surface area without more data.

But wait — perhaps the 6 cm is not the height, but the edge?

Looking again: It shows a triangle with base 8 cm, height 5 cm (so area 20), and then a line from apex to base labeled “h” = 6 cm — but this is vertical, so likely height of pyramid.

But for lateral surface area, we need slant heights or side lengths.

Alternatively, perhaps the lateral edges are given? No.

This shape is ambiguous.

But note: In some diagrams, if the base is a triangle and the apex is directly above the centroid, and we’re given the height, we still need side lengths.

But here, we only have base triangle dimensions and pyramid height.

We can't compute lateral face areas without knowing how the apex connects.

So we cannot compute surface area unless more info.

But maybe the figure is a regular tetrahedron? Unlikely — base has height 5 cm, base 8 cm → not equilateral.

So perhaps the problem expects us to only compute volume, or there's an error.

Wait — another interpretation: Maybe the triangle shown is the cross-section, and the full pyramid has base 8 cm and height 6 cm?

No — the diagram shows a triangle with base 8 cm, height 5 cm, and then a perpendicular from apex down to base = 6 cm? That doesn’t make sense.

Wait — the label “h” is on the side, pointing to the vertical height of the pyramid, which is 6 cm.

So the height of the pyramid is 6 cm, and the base triangle has base 8 cm and height 5 cm.

So volume is fine: $ \frac{1}{3} \times 20 \times 6 = 40\ \text{cm}^3 $

But surface area: we need areas of the three lateral faces.

Each lateral face is a triangle with base 8 cm, but what is its height (slant height)?

We don’t know the distance from apex to base edge.

But if we assume the apex is directly above the centroid of the base triangle, we could compute.

Centroid divides median in 2:1.

Median of base triangle: from vertex to midpoint of base.

Since base triangle has base 8 cm, height 5 cm, the median from apex to base is 5 cm.

Centroid is $ \frac{2}{3} \times 5 = \frac{10}{3} $ cm from base.

But the pyramid height is 6 cm — that’s from apex to base plane.

So the apex is 6 cm above the base.

Then the slant height (distance from apex to a base vertex) can be found via Pythagoras.

But we need the horizontal distance from centroid to vertex.

In a triangle, distance from centroid to vertex is $ \frac{2}{3} $ of median.

But we don’t know the side lengths.

This is getting too complex.

Perhaps the diagram is simplified — maybe the lateral faces are congruent?

But no info.

So likely, surface area cannot be determined with current data.

But perhaps the lateral edges are equal?

Alternatively, maybe the pyramid is right and symmetric.

But without more, we cannot compute surface area.

So we’ll skip surface area or leave blank.

But wait — perhaps the label “h = 6 cm” is the height of the lateral face, not the pyramid?

Look: The diagram shows a triangle with base 8 cm, height 5 cm (for the base), and then a line from apex to base labeled “h = 6 cm”.

Wait — maybe “h” is the height of the lateral face, meaning the slant height?

But then the pyramid height would be different.

But the notation suggests “h” is the vertical height of the pyramid.

Given ambiguity, let's assume:

- Base triangle: base = 8 cm, height = 5 cm → area = 20 cm²
- Height of pyramid = 6 cm → volume = $ \frac{1}{3} \times 20 \times 6 = 40 $ cm³
- Surface area: Not enough data

But maybe the pyramid is such that all lateral edges are equal? Or the lateral faces are isosceles?

Still, we need more.

Alternatively, perhaps the 6 cm is the slant height?

But it's drawn vertically — so likely height of pyramid.

So we’ll proceed with volume only.

✔ Volume = 40 cm³, Surface Area = ? (insufficient data)

But let’s look at the next one.

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7. Triangular Prism (Oblique?)

Base triangle: sides 3 m, 5 m, 8 m? Wait: 3 m, 5 m, 6 m, 8 m?

Wait — diagram shows a triangle with base 8 m, height 3 m, and another side 5 m, and length 6 m?

Wait: it’s a triangular prism with:

- Triangle base: base = 8 m, height = 3 m → area = $ \frac{1}{2} \times 8 \times 3 = 12\ \text{m}^2 $
- Length of prism = 6 m

So:

#### Volume:
$$
V = \text{Base Area} \times \text{Length} = 12 \times 6 = 72\ \text{m}^3
$$

#### Surface Area:
- Two triangular bases: $ 2 \times 12 = 24 $
- Three rectangular faces:
- 8 m × 6 m = 48
- 5 m × 6 m = 30
- 6 m × 6 m = 36? Wait — what are the side lengths?

Wait — the triangle has base 8 m, height 3 m, and another side 5 m? Then third side?

From diagram: the triangle has sides 5 m, 6 m, and base 8 m?

Wait — the diagram shows a triangle with base 8 m, and two other sides: 5 m and 6 m?

But also a height of 3 m — let’s verify:

Area = $ \frac{1}{2} \times 8 \times 3 = 12 $, correct.

So triangle has sides: 5 m, 6 m, 8 m? Let’s check if possible.

Yes — triangle inequality holds.

So lateral faces:
- Rectangle with side 5 m and length 6 m → area = 30
- Rectangle with side 6 m and length 6 m → area = 36
- Rectangle with side 8 m and length 6 m → area = 48

Total lateral area = 30 + 36 + 48 = 114

Total SA = 24 + 114 = 138 m²

✔ Volume = 72 m³, Surface Area = 138 m²

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8. Cylinder

Diameter = 10 cm, so radius = 5 cm, Height = 8 cm

#### Volume:
$$
V = \pi r^2 h = \pi \times 5^2 \times 8 = \pi \times 25 \times 8 = 200\pi \approx 628.32\ \text{cm}^3
$$

#### Surface Area:
$$
SA = 2\pi r^2 + 2\pi r h = 2\pi(25) + 2\pi(5)(8) = 50\pi + 80\pi = 130\pi \approx 408.41\ \text{cm}^2
$$

✔ Volume = $ 200\pi $ cm³ ≈ 628.32 cm³, Surface Area = $ 130\pi $ cm² ≈ 408.41 cm²

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✔ Final Answers Summary:

| Shape | Volume | Surface Area |
|------|--------|--------------|
| 1. Rectangular Prism | 180 mm³ | 216 mm² |
| 2. Cube | 64 mm³ | 96 mm² |
| 3. Cylinder (r=7 cm, h=10 cm) | $ 490\pi $ cm³ ≈ 1539.38 cm³ | $ 238\pi $ cm² ≈ 747.66 cm² |
| 4. Triangular Prism (base 15×8 cm, height 10 cm) | 600 cm³ | 520 cm² |
| 5. Square Pyramid (base 4 yd, slant height 10 yd) | $ \frac{64\sqrt{6}}{3} $ yd³ ≈ 52.27 yd³ | 96 yd² |
| 6. Triangular Pyramid | 40 cm³ | Insufficient data |
| 7. Triangular Prism (base 8×3 m, length 6 m) | 72 m³ | 138 m² |
| 8. Cylinder (d=10 cm, h=8 cm) | $ 200\pi $ cm³ ≈ 628.32 cm³ | $ 130\pi $ cm² ≈ 408.41 cm² |

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Notes:

- For Shape 6, surface area cannot be computed without knowing the lateral edge lengths or slant heights.
- All other values are calculated based on standard formulas.

Let me know if you'd like these answers filled into the worksheet format!