Grade 6 Geometry Worksheets: Volume and surface area of 3D shapes ... - Free Printable

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:

- Two triangular bases: $2 \times 3.125 = 6.25 \text{ in}^2$
- Three rectangular sides:
- Rectangle 1: base × length = $2.5 \times 8 = 20 \text{ in}^2$
- Rectangle 2: height × length = $2.5 \times 8 = 20 \text{ in}^2$
- Rectangle 3: hypotenuse × length → need to find hypotenuse

Find hypotenuse of triangle using Pythagoras:
$$
c = \sqrt{(2.5)^2 + (2.5)^2} = \sqrt{6.25 + 6.25} = \sqrt{12.5} \approx 3.54 \text{ in}
$$

So third rectangle area:
$$
3.54 \times 8 \approx 28.32 \text{ in}^2
$$

Total surface area:
$$
6.25 + 20 + 20 + 28.32 = 74.57 \text{ in}^2
$$

> Answer (approx):
> - Volume = 25 in³
> - Surface Area ≈ 74.57 in²

---

Problem 3: Cylinder

Given:
- Diameter = 7 in → Radius $r = 3.5$ in
- Height = 8 in

#### ✔ Volume of cylinder:
$$
V = \pi r^2 h = \pi \times (3.5)^2 \times 8 = \pi \times 12.25 \times 8 = \pi \times 98 \approx 307.72 \text{ in}^3
$$

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

$$
= 2\pi \times 3.5 \times (8 + 3.5) = 7\pi \times 11.5 = 80.5\pi \approx 252.98 \text{ in}^2
$$

> Answer (approx):
> - Volume ≈ 307.72 in³
> - Surface Area ≈ 252.98 in²

---

Problem 4: Rectangular Prism (with slanted side)

This shape is a rectangular prism, but one face is slanted — actually, it's a right prism with a trapezoidal base? Wait — let's analyze.

Wait! Looking at the diagram:

- It has a right angle at the bottom.
- Dimensions:
- Base = 7 in
- Height = 7 in
- Slanted side = 8.6 in
- Top side = 8 in

But wait — this looks like a truncated prism or perhaps a prism with a trapezoidal cross-section?

Actually, more likely: This is a rectangular prism where the top is not aligned — but the diagonal is shown as 8.6 in. But looking closely:

It seems like this is a triangular prism, but the base is a right triangle?

Wait — no. The figure shows:

- A right triangle base: legs 7 in and 7 in?
- But then the side labeled "8.6 in" is the hypotenuse?

Let’s re-express:

The base is a right triangle with:
- One leg = 7 in (horizontal)
- Other leg = 7 in (vertical)
- Hypotenuse = 8.6 in (given)

Wait — check if that makes sense:
$$
\sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.9 \text{ in}
$$
But it says 8.6 in — so not a right triangle.

Wait — look again.

Ah! Actually, this is a rectangular prism with a slanted top? Or is it a parallelepiped?

No — better interpretation:

Looking at the figure: It has two vertical sides (7 in), a base of 7 in, and a top of 8 in, and a slant edge of 8.6 in.

Wait — this appears to be a prism with a trapezoidal base?

But no — actually, it's a triangular prism with a right triangle base, but the dimensions are misread.

Wait — here's the correct way:

Let me interpret based on standard problems.

Actually, this is a rectangular prism with a right triangle base?

Wait — the figure shows:

- A triangle with:
- One leg = 7 in (vertical)
- Base = 7 in
- Hypotenuse = 8.6 in?

But $\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.9$, not 8.6.

So maybe the triangle is not right-angled at the corner?

Wait — there's a right angle symbol between the 7 in and 7 in sides.

Yes! There is a right angle between the 7 in vertical and 7 in horizontal.

So the triangle has:
- Legs: 7 in and 7 in
- Hypotenuse should be $\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.9$

But the diagram labels the slanted side as 8.6 in — which contradicts.

Wait — unless the 8.6 in is not the hypotenuse?

Let’s look carefully.

Actually, the 8.6 in is the length of the diagonal from bottom-left to top-right, which would be the space diagonal?

But no — the figure shows a 3D shape — possibly a triangular prism with triangular base.

Wait — the shape has:
- A triangular base with:
- Base = 7 in
- Height = 7 in
- Right angle between them
- So it's a right triangle with legs 7 and 7
- Then the prism extends along the 8 in side?

Wait — the label 8 in is along the top, and 8.6 in is the slanted edge.

Ah! Now I see: This is a triangular prism, but the lateral face is a parallelogram?

No — better: This is a right triangular prism with:

- Triangle base: legs = 7 in and 7 in
- Height of prism = 8 in (the depth)

But then the slanted edge of 8.6 in must be the hypotenuse of the triangle formed by the height and the depth?

Wait — no.

Wait — perhaps it's a rectangular prism with a cut-off corner?

Alternatively, this might be a truncated pyramid — but no.

Another possibility: It's a right prism with a right triangle base, and the length of the prism is 8 in.

But then why is there an 8.6 in?

Wait — perhaps the 8.6 in is the hypotenuse of the triangle formed by the height and the base?

No — the triangle has legs 7 and 7 — hypotenuse should be ~9.9.

But the diagram shows 8.6 in — so maybe it's not a right triangle?

Wait — the right angle symbol is present between the 7 in and 7 in sides — so it is a right triangle.

But then hypotenuse should be $\sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.9$ — but it's labeled 8.6 — contradiction.

Wait — maybe the 8.6 in is not the hypotenuse of the triangle?

Looking at the diagram: The 8.6 in is drawn from the bottom-left corner to the top-right corner — but in 3D space.

Ah! So it's the space diagonal of the prism?

Let’s assume:

- The base is a right triangle: legs = 7 in and 7 in
- The prism extends 8 in in depth (the "length")
- Then the space diagonal from one vertex to the opposite is given as 8.6 in?

But that doesn't make sense — because the space diagonal would be:

$$
d = \sqrt{7^2 + 7^2 + 8^2} = \sqrt{49 + 49 + 64} = \sqrt{162} \approx 12.7 \text{ in}
$$

Not 8.6.

So clearly, something is wrong.

Wait — perhaps the 8.6 in is the hypotenuse of the triangular face?

But the triangle has legs 7 and 7 — hypotenuse ≈ 9.9 — not 8.6.

Unless the triangle is not right-angled?

But the right angle symbol is there.

Wait — maybe the 7 in is not both legs?

Let me re-analyze.

Looking at the diagram:

- Vertical side: 7 in
- Horizontal base: 7 in
- Right angle between them
- Then a slanted side of 8.6 in
- And the top is labeled 8 in

Ah! Wait — this is a trapezoidal prism? No.

Actually, now I think this is a triangular prism with:

- Base triangle: right triangle with legs 7 in and 7 in → hypotenuse ≈ 9.9 in
- But the prism length is 8 in

Then the surface area includes:
- Two triangular bases
- Three rectangular faces

But the 8.6 in is probably a typo or mislabeling.

Wait — perhaps the 8.6 in is the length of the prism?

But it's labeled as 8 in.

Wait — the 8 in is the top horizontal side, and the 8.6 in is the slanted edge connecting the bottom-left to top-right.

But in 3D, this could be a face diagonal.

Wait — perhaps this is a rectangular prism with a missing part?

Alternatively, perhaps it's a right prism with a right triangle base, and the 8.6 in is the hypotenuse of the triangle, but the triangle has legs 7 in and something else.

But the right angle is between the 7 in and 7 in — so it must be legs.

Unless the 7 in is the height, and the 8 in is the base, and the 8.6 in is the hypotenuse?

Wait — let’s try that.

Suppose the triangle has:
- Base = 8 in
- Height = 7 in
- Right angle between them
- Then hypotenuse = $\sqrt{8^2 + 7^2} = \sqrt{64 + 49} = \sqrt{113} \approx 10.63$ — still not 8.6

Still not matching.

Wait — what if the 8.6 in is the hypotenuse, and the legs are 7 in and x?

Try: $x = \sqrt{8.6^2 - 7^2} = \sqrt{73.96 - 49} = \sqrt{24.96} \approx 5$ in — not 7.

No.

Wait — perhaps the 8.6 in is not related to the triangle?

Maybe the 8.6 in is the length of the prism, and the 8 in is the top, and the 7 in is the height?

Wait — the diagram shows:

- A quadrilateral side with:
- Bottom: 7 in
- Top: 8 in
- Vertical side: 7 in
- Slanted side: 8.6 in

Ah! Now I see: This is a trapezoidal prism?

No — it's a prism with a trapezoidal base?

But the figure has a right angle, so it's a right trapezoid.

Wait — the base is a trapezoid with:
- One leg = 7 in (vertical)
- Base = 7 in
- Top = 8 in
- Slanted side = 8.6 in

And the height of the prism is not given — but we have a right angle, so the vertical side is perpendicular.

But the other dimension is missing — how deep is the prism?

Wait — the figure only shows a side view — but for volume, we need three dimensions.

Wait — perhaps this is a triangular prism with a right triangle base, and the 8.6 in is a mistake.

Alternatively, perhaps it's a rectangular prism with a cut?

I think there's confusion.

Let me search for similar problems.

Wait — actually, upon closer inspection, this is a triangular prism with:

- Right triangle base: legs = 7 in and 7 in
- Prism length = 8 in

Then the volume is:
$$
V = \frac{1}{2} \times 7 \times 7 \times 8 = \frac{1}{2} \times 49 \times 8 = 24.5 \times 8 = 196 \text{ in}^3
$$

Surface area:
- Two triangular bases: $2 \times \frac{1}{2} \times 7 \times 7 = 49 \text{ in}^2$
- Three rectangular faces:
- 7 in × 8 in = 56 in²
- 7 in × 8 in = 56 in²
- Hypotenuse × 8 in = $\sqrt{7^2 + 7^2} \times 8 = \sqrt{98} \times 8 \approx 9.899 \times 8 \approx 79.19 \text{ in}^2$

Total SA:
$$
49 + 56 + 56 + 79.19 = 240.19 \text{ in}^2
$$

But the 8.6 in is not used — so likely it's a typo or mislabel.

Alternatively, perhaps the 8.6 in is the hypotenuse, and the legs are different.

Wait — suppose the triangle has:
- One leg = 7 in
- Hypotenuse = 8.6 in
- Find other leg: $b = \sqrt{8.6^2 - 7^2} = \sqrt{73.96 - 49} = \sqrt{24.96} \approx 5$ in

So maybe the base is 5 in, height 7 in, hypotenuse 8.6 in.

Then area of triangle:
$$
A = \frac{1}{2} \times 5 \times 7 = 17.5 \text{ in}^2
$$

Prism length = 8 in

Volume:
$$
V = 17.5 \times 8 = 140 \text{ in}^3
$$

Surface area:
- Two triangles: $2 \times 17.5 = 35 \text{ in}^2$
- Three rectangles:
- 5 in × 8 in = 40
- 7 in × 8 in = 56
- 8.6 in × 8 in = 68.8
- Total SA = 35 + 40 + 56 + 68.8 = 199.8 in²

But the diagram shows 7 in as the base — not 5 in.

So conflict.

Wait — the 7 in is the horizontal base, and the 8.6 in is the slanted edge — but the vertical is also 7 in?

Then it's a right triangle with legs 7 and 7, hypotenuse should be ~9.9 — but labeled 8.6 — so inconsistent.

Therefore, likely the 8.6 in is the length of the prism, and the 8 in is the top, but that doesn't help.

Perhaps it's a rectangular prism with a triangular cut, but without more info, it's ambiguous.

Wait — another idea: This is a right prism with a right triangle base, and the 8.6 in is the length of the prism?

But it's labeled 8 in.

I think there's a mislabeling in the diagram.

But looking at common worksheets, this is often a triangular prism with:

- Base: right triangle with legs 7 in and 7 in
- Prism length: 8 in

So despite the 8.6 in, it's likely a typo.

So proceed with that.

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Final Answers:

#### Problem 1: Cube
- Volume: $3^3 = 27 \text{ in}^3$
- Surface Area: $6 \times 3^2 = 54 \text{ in}^2$

#### Problem 2: Triangular Prism
- Base area: $\frac{1}{2} \times 2.5 \times 2.5 = 3.125 \text{ in}^2$
- Volume: $3.125 \times 8 = 25 \text{ in}^3$
- Surface Area: $2 \times 3.125 + (2.5+2.5+3.54)\times8 = 6.25 + (8.54)\times8 = 6.25 + 68.32 = 74.57 \text{ in}^2$

#### Problem 3: Cylinder
- Radius = 3.5 in, height = 8 in
- Volume: $\pi \times 3.5^2 \times 8 = 98\pi \approx 307.72 \text{ in}^3$
- Surface Area: $2\pi \times 3.5 \times (8 + 3.5) = 7\pi \times 11.5 = 80.5\pi \approx 252.98 \text{ in}^2$

#### Problem 4: Triangular Prism
Assume right triangle base with legs 7 in and 7 in, prism length 8 in.
- Volume: $\frac{1}{2} \times 7 \times 7 \times 8 = 196 \text{ in}^3$
- Surface Area: $2 \times \frac{1}{2} \times 7 \times 7 + (7+7+\sqrt{98}) \times 8 = 49 + (14 + 9.899) \times 8 = 49 + 23.899 \times 8 = 49 + 191.19 = 240.19 \text{ in}^2$

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Summary Table:

| Problem | Shape | Volume | Surface Area |
|--------|---------------|--------------|------------------|
| 1 | Cube | 27 in³ | 54 in² |
| 2 | Triangular Prism | 25 in³ | ~74.57 in² |
| 3 | Cylinder | ~307.72 in³ | ~252.98 in² |
| 4 | Triangular Prism | 196 in³ | ~240.19 in² |

> Note: Problem 4 has inconsistent labeling (8.6 in vs expected ~9.9 in). Assuming the triangle has legs 7 in and 7 in, and prism length 8 in.

Let me know if you want exact values or simplified forms.