Surface Area using Nets Worksheets - Free Printable

Let's solve each problem step by step using the nets of the solid shapes. We'll calculate the surface area by finding the area of each individual face in the net and then adding them together.

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Problem 1: Triangular Prism

#### Given Net:
- The solid is a triangular prism.
- The net shows:
- Two triangular faces (J and L)
- Three rectangular faces (K, M, H)

#### Dimensions:
From the diagram:
- Triangle base = 6 ft
- Height of triangle = 4 ft
- Side lengths of triangle: 5 ft, 5 ft, 6 ft (isosceles triangle)
- Rectangular sides:
- K: 6 ft × 3 ft
- M: 5 ft × 3 ft
- H: 5 ft × 3 ft

Wait — actually, looking at the net, it appears that the height of the prism is 3 ft, and the triangular bases have side lengths 5 ft, 5 ft, 6 ft, with height 4 ft.

But let’s double-check:

The triangle has:
- Base = 6 ft
- Height = 4 ft → so area of triangle = (1/2) × base × height = (1/2)(6)(4) = 12 ft²

Now, the three rectangles are:
- One rectangle with width = 6 ft, height = 3 ft → Area = 6 × 3 = 18 ft²
- Two rectangles with width = 5 ft, height = 3 ft → Each = 5 × 3 = 15 ft²

So total surface area = 2 × (triangle area) + sum of rectangle areas

Let’s label the parts as per the net:

- Area of J = triangle = (1/2) × 6 × 4 = 12 ft²
- Area of K = rectangle = 6 × 3 = 18 ft²
- Area of L = triangle = same as J = 12 ft²
- Area of M = rectangle = 5 × 3 = 15 ft²
- Area of N = rectangle = 5 × 3 = 15 ft²

> Wait — but in the net, the rectangles are labeled K, M, H. So likely:
> - K = 6×3 = 18
> - M = 5×3 = 15
> - H = 5×3 = 15

So:

- Area of J = 12
- Area of K = 18
- Area of L = 12
- Area of M = 15
- Area of N = 15 → Wait, N is not labeled on the net.

Looking again: the net shows:
- J (top triangle)
- L (bottom triangle)
- K, M, H (rectangles)

But the question says:
> Area of J = __
> Area of K = __
> Area of L = __
> Area of M = __
> Area of N = __

But there is no N in the net. Possibly a typo?

Wait — perhaps H is meant to be N? Or maybe N is missing.

Actually, looking at the net: it has:
- Two triangles: J and L
- Three rectangles: K, M, H

But the labels ask for:
- Area of J
- Area of K
- Area of L
- Area of M
- Area of N ← This is suspicious

Possibly H is labeled as N? But it's clearly labeled H.

Alternatively, maybe N is a typo and should be H.

But since the problem says "Area of N =", and there is no N, I suspect it might be a labeling error.

Alternatively, perhaps the rectangles are labeled K, L, M, H — but L is already a triangle.

Wait — re-examining the image description: The net shows:
- A diamond-shaped triangle (J) connected to rectangles K, M, and H.
- Another triangle (L) opposite J.

So:
- J and L are the two triangular bases.
- K, M, H are the three lateral faces.

But the question asks for:
- Area of J
- Area of K
- Area of L
- Area of M
- Area of N

But N is not shown. Unless H is called N? That seems unlikely.

Perhaps the label N is meant to be H? Maybe a typo in the worksheet.

Alternatively, maybe N is one of the rectangles, but only three rectangles are present.

Wait — maybe N is a typo and should be H?

Assuming N = H, we proceed.

So:

- Area of J = area of triangle = (1/2) × 6 × 4 = 12 ft²
- Area of K = rectangle = 6 ft × 3 ft = 18 ft²
- Area of L = triangle = same as J = 12 ft²
- Area of M = rectangle = 5 ft × 3 ft = 15 ft²
- Area of N = rectangle = 5 ft × 3 ft = 15 ft² (assuming N = H)

Then:
- Surface Area = 12 + 18 + 12 + 15 + 15 = 72 ft²

✔ Answer for Problem 1:
- Area of J = 12
- Area of K = 18
- Area of L = 12
- Area of M = 15
- Area of N = 15
- Surface Area = 72 ft²

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Problem 2: Square Pyramid

#### Given Net:
- A square pyramid with:
- One square base (W)
- Four triangular faces (U, V, X, Y)

#### Dimensions:
- Base edge = 10 yd
- Slant height of triangular faces = 12 yd (from apex to base edge)
- Height of triangle = 12 yd (as labeled)

Wait — the diagram shows:
- The base is 10 yd × 10 yd
- Each triangular face has a base of 10 yd and height of 12 yd

Note: The slant height is given as 12 yd (the height of the triangle), not the vertical height of the pyramid.

So:

- Area of W (square base) = 10 × 10 = 100 yd²
- Area of each triangle = (1/2) × base × height = (1/2) × 10 × 12 = 60 yd²
- There are four such triangles: U, V, X, Y

So:
- Area of U = 60
- Area of V = 60
- Area of W = 100
- Area of X = 60
- Area of Y = 60

Total surface area = 4 × 60 + 100 = 240 + 100 = 340 yd²

✔ Answer for Problem 2:
- Area of U = 60
- Area of V = 60
- Area of W = 100
- Area of X = 60
- Area of Y = 60
- Surface Area = 340 yd²

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Problem 3: Rectangular Prism (Cube-like)

#### Given Net:
- A rectangular prism with dimensions:
- Length = 10 m
- Width = 4 m
- Height = 12 m

Net shows:
- Six rectangles: A, B, C, D, E, F

We need to identify which faces are which.

From the net:
- It's a cross-shaped net typical of a rectangular prism.
- The central rectangle is D.
- Up: A
- Down: F
- Left: C
- Right: E
- Back: B (or possibly another)

Wait — standard labeling:
- A = top
- B = front
- C = left
- D = bottom
- E = right
- F = back

But from the diagram:
- Dimensions: 10 m (length), 4 m (width), 12 m (height)

So:
- The faces:
- Top and bottom: 10 × 4 = 40 m²
- Front and back: 10 × 12 = 120 m²
- Left and right: 4 × 12 = 48 m²

Now assign to letters:

Looking at the net:
- Rectangle D is the center (likely front or back)
- A and F are above and below → could be top and bottom
- C and E are left and right → side faces
- B is probably the other side

But from the net:
- A and F are both 4 m × 10 m? Wait — look at the dimensions.

Wait — the cube shows:
- 10 m (length)
- 4 m (width)
- 12 m (height)

In the net:
- Rectangle A: height = 4 m, width = 10 m → so 4×10 = 40 m² → top/bottom
- Rectangle B: height = 12 m, width = 4 m → 12×4 = 48 m² → side
- Rectangle C: 12×10 = 120 m² → front/back
- Rectangle D: 12×10 = 120 m² → front/back
- Rectangle E: 12×4 = 48 m² → side
- Rectangle F: 4×10 = 40 m² → top/bottom

So:

- A = 4×10 = 40 m² (top)
- B = 12×4 = 48 m² (left)
- C = 12×10 = 120 m² (front)
- D = 12×10 = 120 m² (back)
- E = 12×4 = 48 m² (right)
- F = 4×10 = 40 m² (bottom)

Wait — but A and F are both 40 → top and bottom
- B and E are 48 → left and right
- C and D are 120 → front and back

So:

- Area of A = 40
- Area of B = 48
- Area of C = 120
- Area of D = 120
- Area of E = 48
- Area of F = 40

Surface area = 2×(40 + 48 + 120) = 2×208 = 416 m²

But let's add directly:
- A + F = 40 + 40 = 80
- B + E = 48 + 48 = 96
- C + D = 120 + 120 = 240
- Total = 80 + 96 + 240 = 416 m²

✔ Answer for Problem 3:
- Area of A = 40
- Area of B = 48
- Area of C = 120
- Area of D = 120
- Area of E = 48
- Area of F = 40
- Surface Area = 416 m²

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✔ Final Answers:

#### Problem 1: Triangular Prism
- Area of J = 12 ft²
- Area of K = 18 ft²
- Area of L = 12 ft²
- Area of M = 15 ft²
- Area of N = 15 ft² (assuming N = H)
- Surface Area = 72 ft²

#### Problem 2: Square Pyramid
- Area of U = 60 yd²
- Area of V = 60 yd²
- Area of W = 100 yd²
- Area of X = 60 yd²
- Area of Y = 60 yd²
- Surface Area = 340 yd²

#### Problem 3: Rectangular Prism
- Area of A = 40 m²
- Area of B = 48 m²
- Area of C = 120 m²
- Area of D = 120 m²
- Area of E = 48 m²
- Area of F = 40 m²
- Surface Area = 416 m²

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📝 Summary of Steps:

1. Identify the shape from the net.
2. Find the area of each individual face.
3. Add all areas to get total surface area.
4. Use appropriate formulas:
- Triangle: (1/2) × base × height
- Rectangle: length × width
- Square: side²

Let me know if you'd like this formatted as a printable answer sheet!