Printable math worksheet for calculating surface area of solids using their nets.

Let's solve each problem step by step using the nets of the solid shapes to find their surface areas.

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Problem 1: Triangular Prism

We are given a triangular prism with:
- Triangle base (N): right triangle with legs 6 ft and 8 ft, hypotenuse 10 ft
- Rectangle sides: one side is 9 ft long (height of prism)

#### Step 1: Identify the faces in the net
The net has:
- Two triangular faces: J and N
- Three rectangular faces: K, L, and M

But note: In the diagram:
- J and N are congruent triangles (same size)
- K, L, M are rectangles

From the 3D shape:
- The two triangular bases: N and J
- Three rectangular lateral faces:
- One rectangle with width = 6 ft (side of triangle), height = 9 ft → face K
- One rectangle with width = 8 ft, height = 9 ft → face L
- One rectangle with width = 10 ft (hypotenuse), height = 9 ft → face M

#### Area Calculations:

1. Area of J = Area of triangle =
$$
\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24\ \text{ft}^2
$$

2. Area of K = rectangle = $6 \times 9 = 54\ \text{ft}^2$

3. Area of L = rectangle = $8 \times 9 = 72\ \text{ft}^2$

4. Area of M = rectangle = $10 \times 9 = 90\ \text{ft}^2$

5. Area of N = same as J = $24\ \text{ft}^2$

#### Surface Area:
$$
\text{Surface Area} = 24 + 54 + 72 + 90 + 24 = 264\ \text{ft}^2
$$

✔ Answer for Problem 1:
- Area of J = 24 ft²
- Area of K = 54 ft²
- Area of L = 72 ft²
- Area of M = 90 ft²
- Area of N = 24 ft²
- Surface Area = 264 ft²

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Problem 2: Square Pyramid

Given:
- Base is a square with side 16 yd
- Slant height of triangular faces = 15 yd
- Net shows:
- One square base: W
- Four triangular faces: U, V, X, Y

All triangular faces are congruent (since pyramid is regular).

#### Step 1: Area of W (square base)
$$
\text{Area of W} = 16 \times 16 = 256\ \text{yd}^2
$$

#### Step 2: Area of one triangular face (e.g., U)
Each triangle has:
- Base = 16 yd
- Height = 15 yd (slant height)

$$
\text{Area of one triangle} = \frac{1}{2} \times 16 \times 15 = 120\ \text{yd}^2
$$

So:
- Area of U = 120 yd²
- Area of V = 120 yd²
- Area of X = 120 yd²
- Area of Y = 120 yd²

#### Total Surface Area:
$$
\text{Surface Area} = 256 + 4 \times 120 = 256 + 480 = 736\ \text{yd}^2
$$

✔ Answer for Problem 2:
- Area of U = 120 yd²
- Area of V = 120 yd²
- Area of W = 256 yd²
- Area of X = 120 yd²
- Area of Y = 120 yd²
- Surface Area = 736 yd²

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Problem 3: Rectangular Prism (Box)

Dimensions:
- Length = 10 in
- Width = 5 in
- Height = 12 in

Net shows six faces labeled A, B, C, D, E, F

From the net:
- A and F are top and bottom → both are $10 \times 5$ → area = $50\ \text{in}^2$
- B and D are front/back → $10 \times 12$ → area = $120\ \text{in}^2$
- C and E are left/right → $5 \times 12$ → area = $60\ \text{in}^2$

Let’s assign based on labels:
- A: top → $10 \times 5 = 50$
- B: front → $10 \times 12 = 120$
- C: left → $5 \times 12 = 60$
- D: back → $10 \times 12 = 120$
- E: right → $5 \times 12 = 60$
- F: bottom → $10 \times 5 = 50$

#### Areas:
- Area of A = 50 in²
- Area of B = 120 in²
- Area of C = 60 in²
- Area of D = 120 in²
- Area of E = 60 in²
- Area of F = 50 in²

#### Surface Area:
$$
\text{Surface Area} = 50 + 120 + 60 + 120 + 60 + 50 = 460\ \text{in}^2
$$

✔ Answer for Problem 3:
- Area of A = 50 in²
- Area of B = 120 in²
- Area of C = 60 in²
- Area of D = 120 in²
- Area of E = 60 in²
- Area of F = 50 in²
- Surface Area = 460 in²

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✔ Final Answers Summary:

#### Problem 1: Triangular Prism
- Area of J = 24 ft²
- Area of K = 54 ft²
- Area of L = 72 ft²
- Area of M = 90 ft²
- Area of N = 24 ft²
- Surface Area = 264 ft²

#### Problem 2: Square Pyramid
- Area of U = 120 yd²
- Area of V = 120 yd²
- Area of W = 256 yd²
- Area of X = 120 yd²
- Area of Y = 120 yd²
- Surface Area = 736 yd²

#### Problem 3: Rectangular Prism
- Area of A = 50 in²
- Area of B = 120 in²
- Area of C = 60 in²
- Area of D = 120 in²
- Area of E = 60 in²
- Area of F = 50 in²
- Surface Area = 460 in²

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