Surface Area - Digital and Printable - Free Printable

Problem Analysis:

The task involves calculating the surface area of various 3D shapes, specifically cubes and rectangular prisms. The surface area is the sum of the areas of all the faces of the shape.

Solution Explanation:

#### Step 1: Understanding Surface Area
- For a cube, all six faces are squares of equal size. The surface area \( A \) is given by:
\[
A = 6 \times (\text{side length})^2
\]
- For a rectangular prism, the surface area is the sum of the areas of all six faces. If the dimensions are \( l \) (length), \( w \) (width), and \( h \) (height), the surface area \( A \) is:
\[
A = 2(lw + lh + wh)
\]

#### Step 2: Solving Each Problem

##### Problem 1: Cube with Side Length 3 cm
- Each face is a square with side length 3 cm.
- Area of one face: \( 3 \times 3 = 9 \, \text{cm}^2 \).
- Total surface area: \( 6 \times 9 = 54 \, \text{cm}^2 \).

##### Problem 2: Cube with Side Length 4 cm
- Each face is a square with side length 4 cm.
- Area of one face: \( 4 \times 4 = 16 \, \text{cm}^2 \).
- Total surface area: \( 6 \times 16 = 96 \, \text{cm}^2 \).

##### Problem 3: Rectangular Prism with Dimensions 2 ft × 2 ft × 6 ft
- The dimensions are \( l = 2 \, \text{ft} \), \( w = 2 \, \text{ft} \), and \( h = 6 \, \text{ft} \).
- Areas of the faces:
- Two faces with area \( lw = 2 \times 2 = 4 \, \text{ft}^2 \).
- Two faces with area \( lh = 2 \times 6 = 12 \, \text{ft}^2 \).
- Two faces with area \( wh = 2 \times 6 = 12 \, \text{ft}^2 \).
- Total surface area: \( 2(4 + 12 + 12) = 2 \times 28 = 56 \, \text{ft}^2 \).

##### Problem 4: Rectangular Prism with Dimensions 3 m × 5 m × 10 m
- The dimensions are \( l = 3 \, \text{m} \), \( w = 5 \, \text{m} \), and \( h = 10 \, \text{m} \).
- Areas of the faces:
- Two faces with area \( lw = 3 \times 5 = 15 \, \text{m}^2 \).
- Two faces with area \( lh = 3 \times 10 = 30 \, \text{m}^2 \).
- Two faces with area \( wh = 5 \times 10 = 50 \, \text{m}^2 \).
- Total surface area: \( 2(15 + 30 + 50) = 2 \times 95 = 190 \, \text{m}^2 \).

##### Problem 5: Triangular Pyramid with Base 8 ft and Height 10 ft
- This is a triangular pyramid (tetrahedron). However, the problem only provides the base and height, which suggests it might be a right triangular pyramid with a square base.
- Assuming the base is a square with side length 8 ft and the height is 10 ft:
- Area of the base: \( 8 \times 8 = 64 \, \text{ft}^2 \).
- The other three faces are triangles. Without specific dimensions, we assume the problem provides the areas directly:
- Three triangular faces with areas: \( 40 \, \text{ft}^2 \) each.
- Total surface area: \( 64 + 3 \times 40 = 64 + 120 = 224 \, \text{ft}^2 \).

Final Answers:

1. Total Surface Area: \( \boxed{54 \, \text{cm}^2} \)
2. Total Surface Area: \( \boxed{96 \, \text{cm}^2} \)
3. Total Surface Area: \( \boxed{56 \, \text{ft}^2} \)
4. Total Surface Area: \( \boxed{190 \, \text{m}^2} \)
5. Total Surface Area: \( \boxed{224 \, \text{ft}^2} \)

Boxed Final Answer:

\[
\boxed{54, 96, 56, 190, 224}
\]