Surface Area & Volume of Rectangular & Triangular Prisms Worksheet - Free Printable

To solve the problem of finding the surface area of each figure, we need to calculate the surface area for prisms and pyramids based on their respective formulas. Let's go through each figure step by step.

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1) Square Pyramid

- Base side: \(2 \, \text{cm}\)
- Slant height: \(10 \, \text{cm}\)

The surface area of a square pyramid is given by:
\[
\text{Surface Area} = \text{Base Area} + \text{Lateral Area}
\]
- Base Area:
\[
\text{Base Area} = s^2 = 2^2 = 4 \, \text{cm}^2
\]
- Lateral Area:
\[
\text{Lateral Area} = 4 \times \left( \frac{1}{2} \times \text{base side} \times \text{slant height} \right) = 4 \times \left( \frac{1}{2} \times 2 \times 10 \right) = 4 \times 10 = 40 \, \text{cm}^2
\]
- Total Surface Area:
\[
\text{Surface Area} = 4 + 40 = 44 \, \text{cm}^2
\]

Answer: \(44 \, \text{cm}^2\)

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2) Triangular Pyramid

- Base sides: \(3 \, \text{in}\), \(6 \, \text{in}\), \(10 \, \text{in}\)
- Slant heights: Not directly given, but we can use the formula for the lateral area of a triangular pyramid.

The surface area of a triangular pyramid is:
\[
\text{Surface Area} = \text{Base Area} + \text{Lateral Area}
\]
- Base Area (using Heron's formula):
\[
s = \frac{3 + 6 + 10}{2} = 9.5 \, \text{in}
\]
\[
\text{Base Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9.5(9.5-3)(9.5-6)(9.5-10)} = \sqrt{9.5 \times 6.5 \times 3.5 \times (-0.5)}
\]
Since the base sides do not form a valid triangle (sum of two sides must be greater than the third side), this figure is not possible. Let's assume it's a typo and proceed with the next figures.

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3) Rectangular Prism

- Dimensions: \(10 \, \text{mm} \times 4 \, \text{mm} \times 3 \, \text{mm}\)

The surface area of a rectangular prism is:
\[
\text{Surface Area} = 2(lw + lh + wh)
\]
\[
\text{Surface Area} = 2(10 \times 4 + 10 \times 3 + 4 \times 3) = 2(40 + 30 + 12) = 2 \times 82 = 164 \, \text{mm}^2
\]

Answer: \(164 \, \text{mm}^2\)

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4) Square Pyramid

- Base side: \(4 \, \text{yd}\)
- Slant height: \(12 \, \text{yd}\)

Using the same formula as in Figure 1:
- Base Area:
\[
\text{Base Area} = 4^2 = 16 \, \text{yd}^2
\]
- Lateral Area:
\[
\text{Lateral Area} = 4 \times \left( \frac{1}{2} \times 4 \times 12 \right) = 4 \times 24 = 96 \, \text{yd}^2
\]
- Total Surface Area:
\[
\text{Surface Area} = 16 + 96 = 112 \, \text{yd}^2
\]

Answer: \(112 \, \text{yd}^2\)

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5) Cube

- Side length: \(2 \, \text{mm}\)

The surface area of a cube is:
\[
\text{Surface Area} = 6s^2
\]
\[
\text{Surface Area} = 6 \times 2^2 = 6 \times 4 = 24 \, \text{mm}^2
\]

Answer: \(24 \, \text{mm}^2\)

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6) Triangular Prism

- Base sides: \(4 \, \text{yd}\), \(4 \, \text{yd}\), \(11 \, \text{yd}\)
- Height: \(3 \, \text{yd}\)

The surface area of a triangular prism is:
\[
\text{Surface Area} = 2 \times (\text{Base Area}) + \text{Perimeter of Base} \times \text{Height}
\]
- Base Area (using Heron's formula):
\[
s = \frac{4 + 4 + 11}{2} = 9.5 \, \text{yd}
\]
\[
\text{Base Area} = \sqrt{9.5(9.5-4)(9.5-4)(9.5-11)} = \sqrt{9.5 \times 5.5 \times 5.5 \times (-1.5)}
\]
This is not possible as the sides do not form a valid triangle. Let's assume it's a typo and proceed.

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7) Pentagonal Prism

- Base side: \(2 \, \text{cm}\)
- Height: \(10 \, \text{cm}\)
- Apex to center: \(3 \, \text{cm}\)

The surface area of a pentagonal prism is:
\[
\text{Surface Area} = 2 \times (\text{Base Area}) + \text{Perimeter of Base} \times \text{Height}
\]
- Base Area (approximation for a regular pentagon):
\[
\text{Base Area} \approx \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} \times s^2
\]
\[
\text{Base Area} \approx \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} \times 2^2 \approx 3.1 \, \text{cm}^2
\]
- Perimeter of Base:
\[
\text{Perimeter} = 5 \times 2 = 10 \, \text{cm}
\]
- Lateral Area:
\[
\text{Lateral Area} = 10 \times 10 = 100 \, \text{cm}^2
\]
- Total Surface Area:
\[
\text{Surface Area} \approx 2 \times 3.1 + 100 = 6.2 + 100 = 106.2 \, \text{cm}^2
\]

Answer: \(106.2 \, \text{cm}^2\)

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8) Triangular Pyramid

- Base sides: \(7 \, \text{ft}\), \(12 \, \text{ft}\), \(13 \, \text{ft}\)
- Slant heights: Not given, so we cannot calculate the surface area without additional information.

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9) Cone

- Radius: \(5 \, \text{in}\)
- Slant height: \(13 \, \text{in}\)

The surface area of a cone is:
\[
\text{Surface Area} = \pi r(r + l)
\]
\[
\text{Surface Area} = \pi \times 5 \times (5 + 13) = \pi \times 5 \times 18 = 90\pi \, \text{in}^2
\]
\[
\text{Surface Area} \approx 282.74 \, \text{in}^2
\]

Answer: \(282.74 \, \text{in}^2\)

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Final Answers

\[
\boxed{44, 164, 112, 24, 106.2, 282.74}
\]