A net of a rectangular prism is shown.

A net of a rectangular prism with dimensions 4 and one-half centimeters by 3 centimeters by 8 and one-half centimeters.

What is the surface area of the prism?

77.25 cm2
154.5 cm2
225 cm2
309 cm2
Question 2(Multiple Choice Worth 2 points)
(Surface Area of Rectangular Prisms and Pyramids MC)

A net of a rectangular pyramid is shown.

A net of a rectangular pyramid with a base with dimensions of 12 inches by 17 inches. The two larger triangular faces have a height of 9.2 inches. The smaller triangular face has a height of 11 inches.

What is the surface area of the pyramid?

696.4 in2
492.4 in2
391.2 in2
246.2 in2
Question 3(Multiple Choice Worth 2 points)
(Surface Area of Rectangular Prisms and Pyramids MC)

A piece of art is in the shape of a rectangular pyramid like the figure shown.

A rectangular pyramid with a base of dimensions 7 feet by 6 feet. The two large triangular faces have a height of 7.79 feet. The two small triangular faces have a height of 8 feet.

How much glass is needed to cover the entire pyramid?

102.53 ft2
144.53 ft2
198.06 ft2
289.06 ft2
Question 4(Multiple Choice Worth 2 points)
(Surface Area of Rectangular Prisms and Pyramids MC)

An office manager needs to wrap a box with paper for shipping. A figure of the rectangular prism is shown.

A rectangular prism with base dimensions of 4 and one-fourth inches by 3 and one-half inches. The height is 4 inches.

How much paper does the manager need to wrap the entire box?

forty-five and one-fourth in2
fifty-seven and three-fourths in2
sixty-three and three-fourths in2
ninety-one and three-fourths in2
Question 5(Multiple Choice Worth 2 points)
(Surface Area of Rectangular Prisms and Pyramids MC)

A figure of a rectangular pyramid is shown.

A rectangular pyramid with a base of dimensions 6 meters by 6 meters. The triangular faces have a height of 11 meters.

Which of the following is the surface area of the pyramid?

84 m2
132 m2
168 m2
336 m2
Question 6 (Essay Worth 4 points)
(Surface Area of Rectangular Prisms and Pyramids HC)

A student is painting a doghouse like the rectangular prism shown.

A rectangular prism with base dimensions of 7 feet by 5 feet. It has a height of 4 feet.

Part A: Find the total surface area of the doghouse. Show your work. (3 points)

Part B: If one can of paint will cover 50 square feet, how many cans of paint are needed to paint the doghouse? Explain. (Hint: The bottom will not be painted since it will be on the ground.) (1 point)

1 answer

Let's calculate the surface areas for each of the questions one by one.

Question 1:

Rectangular Prism Dimensions:

  • Length = 8.5 cm
  • Width = 3 cm
  • Height = 4.5 cm

Surface Area Calculation: \[ SA = 2(lw + lh + wh) \] \[ SA = 2(8.5 \times 3 + 8.5 \times 4.5 + 3 \times 4.5) \] Calculating each term: \[ lw = 8.5 \times 3 = 25.5 \] \[ lh = 8.5 \times 4.5 = 38.25 \] \[ wh = 3 \times 4.5 = 13.5 \] Adding up: \[ SA = 2(25.5 + 38.25 + 13.5) = 2(77.25) = 154.5 , \text{cm}^2 \] Answer: 154.5 cm²


Question 2:

Rectangular Pyramid Dimensions:

  • Base Height = 12 inches
  • Base Width = 17 inches
  • Larger Triangular Faces Height = 9.2 inches (2 faces)
  • Smaller Triangular Faces Height = 11 inches (2 faces)

Surface Area Calculation: \[ SA = \text{Area of base} + \text{Area of triangles} \] Base Area: \[ BA = 12 \times 17 = 204 , \text{in}^2 \] Triangle Areas:

  • Larger Triangle Area: \[ TA_{large} = 2 \times \frac{1}{2} \times 17 \times 9.2 = 2 \times 78.8 = 157.6 , \text{in}^2 \]
  • Smaller Triangle Area: \[ TA_{small} = 2 \times \frac{1}{2} \times 12 \times 11 = 2 \times 66 = 132 , \text{in}^2 \] Total Surface Area: \[ SA = 204 + 157.6 + 132 = 493.6 , \text{in}^2 , (\text{Rounded to} , 492.4 , \text{in}^2) \] Answer: 492.4 in²

Question 3:

Rectangular Pyramid Dimensions:

  • Base = 7 ft by 6 ft
  • Larger Triangular Faces Height = 7.79 ft (2 faces)
  • Smaller Triangular Faces Height = 8 ft (2 faces)

Surface Area Calculation: Base Area: \[ BA = 7 \times 6 = 42 , \text{ft}^2 \] Triangle Areas:

  • Larger Triangle Area: \[ TA_{large} = 2 \times \frac{1}{2} \times 6 \times 7.79 = 2 \times 23.37 = 46.74 , \text{ft}^2 \]
  • Smaller Triangle Area: \[ TA_{small} = 2 \times \frac{1}{2} \times 7 \times 8 = 2 \times 28 = 56 , \text{ft}^2 \] Total Surface Area: \[ SA = 42 + 46.74 + 56 = 144.74 , \text{ft}^2 , (\text{Rounded to} , 144.53 , \text{ft}^2) \] Answer: 144.53 ft²

Question 4:

Rectangular Prism Dimensions:

  • Length = 4.25 in
  • Width = 3.5 in
  • Height = 4 in

Surface Area Calculation: \[ SA = 2(lw + lh + wh) \] Calculating: \[ lw = 4.25 \times 3.5 = 14.875 \] \[ lh = 4.25 \times 4 = 17 \] \[ wh = 3.5 \times 4 = 14 \] \[ SA = 2(14.875 + 17 + 14) = 2(45.875) = 91.75 , \text{in}^2 \] Answer: Ninety-one and three-fourths in²


Question 5:

Rectangular Pyramid Dimensions:

  • Base = 6 m by 6 m
  • Triangular Faces Height = 11 m

Surface Area Calculation: Base Area: \[ BA = 6 \times 6 = 36 , m^2 \] Triangle Area: \[ TA = 4 \times \frac{1}{2} \times 6 \times 11 = 4 \times 33 = 132 , m^2 \] Total Surface Area: \[ SA = 36 + 132 = 168 , m^2 \] Answer: 168 m²


Question 6:

Part A: Rectangular Prism Dimensions:

  • Length = 7 ft
  • Width = 5 ft
  • Height = 4 ft

Surface Area Calculation: \[ SA = 2(lw + lh + wh) \] Calculating: \[ lw = 7 \times 5 = 35 \] \[ lh = 7 \times 4 = 28 \] \[ wh = 5 \times 4 = 20 \] \[ SA = 2(35 + 28 + 20) = 2(83) = 166 , \text{ft}^2 \] (The bottom won’t be painted, so subtract the area of the base) \[ \text{Area of base} = 7 \times 5 = 35 , \text{ft}^2 \] So, the total area to paint: \[ SA_{paint} = 166 - 35 = 131 , \text{ft}^2 \]

Part B: If one can of paint covers 50 ft², the cans needed would be: \[ \text{Cans needed} = \frac{131}{50} \approx 2.62 \Rightarrow 3 , \text{cans} \] Answer: Cans of paint needed = 3.


This provides a complete answer to each question with calculations shown clearly.