Use the formula for the surface area of a right rectangular prism to find the surface area of a shape with the following measurements: length = 4.5 cm, width = 3.5 cm, and height = 2.5 cm.(1 point)

Question

Responses

71.5 cm2
71.5 cm squared

35.75 cm2
35.75 cm squared

10.5 cm2
10.5 cm squared

63 cm2

Answer 1

To find the surface area of a right rectangular prism, use the formula:

\[ \text{Surface Area} = 2(lw + lh + wh) \]

where \( l \) is the length, \( w \) is the width, and \( h \) is the height.

Given:

• Length \( l = 4.5 \) cm
• Width \( w = 3.5 \) cm
• Height \( h = 2.5 \) cm

Now, we can calculate each part of the formula:

1. \( lw = 4.5 \times 3.5 = 15.75 \) cm²
2. \( lh = 4.5 \times 2.5 = 11.25 \) cm²
3. \( wh = 3.5 \times 2.5 = 8.75 \) cm²

Next, add these areas together:

\[ lw + lh + wh = 15.75 + 11.25 + 8.75 = 35.75 \text{ cm}^2 \]

Finally, multiply by 2 to find the total surface area:

\[ \text{Surface Area} = 2 \times 35.75 = 71.5 \text{ cm}^2 \]

Thus, the surface area of the right rectangular prism is:

71.5 cm²

Answer 2

Use the formula to determine the surface area of a right rectangular prism where each edge length measures 12.75 cm.(1 point)
Responses

5,852.25 cm2
5,852.25 cm squared

76.5 cm2
76.5 cm squared

459 cm2
459 cm squared

975.38 cm2

Answer 3

To determine the surface area of a right rectangular prism where all edge lengths are equal (meaning it is a cube), we can use the formula for the surface area of a cube:

\[ \text{Surface Area} = 6s^2 \]

where \( s \) is the length of each edge.

Given:

• Edge length \( s = 12.75 \) cm

Now, calculate the surface area:

\[ \text{Surface Area} = 6 \times (12.75)^2 \]

First, calculate \( (12.75)^2 \):

\[ (12.75)^2 = 162.5625 \]

Now, multiply by 6:

\[ \text{Surface Area} = 6 \times 162.5625 = 975.375 \text{ cm}^2 \]

Rounding to two decimal places, we have:

975.38 cm²

Thus, the surface area of the right rectangular prism (or cube, in this case) is 975.38 cm².

Answer 4

Find the surface area of a textbook that is 2 in. × 14 in. × 9 in.
(1 point)
Responses

172 in.2
172 in. squared

50 in.2
50 in. squared

252 in.2
252 in. squared

344 in.2

Answer 5

To find the surface area of a rectangular prism (in this case, the textbook), we use the formula:

\[ \text{Surface Area} = 2(lw + lh + wh) \]

where \( l \) is the length, \( w \) is the width, and \( h \) is the height.

Given the dimensions:

• Length \( l = 2 \) in.
• Width \( w = 14 \) in.
• Height \( h = 9 \) in.

Now, calculate each product:

1. \( lw = 2 \times 14 = 28 \) in²
2. \( lh = 2 \times 9 = 18 \) in²
3. \( wh = 14 \times 9 = 126 \) in²

Now, add these areas together:

\[ lw + lh + wh = 28 + 18 + 126 = 172 \text{ in}^2 \]

Finally, multiply by 2 to find the total surface area:

\[ \text{Surface Area} = 2 \times 172 = 344 \text{ in}^2 \]

Thus, the surface area of the textbook is:

344 in²

Answer 6

Sue is wrapping books to give as gifts. Book A has a length of 6.5 inches, a width of 1 inch, and a height of 8.5 inches. Book B has a length of 5.5 inches, a width of 2 inches, and a height of 8 inches. Based on surface area, which book will require less wrapping paper?(1 point)
Responses

Book A will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book B.
Book A will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book B.

Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.
Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.

Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.
Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.

Book B will require less wrapping paper because its surface area is 0.5 in.2
less than the surface area of Book A.

Answer 7

To determine which book will require less wrapping paper, we need to calculate the surface area of both Book A and Book B using the formula for the surface area of a rectangular prism:

\[ \text{Surface Area} = 2(lw + lh + wh) \]

Calculate the Surface Area of Book A:

Dimensions of Book A:

• Length \( l = 6.5 \) in.
• Width \( w = 1 \) in.
• Height \( h = 8.5 \) in.

Calculate the individual areas:

1. \( lw = 6.5 \times 1 = 6.5 \) in²
2. \( lh = 6.5 \times 8.5 = 55.25 \) in²
3. \( wh = 1 \times 8.5 = 8.5 \) in²

Now, add these together:

\[ lw + lh + wh = 6.5 + 55.25 + 8.5 = 70.25 \text{ in}^2 \]

Now, multiply by 2:

\[ \text{Surface Area of Book A} = 2 \times 70.25 = 140.5 \text{ in}^2 \]

Calculate the Surface Area of Book B:

Dimensions of Book B:

• Length \( l = 5.5 \) in.
• Width \( w = 2 \) in.
• Height \( h = 8 \) in.

Calculate the individual areas:

1. \( lw = 5.5 \times 2 = 11 \) in²
2. \( lh = 5.5 \times 8 = 44 \) in²
3. \( wh = 2 \times 8 = 16 \) in²

Now, add these together:

\[ lw + lh + wh = 11 + 44 + 16 = 71 \text{ in}^2 \]

Now, multiply by 2:

\[ \text{Surface Area of Book B} = 2 \times 71 = 142 \text{ in}^2 \]

Summary of Surface Areas:

• Surface Area of Book A: 140.5 in²
• Surface Area of Book B: 142 in²

Conclusion:

Now, let's compare the two surface areas:

• Book A: 140.5 in²
• Book B: 142 in²

Book A has a surface area of 140.5 in², while Book B has a surface area of 142 in².

Thus, Book A will require less wrapping paper because its surface area is 1.5 in² less than the surface area of Book B.

Answer 8

Use the image to answer the question.

An illustration shows a 3 D box with length of 27 centimeters labeled as front, a width of 20 centimeters, and height of 20 centimeters.

A box needs to be decorated to use as a prop in a play. The bottom and the back of the box do not need to be decorated. What is the surface area of the box that needs to be decorated?

(1 point)
Responses

67 cm2
67 cm squared

2,960 cm2
2,960 cm squared

1,080 cm
1,080 cm

1,880 cm2
1,880 cm squared
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Answer 9

To calculate the surface area of the box that needs to be decorated, we first need to determine which surfaces will be decorated. Since the bottom and back of the box do not need to be decorated, we will only consider the top, front, and two sides of the box.

Given the dimensions of the box:

• Length \( l = 27 \) cm
• Width \( w = 20 \) cm
• Height \( h = 20 \) cm

Surfaces to be decorated:

1. Top Surface (Length × Width): \[ \text{Top Surface Area} = l \times w = 27 , \text{cm} \times 20 , \text{cm} = 540 , \text{cm}^2 \]

2. Front Surface (Length × Height): \[ \text{Front Surface Area} = l \times h = 27 , \text{cm} \times 20 , \text{cm} = 540 , \text{cm}^2 \]

3. Two Side Surfaces (Width × Height): \[ \text{Side Surface Area} = w \times h = 20 , \text{cm} \times 20 , \text{cm} = 400 , \text{cm}^2 \] Since there are two side surfaces, we multiply by 2: \[ \text{Total Side Surface Area} = 2 \times 400 , \text{cm}^2 = 800 , \text{cm}^2 \]

Total Surface Area to be Decorated:

Now, add the areas of the top, front, and both sides:

\[ \text{Total Surface Area} = \text{Top Surface Area} + \text{Front Surface Area} + \text{Total Side Surface Area} \] \[ \text{Total Surface Area} = 540 , \text{cm}^2 + 540 , \text{cm}^2 + 800 , \text{cm}^2 = 1,880 , \text{cm}^2 \]

Thus, the surface area of the box that needs to be decorated is:

1,880 cm²