Question
Calculate the surface area of a right rectangular prism with dimensions w=12 in.
𝑤
=
12
in.
, l=18 in.
đť‘™
=
18
in.
, and h=8 in.
â„Ž
=
8
in.
. How many square inches is the surface area?(1 point)
in.^2
2. Calculate the surface area of a cube with 12-inch edges. How many square inches is the surface area?(1 point)
in.^2
3. Calculate the surface area of a right rectangular prism. Each edge length is 15 ft.(1 point)
ft.^2
4. Caroline needs to find the surface area of a cereal box for an art project. The box is a rectangular prism with a length of 8 inches, a height of 12 inches, and a width of 2 inches. What is its surface area?(1 point)
in.^2
5. A roll of wrapping paper measures 25 in.Ă—405 in.
25
in.
Ă—
405
in.
The boxes being wrapped are 16 in.Ă—14 in.Ă—6 in.
16
in.
Ă—
14
in.
Ă—
6
in.
How many complete boxes can be wrapped with one roll of wrapping paper?(1 point)
𝑤
=
12
in.
, l=18 in.
đť‘™
=
18
in.
, and h=8 in.
â„Ž
=
8
in.
. How many square inches is the surface area?(1 point)
in.^2
2. Calculate the surface area of a cube with 12-inch edges. How many square inches is the surface area?(1 point)
in.^2
3. Calculate the surface area of a right rectangular prism. Each edge length is 15 ft.(1 point)
ft.^2
4. Caroline needs to find the surface area of a cereal box for an art project. The box is a rectangular prism with a length of 8 inches, a height of 12 inches, and a width of 2 inches. What is its surface area?(1 point)
in.^2
5. A roll of wrapping paper measures 25 in.Ă—405 in.
25
in.
Ă—
405
in.
The boxes being wrapped are 16 in.Ă—14 in.Ă—6 in.
16
in.
Ă—
14
in.
Ă—
6
in.
How many complete boxes can be wrapped with one roll of wrapping paper?(1 point)
Answers
GPT-4o mini
Let's calculate each part step by step.
### 1. Surface area of a right rectangular prism (w=12 in., l=18 in., h=8 in.)
The formula for the surface area \( S \) of a rectangular prism is given by:
\[
S = 2(lw + lh + wh)
\]
Plugging in the given dimensions:
- \( w = 12 \) in.
- \( l = 18 \) in.
- \( h = 8 \) in.
\[
S = 2(18 \cdot 12 + 18 \cdot 8 + 12 \cdot 8)
\]
Calculating each term:
- \( 18 \cdot 12 = 216 \)
- \( 18 \cdot 8 = 144 \)
- \( 12 \cdot 8 = 96 \)
Now adding them together:
\[
216 + 144 + 96 = 456
\]
Multiply by 2:
\[
S = 2 \cdot 456 = 912 \text{ in}^2
\]
### 2. Surface area of a cube with 12-inch edges
The formula for the surface area \( S \) of a cube is given by:
\[
S = 6s^2
\]
where \( s \) is the length of an edge. Given that \( s = 12 \) in., we calculate:
\[
S = 6 \cdot 12^2 = 6 \cdot 144 = 864 \text{ in}^2
\]
### 3. Surface area of a right rectangular prism with each edge length of 15 ft
Since it's a cube (all edges are the same), we use the same cube formula:
\[
S = 6s^2
\]
with \( s = 15 \) ft:
\[
S = 6 \cdot 15^2 = 6 \cdot 225 = 1350 \text{ ft}^2
\]
### 4. Surface area of a cereal box (rectangular prism) with dimensions:
- Length = 8 in.
- Height = 12 in.
- Width = 2 in.
Using the same formula for surface area:
\[
S = 2(lw + lh + wh)
\]
Plugging in the dimensions:
\[
S = 2(8 \cdot 2 + 8 \cdot 12 + 2 \cdot 12)
\]
Calculating each term:
- \( 8 \cdot 2 = 16 \)
- \( 8 \cdot 12 = 96 \)
- \( 2 \cdot 12 = 24 \)
Adding these together:
\[
16 + 96 + 24 = 136
\]
Multiply by 2:
\[
S = 2 \cdot 136 = 272 \text{ in}^2
\]
### 5. How many complete boxes can be wrapped with one roll of wrapping paper?
First, we need to find the surface area of one box with dimensions:
- Length = 16 in.
- Width = 14 in.
- Height = 6 in.
Using the surface area formula for a rectangular prism:
\[
S = 2(lw + lh + wh)
\]
Plugging in the dimensions:
\[
S = 2(16 \cdot 14 + 16 \cdot 6 + 14 \cdot 6)
\]
Calculating each term:
- \( 16 \cdot 14 = 224 \)
- \( 16 \cdot 6 = 96 \)
- \( 14 \cdot 6 = 84 \)
Adding them together:
\[
224 + 96 + 84 = 404
\]
Multiply by 2:
\[
S = 2 \cdot 404 = 808 \text{ in}^2
\]
Next, calculate the area of the roll of wrapping paper:
\[
\text{Area} = 25 \cdot 405 = 10125 \text{ in}^2
\]
Now, divide the total area of the wrapping paper by the surface area of one box:
\[
\text{Number of boxes} = \frac{10125}{808} \approx 12.5
\]
Since only complete boxes can be wrapped, we take the floor of this number:
\[
\text{Complete boxes} = 12
\]
### Summary of Answers:
1. **Surface area of rectangular prism**: 912 in²
2. **Surface area of cube**: 864 in²
3. **Surface area of rectangular prism (15 ft edges)**: 1350 ft²
4. **Surface area of cereal box**: 272 in²
5. **Complete boxes that can be wrapped**: 12
### 1. Surface area of a right rectangular prism (w=12 in., l=18 in., h=8 in.)
The formula for the surface area \( S \) of a rectangular prism is given by:
\[
S = 2(lw + lh + wh)
\]
Plugging in the given dimensions:
- \( w = 12 \) in.
- \( l = 18 \) in.
- \( h = 8 \) in.
\[
S = 2(18 \cdot 12 + 18 \cdot 8 + 12 \cdot 8)
\]
Calculating each term:
- \( 18 \cdot 12 = 216 \)
- \( 18 \cdot 8 = 144 \)
- \( 12 \cdot 8 = 96 \)
Now adding them together:
\[
216 + 144 + 96 = 456
\]
Multiply by 2:
\[
S = 2 \cdot 456 = 912 \text{ in}^2
\]
### 2. Surface area of a cube with 12-inch edges
The formula for the surface area \( S \) of a cube is given by:
\[
S = 6s^2
\]
where \( s \) is the length of an edge. Given that \( s = 12 \) in., we calculate:
\[
S = 6 \cdot 12^2 = 6 \cdot 144 = 864 \text{ in}^2
\]
### 3. Surface area of a right rectangular prism with each edge length of 15 ft
Since it's a cube (all edges are the same), we use the same cube formula:
\[
S = 6s^2
\]
with \( s = 15 \) ft:
\[
S = 6 \cdot 15^2 = 6 \cdot 225 = 1350 \text{ ft}^2
\]
### 4. Surface area of a cereal box (rectangular prism) with dimensions:
- Length = 8 in.
- Height = 12 in.
- Width = 2 in.
Using the same formula for surface area:
\[
S = 2(lw + lh + wh)
\]
Plugging in the dimensions:
\[
S = 2(8 \cdot 2 + 8 \cdot 12 + 2 \cdot 12)
\]
Calculating each term:
- \( 8 \cdot 2 = 16 \)
- \( 8 \cdot 12 = 96 \)
- \( 2 \cdot 12 = 24 \)
Adding these together:
\[
16 + 96 + 24 = 136
\]
Multiply by 2:
\[
S = 2 \cdot 136 = 272 \text{ in}^2
\]
### 5. How many complete boxes can be wrapped with one roll of wrapping paper?
First, we need to find the surface area of one box with dimensions:
- Length = 16 in.
- Width = 14 in.
- Height = 6 in.
Using the surface area formula for a rectangular prism:
\[
S = 2(lw + lh + wh)
\]
Plugging in the dimensions:
\[
S = 2(16 \cdot 14 + 16 \cdot 6 + 14 \cdot 6)
\]
Calculating each term:
- \( 16 \cdot 14 = 224 \)
- \( 16 \cdot 6 = 96 \)
- \( 14 \cdot 6 = 84 \)
Adding them together:
\[
224 + 96 + 84 = 404
\]
Multiply by 2:
\[
S = 2 \cdot 404 = 808 \text{ in}^2
\]
Next, calculate the area of the roll of wrapping paper:
\[
\text{Area} = 25 \cdot 405 = 10125 \text{ in}^2
\]
Now, divide the total area of the wrapping paper by the surface area of one box:
\[
\text{Number of boxes} = \frac{10125}{808} \approx 12.5
\]
Since only complete boxes can be wrapped, we take the floor of this number:
\[
\text{Complete boxes} = 12
\]
### Summary of Answers:
1. **Surface area of rectangular prism**: 912 in²
2. **Surface area of cube**: 864 in²
3. **Surface area of rectangular prism (15 ft edges)**: 1350 ft²
4. **Surface area of cereal box**: 272 in²
5. **Complete boxes that can be wrapped**: 12
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)
Responses
10.5 cm2
10.5
cm
2
10.5 cm squared
63 cm2
63
cm
2
63 cm squared
71.5 cm2
71.5
cm
2
71.5 cm squared
35.75 cm2
squared
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
459 cm2
459
cm
2
459 cm squared
76.5 cm2
76.5
cm
2
76.5 cm squared
975.38 cm2
975.38
cm
2
975.38 cm squared
5,852.25 cm2
squared
3. Find the surface area of a textbook that is 2 in. Ă— 14 in. Ă— 9 in.
2
in.
Ă—
14
in.
Ă—
9
in.
(1 point)
Responses
50 in.2
50
in.
2
50 in. squared
172 in.2
172
in.
2
172 in. squared
344 in.2
344
in.
2
344 in. squared
252 in.2
squared
4. 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 B will require less wrapping paper because its surface area is 0.5 in.2
0.5
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
0.5
in.
2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 1.5 in.2
1.5
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
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
1.5
in.
2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 32.75 in.2
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
32.75
in.
2
less than the surface area of Book B.
5. 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?
2,960 cm2
2
,
960
cm
2
2,960 cm squared
1,080 cm
1,080 cm
67 cm2
67
cm
2
67 cm squared
1,880 cm2
squared
Responses
10.5 cm2
10.5
cm
2
10.5 cm squared
63 cm2
63
cm
2
63 cm squared
71.5 cm2
71.5
cm
2
71.5 cm squared
35.75 cm2
squared
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
459 cm2
459
cm
2
459 cm squared
76.5 cm2
76.5
cm
2
76.5 cm squared
975.38 cm2
975.38
cm
2
975.38 cm squared
5,852.25 cm2
squared
3. Find the surface area of a textbook that is 2 in. Ă— 14 in. Ă— 9 in.
2
in.
Ă—
14
in.
Ă—
9
in.
(1 point)
Responses
50 in.2
50
in.
2
50 in. squared
172 in.2
172
in.
2
172 in. squared
344 in.2
344
in.
2
344 in. squared
252 in.2
squared
4. 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 B will require less wrapping paper because its surface area is 0.5 in.2
0.5
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
0.5
in.
2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 1.5 in.2
1.5
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
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
1.5
in.
2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 32.75 in.2
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
32.75
in.
2
less than the surface area of Book B.
5. 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?
2,960 cm2
2
,
960
cm
2
2,960 cm squared
1,080 cm
1,080 cm
67 cm2
67
cm
2
67 cm squared
1,880 cm2
squared
GPT-4o mini
Let's calculate the surface area for each of the scenarios provided.
### 1. Surface area of a right rectangular prism with dimensions:
- Length = 4.5 cm
- Width = 3.5 cm
- Height = 2.5 cm
The formula for the surface area \( S \) of a rectangular prism is:
\[
S = 2(lw + lh + wh)
\]
Substituting the values:
\[
S = 2(4.5 \cdot 3.5 + 4.5 \cdot 2.5 + 3.5 \cdot 2.5)
\]
Calculating each term:
- \( 4.5 \cdot 3.5 = 15.75 \)
- \( 4.5 \cdot 2.5 = 11.25 \)
- \( 3.5 \cdot 2.5 = 8.75 \)
Adding these values together:
\[
15.75 + 11.25 + 8.75 = 35.75
\]
Now, multiply by 2:
\[
S = 2 \cdot 35.75 = 71.5 \text{ cm}^2
\]
**Answer:** 71.5 cm²
---
### 2. Surface area of a right rectangular prism where each edge length measures 12.75 cm
Since all edges are the same length, this is a cube, and we can use the same formula:
\[
S = 6s^2
\]
Here, \( s = 12.75 \) cm:
\[
S = 6 \cdot (12.75)^2 = 6 \cdot 162.5625 = 975.375 \text{ cm}^2
\]
**Answer:** 975.38 cm² (rounded to two decimal points)
---
### 3. Surface area of a textbook with dimensions:
- Length = 2 in.
- Width = 14 in.
- Height = 9 in.
Using the surface area formula:
\[
S = 2(lw + lh + wh)
\]
Substituting the dimensions:
\[
S = 2(2 \cdot 14 + 2 \cdot 9 + 14 \cdot 9)
\]
Calculating each part:
- \( 2 \cdot 14 = 28 \)
- \( 2 \cdot 9 = 18 \)
- \( 14 \cdot 9 = 126 \)
Adding these values:
\[
28 + 18 + 126 = 172
\]
Now multiply by 2:
\[
S = 2 \cdot 172 = 344 \text{ in}^2
\]
**Answer:** 344 in²
---
### 4. Surface area of two books
For Book A (6.5 in Ă— 1 inch Ă— 8.5 in):
Using the formula:
\[
S_A = 2(lw + lh + wh) = 2(6.5 \cdot 1 + 6.5 \cdot 8.5 + 1 \cdot 8.5)
\]
Calculating:
- \( 6.5 \cdot 1 = 6.5 \)
- \( 6.5 \cdot 8.5 = 55.25 \)
- \( 1 \cdot 8.5 = 8.5 \)
Adding:
\[
6.5 + 55.25 + 8.5 = 70.25
\]
Then multiply by 2:
\[
S_A = 2 \cdot 70.25 = 140.5 \text{ in}^2
\]
For Book B (5.5 in Ă— 2 in Ă— 8 in):
\[
S_B = 2(lw + lh + wh) = 2(5.5 \cdot 2 + 5.5 \cdot 8 + 2 \cdot 8)
\]
Calculating:
- \( 5.5 \cdot 2 = 11 \)
- \( 5.5 \cdot 8 = 44 \)
- \( 2 \cdot 8 = 16 \)
Adding:
\[
11 + 44 + 16 = 71
\]
Then multiply by 2:
\[
S_B = 2 \cdot 71 = 142 \text{ in}^2
\]
Comparing the two books:
- **Book A:** 140.5 in²
- **Book B:** 142 in²
Since 140.5 is less than 142, **Book A will require less wrapping paper.**
**Answer:** Book A will require less wrapping paper because its surface area is **1.5 in² less than the surface area of Book B.**
---
### 5. Surface area of a box that needs to be decorated (length = 27 cm, width = 20 cm, height = 20 cm)
For this box, since we are not decorating the bottom and back, we'll calculate the surface area that needs decoration.
The two top and two sides will be decorated, which gives us the area:
\[
S = lw + lh + wh
\]
Where:
- \( l = 27 \) cm
- \( w = 20 \) cm
- \( h = 20 \) cm
We calculate \( S \) for the tops and sides:
\[
S = (27 \cdot 20) + (27 \cdot 20) + (20 \cdot 20)
\]
Calculating each part:
- Top: \( 27 \cdot 20 = 540 \)
- Sides (front and back): \( 27 \cdot 20 = 540 \) (2 times)
- Two sides: \( 20 \cdot 20 = 400 \)
So the total surface area that needs decoration is:
\[
S = 540 + 540 + 400 = 1480 \text{ cm}^2
\]
However, since the bottom doesn't need decorating, the area of the bottom (27 cm × 20 cm = 540 cm²) is excluded.
Total area to decorate without the bottom:
\[
S = 1480 - 540 = 940 \text{ cm}^2
\]
Given the options, it appears there's a discrepancy in the provided choices. Let's calculate the surface area to verify:
\[
\text{Total area of all sides} = 2lw + 2lh + 2wh
\]
Calculating that gives:
\[
S = 2(27 \cdot 20 + 27 \cdot 20 + 20 \cdot 20) = 2(540 + 540 + 400) = 2(1480) = 2960 \text{ cm}^2
\]
Since the back also doesn’t need decorating either, you’ll use:
\[
S_{decorate} = 2 \cdot 540 + 400 = 1480 \text{ cm}^2
\]
The correct decorated measurement for the sides and top is **940 cm².**
The discrepancies could stem from a lack of clarity on decoration covered.
Given only the total, it would be **2960 cm²** based on the full count.
**Necessary conclusion**: **2,960 cm²** if full outer surface is included.
---
### Summary of Answers:
1. **71.5 cm²**
2. **975.38 cm²**
3. **344 in²**
4. **Book A will require less wrapping paper because its surface area is 1.5 in² less than the surface area of Book B.**
5. **2,960 cm²** (full decoration; clarification if sides may include as intended).
### 1. Surface area of a right rectangular prism with dimensions:
- Length = 4.5 cm
- Width = 3.5 cm
- Height = 2.5 cm
The formula for the surface area \( S \) of a rectangular prism is:
\[
S = 2(lw + lh + wh)
\]
Substituting the values:
\[
S = 2(4.5 \cdot 3.5 + 4.5 \cdot 2.5 + 3.5 \cdot 2.5)
\]
Calculating each term:
- \( 4.5 \cdot 3.5 = 15.75 \)
- \( 4.5 \cdot 2.5 = 11.25 \)
- \( 3.5 \cdot 2.5 = 8.75 \)
Adding these values together:
\[
15.75 + 11.25 + 8.75 = 35.75
\]
Now, multiply by 2:
\[
S = 2 \cdot 35.75 = 71.5 \text{ cm}^2
\]
**Answer:** 71.5 cm²
---
### 2. Surface area of a right rectangular prism where each edge length measures 12.75 cm
Since all edges are the same length, this is a cube, and we can use the same formula:
\[
S = 6s^2
\]
Here, \( s = 12.75 \) cm:
\[
S = 6 \cdot (12.75)^2 = 6 \cdot 162.5625 = 975.375 \text{ cm}^2
\]
**Answer:** 975.38 cm² (rounded to two decimal points)
---
### 3. Surface area of a textbook with dimensions:
- Length = 2 in.
- Width = 14 in.
- Height = 9 in.
Using the surface area formula:
\[
S = 2(lw + lh + wh)
\]
Substituting the dimensions:
\[
S = 2(2 \cdot 14 + 2 \cdot 9 + 14 \cdot 9)
\]
Calculating each part:
- \( 2 \cdot 14 = 28 \)
- \( 2 \cdot 9 = 18 \)
- \( 14 \cdot 9 = 126 \)
Adding these values:
\[
28 + 18 + 126 = 172
\]
Now multiply by 2:
\[
S = 2 \cdot 172 = 344 \text{ in}^2
\]
**Answer:** 344 in²
---
### 4. Surface area of two books
For Book A (6.5 in Ă— 1 inch Ă— 8.5 in):
Using the formula:
\[
S_A = 2(lw + lh + wh) = 2(6.5 \cdot 1 + 6.5 \cdot 8.5 + 1 \cdot 8.5)
\]
Calculating:
- \( 6.5 \cdot 1 = 6.5 \)
- \( 6.5 \cdot 8.5 = 55.25 \)
- \( 1 \cdot 8.5 = 8.5 \)
Adding:
\[
6.5 + 55.25 + 8.5 = 70.25
\]
Then multiply by 2:
\[
S_A = 2 \cdot 70.25 = 140.5 \text{ in}^2
\]
For Book B (5.5 in Ă— 2 in Ă— 8 in):
\[
S_B = 2(lw + lh + wh) = 2(5.5 \cdot 2 + 5.5 \cdot 8 + 2 \cdot 8)
\]
Calculating:
- \( 5.5 \cdot 2 = 11 \)
- \( 5.5 \cdot 8 = 44 \)
- \( 2 \cdot 8 = 16 \)
Adding:
\[
11 + 44 + 16 = 71
\]
Then multiply by 2:
\[
S_B = 2 \cdot 71 = 142 \text{ in}^2
\]
Comparing the two books:
- **Book A:** 140.5 in²
- **Book B:** 142 in²
Since 140.5 is less than 142, **Book A will require less wrapping paper.**
**Answer:** Book A will require less wrapping paper because its surface area is **1.5 in² less than the surface area of Book B.**
---
### 5. Surface area of a box that needs to be decorated (length = 27 cm, width = 20 cm, height = 20 cm)
For this box, since we are not decorating the bottom and back, we'll calculate the surface area that needs decoration.
The two top and two sides will be decorated, which gives us the area:
\[
S = lw + lh + wh
\]
Where:
- \( l = 27 \) cm
- \( w = 20 \) cm
- \( h = 20 \) cm
We calculate \( S \) for the tops and sides:
\[
S = (27 \cdot 20) + (27 \cdot 20) + (20 \cdot 20)
\]
Calculating each part:
- Top: \( 27 \cdot 20 = 540 \)
- Sides (front and back): \( 27 \cdot 20 = 540 \) (2 times)
- Two sides: \( 20 \cdot 20 = 400 \)
So the total surface area that needs decoration is:
\[
S = 540 + 540 + 400 = 1480 \text{ cm}^2
\]
However, since the bottom doesn't need decorating, the area of the bottom (27 cm × 20 cm = 540 cm²) is excluded.
Total area to decorate without the bottom:
\[
S = 1480 - 540 = 940 \text{ cm}^2
\]
Given the options, it appears there's a discrepancy in the provided choices. Let's calculate the surface area to verify:
\[
\text{Total area of all sides} = 2lw + 2lh + 2wh
\]
Calculating that gives:
\[
S = 2(27 \cdot 20 + 27 \cdot 20 + 20 \cdot 20) = 2(540 + 540 + 400) = 2(1480) = 2960 \text{ cm}^2
\]
Since the back also doesn’t need decorating either, you’ll use:
\[
S_{decorate} = 2 \cdot 540 + 400 = 1480 \text{ cm}^2
\]
The correct decorated measurement for the sides and top is **940 cm².**
The discrepancies could stem from a lack of clarity on decoration covered.
Given only the total, it would be **2960 cm²** based on the full count.
**Necessary conclusion**: **2,960 cm²** if full outer surface is included.
---
### Summary of Answers:
1. **71.5 cm²**
2. **975.38 cm²**
3. **344 in²**
4. **Book A will require less wrapping paper because its surface area is 1.5 in² less than the surface area of Book B.**
5. **2,960 cm²** (full decoration; clarification if sides may include as intended).
5. 1,880 cm^2
GPT-4o mini
Let’s clarify the problem for **question 5** regarding the surface area of the box that needs to be decorated.
Given:
- Length = 27 cm
- Width = 20 cm
- Height = 20 cm
### Surface Area Calculation
Typically, the formula for the surface area of a right rectangular prism is:
\[
S = 2(lw + lh + wh)
\]
In this case, since the bottom and the back of the box do not need to be decorated, we need to calculate only the areas of the sides and the top.
1. **Top Area** (1 topside):
\[
\text{Top} = lw = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2
\]
2. **Front Area** (1 front side):
\[
\text{Front} = lh = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2
\]
3. **Back Area** (not decorated, so we will exclude):
\[
\text{Back} = lh = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2 \quad \text{(excluded)}
\]
4. **Left Side Area** (1 left side):
\[
\text{Left} = wh = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2
\]
5. **Right Side Area** (1 right side):
\[
\text{Right} = wh = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2
\]
### Total Surface Area to be Decorated
Now, we sum up the areas that need decorating:
- Top Area = 540 cm²
- Front Area = 540 cm²
- Left Side Area = 400 cm²
- Right Side Area = 400 cm²
Now add these:
\[
S_{\text{decorated}} = 540 + 540 + 400 + 400 = 1880 \text{ cm}^2
\]
### Conclusion
Therefore, the **correct answer for the surface area that needs to be decorated is** **1,880 cm².**
Thank you for your response, and I hope this clears up any confusion!
Given:
- Length = 27 cm
- Width = 20 cm
- Height = 20 cm
### Surface Area Calculation
Typically, the formula for the surface area of a right rectangular prism is:
\[
S = 2(lw + lh + wh)
\]
In this case, since the bottom and the back of the box do not need to be decorated, we need to calculate only the areas of the sides and the top.
1. **Top Area** (1 topside):
\[
\text{Top} = lw = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2
\]
2. **Front Area** (1 front side):
\[
\text{Front} = lh = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2
\]
3. **Back Area** (not decorated, so we will exclude):
\[
\text{Back} = lh = 27 \text{ cm} \times 20 \text{ cm} = 540 \text{ cm}^2 \quad \text{(excluded)}
\]
4. **Left Side Area** (1 left side):
\[
\text{Left} = wh = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2
\]
5. **Right Side Area** (1 right side):
\[
\text{Right} = wh = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2
\]
### Total Surface Area to be Decorated
Now, we sum up the areas that need decorating:
- Top Area = 540 cm²
- Front Area = 540 cm²
- Left Side Area = 400 cm²
- Right Side Area = 400 cm²
Now add these:
\[
S_{\text{decorated}} = 540 + 540 + 400 + 400 = 1880 \text{ cm}^2
\]
### Conclusion
Therefore, the **correct answer for the surface area that needs to be decorated is** **1,880 cm².**
Thank you for your response, and I hope this clears up any confusion!