Question

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 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. 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 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. Skip to navigation

Answer 1

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 2,960 cm2 2,960 cm squared 67 cm2 67 cm squared 1,080 cm 1,080 cm 1,880 cm2 1,880 cm squared Skip to navigation

Answer 2

To determine which book will require less wrapping paper, we need to calculate the surface area of both Book A and Book B.

The formula for the surface area \( SA \) of a rectangular prism is:

\[
SA = 2lw + 2lh + 2wh
\]

where:
- \( l \) = length
- \( w \) = width
- \( h \) = height

### Calculation for Book A:
- Length = 6.5 inches
- Width = 1 inch
- Height = 8.5 inches

\[
SA_A = 2(6.5 \times 1) + 2(6.5 \times 8.5) + 2(1 \times 8.5)
\]

Calculating each term:
- \( 2(6.5 \times 1) = 2(6.5) = 13 \)
- \( 2(6.5 \times 8.5) = 2(55.25) = 110.5 \)
- \( 2(1 \times 8.5) = 2(8.5) = 17 \)

Now sum these values:

\[
SA_A = 13 + 110.5 + 17 = 140.5 \text{ in}^2
\]

### Calculation for Book B:
- Length = 5.5 inches
- Width = 2 inches
- Height = 8 inches

\[
SA_B = 2(5.5 \times 2) + 2(5.5 \times 8) + 2(2 \times 8)
\]

Calculating each term:
- \( 2(5.5 \times 2) = 2(11) = 22 \)
- \( 2(5.5 \times 8) = 2(44) = 88 \)
- \( 2(2 \times 8) = 2(16) = 32 \)

Now sum these values:

\[
SA_B = 22 + 88 + 32 = 142 \text{ in}^2
\]

### Comparing Surface Areas:
- Surface area of Book A: \( 140.5 \text{ in}^2 \)
- Surface area of Book B: \( 142 \text{ in}^2 \)

Now, we subtract the surface areas:

\[
SA_B - SA_A = 142 - 140.5 = 1.5 \text{ in}^2
\]

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

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