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)

Question

Responses

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

Answer 1

To determine which book will require less wrapping paper, we need to calculate the surface area of each book.

The surface area \( S \) of a rectangular prism (which a book can be approximated as) can be calculated using the formula: \[ S = 2(lw + lh + wh) \] where \( l \) = length, \( w \) = width, and \( h \) = height.

For Book A:

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

Calculating the surface area of Book A: \[ S_A = 2(6.5 \times 1 + 6.5 \times 8.5 + 1 \times 8.5) \]

Calculating each term:

1. \( 6.5 \times 1 = 6.5 \)
2. \( 6.5 \times 8.5 = 55.25 \)
3. \( 1 \times 8.5 = 8.5 \)

Adding these results: \[ S_A = 2(6.5 + 55.25 + 8.5) = 2(70.25) = 140.5 \text{ in}^2 \]

For Book B:

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

Calculating the surface area of Book B: \[ S_B = 2(5.5 \times 2 + 5.5 \times 8 + 2 \times 8) \]

Calculating each term:

1. \( 5.5 \times 2 = 11 \)
2. \( 5.5 \times 8 = 44 \)
3. \( 2 \times 8 = 16 \)

Adding these results: \[ S_B = 2(11 + 44 + 16) = 2(71) = 142 \text{ in}^2 \]

Now we can compare the surface areas:

• Surface area of Book A = 140.5 in²
• Surface area of Book B = 142 in²

Since 140.5 in² is less than 142 in², Book A will require less wrapping paper.

The correct response is: Book A will require less wrapping paper because its surface area is 1.5 in² less than the surface area of Book B.