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.

1 answer

To determine which book requires less wrapping paper, we need to calculate the surface area of both books.

The formula for the surface area \( S \) of a rectangular box (or book, in this case) is: \[ S = 2(lw + lh + wh) \] where \( l \) is length, \( w \) is width, and \( h \) is height.

For Book A:

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

Calculating the surface area: \[ S_A = 2(6.5 \times 1 + 6.5 \times 8.5 + 1 \times 8.5) \] Calculating each term:

  • \( 6.5 \times 1 = 6.5 \)
  • \( 6.5 \times 8.5 = 55.25 \)
  • \( 1 \times 8.5 = 8.5 \)

Now, summing these: \[ 6.5 + 55.25 + 8.5 = 70.25 \] Then, multiplying by 2: \[ S_A = 2 \times 70.25 = 140.5 \text{ in}^2 \]

For Book B:

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

Calculating the surface area: \[ S_B = 2(5.5 \times 2 + 5.5 \times 8 + 2 \times 8) \] Calculating each term:

  • \( 5.5 \times 2 = 11 \)
  • \( 5.5 \times 8 = 44 \)
  • \( 2 \times 8 = 16 \)

Now, summing these: \[ 11 + 44 + 16 = 71 \] Then, multiplying by 2: \[ S_B = 2 \times 71 = 142 \text{ in}^2 \]

Comparison:

  • Surface area of Book A: \( 140.5 \text{ in}^2 \)
  • Surface area of Book B: \( 142 \text{ in}^2 \)

Now, finding the difference in surface area: \[ 142 - 140.5 = 1.5 \text{ in}^2 \]

Thus, Book A has a smaller surface area by \(1.5 \text{ in}^2\).

Based on this calculation, 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.