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)

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 0.5 in. 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.

1 answer

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

The formula for the surface area \( S \) of a rectangular box is:

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

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

For Book A:

  • Length \( l_A = 6.5 \) inches
  • Width \( w_A = 1 \) inch
  • Height \( h_A = 8.5 \) inches

Calculating the surface area:

\[ S_A = 2(6.5)(1) + 2(6.5)(8.5) + 2(1)(8.5) \]

Calculating each term:

  1. \( 2(6.5)(1) = 13 \)
  2. \( 2(6.5)(8.5) = 110.5 \)
  3. \( 2(1)(8.5) = 17 \)

Now, summing these values:

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

For Book B:

  • Length \( l_B = 5.5 \) inches
  • Width \( w_B = 2 \) inches
  • Height \( h_B = 8 \) inches

Calculating the surface area:

\[ S_B = 2(5.5)(2) + 2(5.5)(8) + 2(2)(8) \]

Calculating each term:

  1. \( 2(5.5)(2) = 22 \)
  2. \( 2(5.5)(8) = 88 \)
  3. \( 2(2)(8) = 32 \)

Now, summing these values:

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

Comparison:

  • Surface area of Book A: \( 140.5 \) in²
  • Surface area of Book B: \( 142 \) in²

Since \( 140.5 < 142 \), Book A will require less wrapping paper.

Thus, the correct statement is:

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