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 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.

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 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 0.5 in.2
0.5

in.
2
less than the surface area of Book A.

Answer 1

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

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

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

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

**Calculating the surface area of Book A:**
- Length \( l = 6.5 \) inches
- Width \( w = 1 \) inch
- Height \( h = 8.5 \) inches

\[
S_A = 2(lw + lh + wh) = 2(6.5 \cdot 1 + 6.5 \cdot 8.5 + 1 \cdot 8.5)
\]

Calculating each term:
- \( lw = 6.5 \times 1 = 6.5 \)
- \( lh = 6.5 \times 8.5 = 55.25 \)
- \( wh = 1 \times 8.5 = 8.5 \)

Now summing these values:
\[
S_A = 2(6.5 + 55.25 + 8.5) = 2(70.25) = 140.5 \, \text{in}^2
\]

**Calculating the surface area of Book B:**
- Length \( l = 5.5 \) inches
- Width \( w = 2 \) inches
- Height \( h = 8 \) inches

\[
S_B = 2(lw + lh + wh) = 2(5.5 \cdot 2 + 5.5 \cdot 8 + 2 \cdot 8)
\]

Calculating each term:
- \( lw = 5.5 \times 2 = 11 \)
- \( lh = 5.5 \times 8 = 44 \)
- \( wh = 2 \times 8 = 16 \)

Now summing these values:
\[
S_B = 2(11 + 44 + 16) = 2(71) = 142 \, \text{in}^2
\]

**Comparing the surface areas:**
- Surface area of Book A: \( 140.5 \, \text{in}^2 \)
- Surface area of Book B: \( 142 \, \text{in}^2 \)

Since \( 140.5 \, \text{in}^2 < 142 \, \text{in}^2 \), Book A has a smaller surface area and thus will require less wrapping paper.

The correct response is:

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