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 is: \[ S = 2(lw + lh + wh) \] where \( l \) is the length, \( w \) is the width, and \( h \) is the height.
For Book A:
- Length \( l = 6.5 \) inches
- Width \( w = 1 \) inch
- Height \( h = 8.5 \) inches
Calculating the surface area: \[ S_A = 2(lw + lh + wh) = 2(6.5 \times 1 + 6.5 \times 8.5 + 1 \times 8.5) \] 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, substituting back: \[ S_A = 2(6.5 + 55.25 + 8.5) = 2(70.25) = 140.5 \text{ in}^2 \]
For Book B:
- Length \( l = 5.5 \) inches
- Width \( w = 2 \) inches
- Height \( h = 8 \) inches
Calculating the surface area: \[ S_B = 2(lw + lh + wh) = 2(5.5 \times 2 + 5.5 \times 8 + 2 \times 8) \] Calculating each term:
- \( lw = 5.5 \times 2 = 11 \)
- \( lh = 5.5 \times 8 = 44 \)
- \( wh = 2 \times 8 = 16 \)
Now, substituting back: \[ S_B = 2(11 + 44 + 16) = 2(71) = 142 \text{ in}^2 \]
Conclusion
Now we compare the two surface areas:
- Surface area of Book A: \( S_A = 140.5 \text{ in}^2 \)
- Surface area of Book B: \( S_B = 142 \text{ in}^2 \)
Since \( S_A < S_B \):
- Book A requires less wrapping paper.
So, the correct option is: 4. Book A will require less wrapping paper because its surface area is 32.75 in.² less than the surface area of Book B.
(Note: The reasoning in Option 4 is incorrect about the difference; it should be \( 142 - 140.5 = 1.5 \) rather than 32.75, but the conclusion regarding Book A needing less wrapping paper is correct.)
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