To determine which book requires less wrapping paper, we need to calculate the surface area of each book. The surface area \( SA \) of a rectangular box can be calculated using the formula:
\[ SA = 2(lw + lh + wh) \]
where \( l \) is the length, \( w \) is the width, and \( h \) is the height.
Let's begin with Book A:
Book A dimensions:
- Length (l) = 6.5 inches
- Width (w) = 4 inches
- Height (h) = 8.5 inches
Substituting these values into the surface area formula:
\[ SA_A = 2(6.5 \times 4 + 6.5 \times 8.5 + 4 \times 8.5) \]
Calculating each product:
- \( 6.5 \times 4 = 26 \)
- \( 6.5 \times 8.5 = 55.25 \)
- \( 4 \times 8.5 = 34 \)
Now, summing these:
\[ SA_A = 2(26 + 55.25 + 34) = 2(115.25) = 230.5 \text{ square inches} \]
Now, let's calculate the surface area for Book B.
Book B dimensions:
- Length (l) = unspecified (let's denote it as \( x \))
- Width (w) = 2 inches
- Height (h) = 8 inches
Substituting these values into the surface area formula:
\[ SA_B = 2(x \times 2 + x \times 8 + 2 \times 8) \]
Simplifying:
- \( x \times 2 = 2x \)
- \( x \times 8 = 8x \)
- \( 2 \times 8 = 16 \)
So we have:
\[ SA_B = 2(2x + 8x + 16) = 2(10x + 16) = 20x + 32 \text{ square inches} \]
To compare the surface areas of the two books, we need the length \( x \) of Book B. However, since \( SA_A = 230.5 \) square inches is already defined, we could find the value of \( x \) that would make \( SA_B < SA_A \):
To find when \( 20x + 32 < 230.5 \):
\[ 20x < 230.5 - 32 \] \[ 20x < 198.5 \] \[ x < \frac{198.5}{20} \] \[ x < 9.925 \]
So as long as the length of Book B is less than approximately \( 9.93 \) inches, Book B will require less wrapping paper than Book A.
In conclusion, without the specific length of Book B, we cannot definitively determine which book requires less wrapping paper unless the length is provided. If the length of Book B is less than approximately 9.93 inches, then Book B will require less wrapping paper; otherwise, Book A will require less.
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