Question

Surface Area of Right Rectangular Prisms Quick Check
4 of 54 of 5 Items
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 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 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 32.75 in.2
less than the surface area of Book B.

Answers

GPT-4o mini
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 (SA) of a right rectangular prism is given by:

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

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

**Calculating the surface area of Book A:**

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

\[
SA_A = 2(6.5 \cdot 1) + 2(6.5 \cdot 8.5) + 2(1 \cdot 8.5)
\]

Calculating each term:

1. \( 2(6.5 \cdot 1) = 13 \)
2. \( 2(6.5 \cdot 8.5) = 2(55.25) = 110.5 \)
3. \( 2(1 \cdot 8.5) = 17 \)

Adding these together:
\[
SA_A = 13 + 110.5 + 17 = 140.5 \text{ in}^2
\]

**Calculating the surface area of Book B:**

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

\[
SA_B = 2(5.5 \cdot 2) + 2(5.5 \cdot 8) + 2(2 \cdot 8)
\]

Calculating each term:

1. \( 2(5.5 \cdot 2) = 2(11) = 22 \)
2. \( 2(5.5 \cdot 8) = 2(44) = 88 \)
3. \( 2(2 \cdot 8) = 2(16) = 32 \)

Adding these together:
\[
SA_B = 22 + 88 + 32 = 142 \text{ in}^2
\]

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

Book A has a surface area of \( 140.5 \text{ in}^2 \) and Book B has a surface area of \( 142 \text{ in}^2 \).

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

So 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.**

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