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.

1 answer

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.