Sue is wrapping books to give as gifts a book has a length of 6.5 inches width of 1 inch and a height of 8.5 inches book B has a length of 5.5 inches of width of 2 inches and height of 8 inches based on surface area, which book will require less wrapping paper
To calculate the surface area of a rectangular cuboid (which represents the shape of the books), you need to find the sum of the areas of all its six faces.
For book A:
- The surface area of the first face (length x width) is 6.5 inches x 1 inch = 6.5 square inches.
- The surface area of the second face (length x width) is also 6.5 square inches.
- The surface area of the third face (width x height) is 1 inch x 8.5 inches = 8.5 square inches.
- The surface area of the fourth face (width x height) is also 8.5 square inches.
- The surface area of the fifth face (length x height) is 6.5 inches x 8.5 inches = 55.25 square inches.
- The surface area of the sixth face (length x height) is also 55.25 square inches.
Thus, the total surface area of book A is 6.5 + 6.5 + 8.5 + 8.5 + 55.25 + 55.25 = 140 inches.
For book B:
- The surface area of the first face (length x width) is 5.5 inches x 2 inches = 11 square inches.
- The surface area of the second face (length x width) is also 11 square inches.
- The surface area of the third face (width x height) is 2 inches x 8 inches = 16 square inches.
- The surface area of the fourth face (width x height) is also 16 square inches.
- The surface area of the fifth face (length x height) is 5.5 inches x 8 inches = 44 square inches.
- The surface area of the sixth face (length x height) is also 44 square inches.
So the total surface area of book B is 11 + 11 + 16 + 16 + 44 + 44 = 142 inches.
Therefore, based on surface area, book A requires less wrapping paper than book B.