Sue is wrapping books to give as gifts book A has a length of 6.5 inches a width if 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
To find the surface area of a rectangular object, we need to calculate the area of each of its faces and then add them all together.
For book A:
- The area of the front and back faces is length x height = 6.5 inches x 8.5 inches = 55.25 square inches.
- The area of the top and bottom faces is width x length = 1 inch x 6.5 inches = 6.5 square inches.
- The area of the two side faces is height x width = 8.5 inches x 1 inch = 8.5 square inches.
The total surface area of book A is 2 * (55.25 square inches) + 2 * (6.5 square inches) + 2 * (8.5 square inches) = 110.5 + 13 + 17 = 140.5 square inches.
For book B:
- The area of the front and back faces is length x height = 5.5 inches x 8 inches = 44 square inches.
- The area of the top and bottom faces is width x length = 2 inches x 5.5 inches = 11 square inches.
- The area of the two side faces is height x width = 8 inches x 2 inches = 16 square inches.
The total surface area of book B is 2 * (44 square inches) + 2 * (11 square inches) + 2 * (16 square inches) = 88 + 22 + 32 = 142 square inches.
Comparing the surface areas, we can see that book A requires less wrapping paper as it has a surface area of 140.5 square inches compared to book B's surface area of 142 square inches.