“A pre-algebra book is 8 1 half Click for more options
inches by 9 1 fourth Click for more options inches. Determine the perimeter of the textbook.
Describe the steps you took to solve this problem.
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A pre-algebra book is 8 1 inches by 9 1 (inches). Determine the perimeter of the textbook.
P = length times width.
I'll be glad to check your answer.
Book is 8 1/2 inches and 9 1/4
The perimeter of the textbook?
Area is Length x width
Perimeter is Length + Length + width + width
or 2*L + 2*w
Oops -- I goofed. Ms. Pi is right.
To determine the perimeter of the textbook, we need to find the sum of all four sides of the rectangle.
Step 1: Convert mixed numbers to improper fractions.
The dimensions of the textbook are 8 1/2 inches by 9 1/4 inches. To make calculations easier, we need to convert these mixed numbers to improper fractions.
8 1/2 inches = (8 * 2 + 1)/2 inches = 17/2 inches
9 1/4 inches = (9 * 4 + 1)/4 inches = 37/4 inches
So, the dimensions of the textbook in improper fractions are 17/2 inches by 37/4 inches.
Step 2: Find the sum of all four sides.
To find the perimeter, we add up the lengths of all four sides. Since opposite sides of a rectangle are equal, we can add the lengths of two adjacent sides and then double the result to get the total perimeter.
Perimeter = 2 * (Length + Width)
In this case, the length is 17/2 inches and the width is 37/4 inches.
Perimeter = 2 * (17/2 inches + 37/4 inches)
Step 3: Simplify and calculate.
Now, let's simplify the expression and calculate the final answer.
Perimeter = 2 * (17/2 inches + 37/4 inches)
= 2 * (17/2 inches + (37/4 inches * 2/2))
= 2 * (17/2 inches + 74/4 inches)
= 2 * (17 inches/2 + 74 inches/4)
= 2 * (34 inches/4 + 74 inches/4)
= 2 * (108 inches/4)
= 2 * 27 inches
= 54 inches
Therefore, the perimeter of the pre-algebra textbook is 54 inches.