Question
The length of a rectangle is four times the width. The perimeter of the rectangle is 45 inches. Write a system of equations that represents this problem. What is the area of the rectangle?
Answers
Let's call the width of the rectangle "w" (in inches). According to the problem, the length of the rectangle is four times the width, so we can call the length "4w" (in inches).
The formula for the perimeter of a rectangle is P=2L+2W, where P is the perimeter, L is the length, and W is the width.
We know that the perimeter of this rectangle is 45 inches, so we can set up an equation:
45 = 2(4w) + 2w
Simplifying this equation:
45 = 8w + 2w
45 = 10w
w = 4.5
So the width of the rectangle is 4.5 inches. To find the length, we can use the fact that it is four times the width:
L = 4w = 4(4.5) = 18
So the length of the rectangle is 18 inches.
To find the area, we use the formula A=LW:
A = (4.5)(18) = 81
So the area of the rectangle is 81 square inches.
The formula for the perimeter of a rectangle is P=2L+2W, where P is the perimeter, L is the length, and W is the width.
We know that the perimeter of this rectangle is 45 inches, so we can set up an equation:
45 = 2(4w) + 2w
Simplifying this equation:
45 = 8w + 2w
45 = 10w
w = 4.5
So the width of the rectangle is 4.5 inches. To find the length, we can use the fact that it is four times the width:
L = 4w = 4(4.5) = 18
So the length of the rectangle is 18 inches.
To find the area, we use the formula A=LW:
A = (4.5)(18) = 81
So the area of the rectangle is 81 square inches.
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