the perimeter of a rectangle is 112 inches. The length exceeds the width by 50 inches and width.
please someone help it is due tonight I have tried all I know how to figure it out
Let w=width
Length=w+50
Permeter=2(length+width)
=2(w+50+w)
Solve for w in
2(w+50+w)=112
To find the perimeter of a rectangle, you need to know both the length and the width. However, in this question, only the relationship between the length and width is given, as well as the total perimeter. We'll need to use this information to find the dimensions and calculate the perimeter.
Let's use algebra to solve the problem step by step:
Let's assume the width of the rectangle is "x" inches.
Since the length exceeds the width by 50 inches, the length can be represented as "x + 50".
The formula for the perimeter of the rectangle is P = 2(length + width).
So, in this case, the perimeter (P) is 112 inches. Substituting the values we have:
112 = 2(x + x + 50)
Let's simplify the equation:
112 = 2(2x + 50)
112 = 4x + 100
4x = 112 - 100
4x = 12
Dividing both sides of the equation by 4:
x = 12/4
x = 3
Therefore, the width of the rectangle is 3 inches.
Since the length is given as x + 50:
Length = 3 + 50 = 53 inches.
To calculate the perimeter, we can use the formula:
Perimeter = 2(length + width)
Perimeter = 2(53 + 3)
Perimeter = 2(56)
Perimeter = 112 inches.
So, the dimensions of the rectangle are width = 3 inches and length = 53 inches, and the perimeter is 112 inches.