can someone please help me.
The perimeter of a rectangle is 80m. The length is 7m more than twice and width. Find the dimensions.
I would appreciate help with the steps so I can practice it for the my next problem. Thanks.
Let l be the length.
Let w be the width.
"The length is 7m more than twice and width."
l = 2w+7
The perimeter of a rectangle is P=2(l+w).
Substitute in l=2w+7 for the length.
80 = 2(2w+7+w)
80 = 2(3w+7)
80 = 6w + 14
66 = 6w
w = 11
l = 2w+7
l = 29
i dont know the answer
Of course, I'd be happy to help you. Let's break down the problem step by step.
Step 1: Understand the problem
We are given that the perimeter of a rectangle is 80m. We need to find the dimensions of the rectangle given that the length is 7m more than twice the width.
Step 2: Define the variables
Let's assign variables to the dimensions of the rectangle. Let's say the width is "w" and the length is "l".
Step 3: Set up the equations
The perimeter of a rectangle is given by the formula: Perimeter = 2(length + width).
We know that the perimeter of the rectangle is 80m, so we can set up the equation:
80 = 2(l + w).
We are also given that the length is 7m more than twice the width, so we can set up another equation:
l = 2w + 7.
Step 4: Solve the equations
Now, we have two equations with two variables. We can use these equations to solve for the dimensions of the rectangle.
Using the second equation, we can substitute l in the first equation:
80 = 2((2w + 7) + w).
Simplifying the equation:
80 = 2(3w + 7).
Distributing the 2:
80 = 6w + 14.
Subtracting 14 from both sides:
66 = 6w.
Dividing both sides by 6:
11 = w.
Now that we know the width, we can substitute it back into the second equation to find the length:
l = 2w + 7.
Substituting w = 11 into the equation:
l = 2(11) + 7.
Calculating:
l = 22 + 7,
l = 29.
Step 5: Check the solution
To confirm that our solution is correct, we can substitute the width and length values into the equation for the perimeter:
Perimeter = 2(length + width),
Perimeter = 2(29 + 11),
Perimeter = 2(40),
Perimeter = 80.
The perimeter matches the given value, so our solution is correct.
Therefore, the dimensions of the rectangle are width = 11m and length = 29m.