Final answer:
To find the dimensions of the rectangle, a system of equations is created: 2w + 2l = 38 and l = 2w + 4. By substituting the second equation into the first and solving, the width w is found to be 5 cm and the length l is 14 cm.
Step-by-step explanation:
The student is asked to find the dimensions of a rectangle with a specific area and with the length being 4 more than twice its width. To find these dimensions, we will set up and solve a system of equations.
Let's denote the width of the rectangle as w and the length as l. The problem gives us two key pieces of information:
- The perimeter of the rectangle is 38 cm, which gives us the equation: 2w + 2l = 38.
- The length is 4 more than twice the width, giving us another equation: l = 2w + 4.
We can solve this system of equations by substitution or elimination. Here, substitution is more straightforward. Substituting the expression for l from the second equation into the first gives us:
2w + 2(2w + 4) = 38
Simplifying, we have:
2w + 4w + 8 = 38
6w + 8 = 38
Subtracting 8 from both sides, we get:
6w = 30
Dividing both sides by 6, we find that:
w = 5 cm
Substituting w back into the second equation, we get:
l = 2(5) + 4 = 14 cm
Thus, the dimensions of the rectangle are:
Length = 14 cm
Width = 5 cm