A rectangle is 3 times as long as it is wide and has the same perimeter as a square whose are is 4 sp.ft. larger than that of the rectangle. What are the dimensions of both the rectangle and the square?

rectangle:

width = x
length = 3x
perimeter = 8x
area = 3x^2

square:
side y
perimeter = 4y
area = y^2

perimeters are the same --> 4y = 8x
y = 2x

area of square is 4 more than area of rectangle

y^2 = 3x^2 + 4
4x^2 = 3x^2 + 4
x^2 = 4
x = 2, then y = 4

rectangle:
width = 2
length = 6
perimeter = 16
area = 12

square:
side = 4
perimeter = 16
area = 16 , which is 4 more than that of the rectangle

everything checks!

To solve this problem, let's break it down step by step:

Step 1: Define the variables
Let's define the width of the rectangle as 'w' and the length as 'l.' We need to find the values of 'w' and 'l' for the rectangle, as well as the side length of the square.

Step 2: Perimeter of the rectangle
The perimeter of a rectangle can be calculated by adding all four sides. The formula for the perimeter of a rectangle is: P = 2(l + w)

Given that the rectangle's perimeter is equal to the square's perimeter (which is 4 times the length of one side), we can write the equation:
2(l + w) = 4s, where 's' is the side length of the square.

Step 3: Relationship between the length and width of the rectangle
The problem states that the rectangle's length is 3 times its width. We can write this relationship as: l = 3w.

Step 4: Area of the square
The area of a square is given by the formula: A = s^2.
Given that the square's area is 4 sq.ft greater than the rectangle's area, we can write the equation:
s^2 = l * w + 4

Step 5: Substitute values and solve the system of equations
Now, let's substitute the value of 'l' from the second equation into the first equation:
2(3w + w) = 4s
Simplifying the equation further:
8w = 4s
We can divide both sides of the equation by 4 to simplify it even more:
2w = s

Next, substitute the value of 'l' from the second equation into the third equation:
s^2 = 3w * w + 4
Simplifying the equation further:
s^2 = 3w^2 + 4

Step 6: Solve the quadratic equation
To solve this quadratic equation, we need to set it equal to zero.
s^2 - 3w^2 - 4 = 0

Now we need to factorize this quadratic equation:
(s + 2)(s - 2) - 3w^2 = 0

Step 7: Substitute values and solve for 'w'
From the relationship we found in Step 5, we know that 2w = s, so we can substitute this into the equation:
(2w + 2)(2w - 2) - 3w^2 = 0
(4w^2 - 4) - 3w^2 = 0
w^2 - 4 = 0

Now solve for 'w':
w^2 = 4
w = ±√4
w = ±2

Since the width of a rectangle cannot be negative, we take w = 2.

Step 8: Calculate the length of the rectangle and the side length of the square
Using the relationship l = 3w, we can calculate the length of the rectangle:
l = 3 * 2 = 6

We know that the side length of the square is two times the width of the rectangle:
s = 2 * w = 2 * 2 = 4

So, the dimensions of the rectangle are width = 2 ft and length = 6 ft, while the side length of the square is 4 ft.