The perimeter of a rectangle is 52 feet. The length is 2 feet shorter than three times the width. What is the area of the rectangle in square feet?
width --- x
length ---- 3x - 2
solve for x ...
2x + 2(3x-2) = 52
To find the area of the rectangle, we first need to solve for the length and width of the rectangle.
Let's suppose the width of the rectangle is "w" feet.
According to the problem, the length is 2 feet shorter than three times the width, which can be expressed as: (3w - 2).
Now, let's find the equation for the perimeter of the rectangle.
Perimeter of a rectangle = 2 * (Length + Width)
In this case, the perimeter of the rectangle is given as 52 feet.
So, we have: 2 * ((3w - 2) + w) = 52
Simplifying the equation further, we get: 2 * (4w - 2) = 52
Now, we can solve for "w".
Dividing both sides of the equation by 2, we have: 4w - 2 = 26
Adding 2 to both sides of the equation, we get: 4w = 28
Dividing both sides of the equation by 4, we have: w = 7
So, the width of the rectangle is 7 feet.
Now, let's find the length of the rectangle.
The length of the rectangle is (3w - 2), so substituting the value of "w" as 7, we get: 3 * 7 - 2 = 19
So, the length of the rectangle is 19 feet.
Finally, to find the area of the rectangle, we use the formula:
Area = Length * Width
Substituting the values, we get: Area = 19 * 7
Simplifying the equation, we have: Area = 133
Therefore, the area of the rectangle is 133 square feet.