the lenght of a rectangular room is 6 feet longer than twice the width. If the room's perimeter is 132 feet, what are the room's dimensions?
P = 2L + 2W
132 = 2(2W + 6) + 2W
132 = 6W + 12
120 = 6W
20 = W
To find the dimensions, we need to set up equations based on the given information.
Let's assume the width of the room as "x" feet.
According to the problem, the length of the room is 6 feet longer than twice the width. So, the length can be expressed as 2x + 6.
The perimeter of a rectangle is given by the formula: Perimeter = 2(length + width).
In this case, the perimeter is 132 feet, so we can write the equation:
132 = 2(2x + 6 + x)
Now we can solve this equation to find the value of x.
132 = 2(3x + 6)
132 = 6x + 12
120 = 6x
x = 20
Now that we have the value of x, we can find the length by substituting it back into our equation for the length:
length = 2x + 6
length = 2(20) + 6
length = 40 + 6
length = 46
So, the width of the room is 20 feet, and the length is 46 feet.
Let's represent the width of the room as x feet.
According to the given information, the length of the room is 6 feet longer than twice the width, which can be expressed as 2x + 6 feet.
The perimeter of a rectangle is calculated by adding the lengths of all four sides. For this rectangular room, the perimeter is given as 132 feet.
Perimeter = 2 * (length + width)
Therefore, we can write the equation as:
132 = 2 * (2x + 6 + x)
Simplifying the equation:
132 = 2 * (3x + 6)
Dividing both sides by 2 to solve for x:
66 = 3x + 6
Subtracting 6 from both sides:
60 = 3x
Dividing both sides by 3:
x = 20
So, the width of the room is 20 feet.
Now, we can substitute this value into the expression for the length:
Length = 2x + 6
Length = 2 * 20 + 6
Length = 40 + 6
Length = 46
Thus, the length of the room is 46 feet.
In conclusion, the room has dimensions of 20 feet width and 46 feet length.