The width of a rectangle is 12 units less than its length. If you add 30 units to both length and width, you double the perimeter. Find the length and with of the original rectangle.
W+12=L
2(L+30)+ 2(W+30)=2(2L+2W)
surely you can do it from here. I will be happy to check your work.
Let's assume the original length of the rectangle is L units.
According to the given information, the width of the rectangle is 12 units less than its length. Therefore, the original width can be expressed as (L - 12) units.
When you add 30 units to both the length and width, the new length and width become (L + 30) units and ((L - 12) + 30) units, respectively.
According to the given condition, when we double the perimeter, it means that the new perimeter is twice the original perimeter.
The formula for the perimeter of a rectangle is P = 2(L + W).
Let's calculate the original perimeter of the rectangle:
Original Perimeter = 2(L + (L - 12)) = 2(2L - 12) = 4L - 24
Now, let's calculate the new perimeter after adding 30 units to both length and width:
New Perimeter = 2((L + 30) + ((L - 12) + 30)) = 2(2L + 48) = 4L + 96
Since the new perimeter is twice the original perimeter, we can set up the following equation:
4L + 96 = 2(4L - 24)
Now, let's solve this equation to find the value of L:
4L + 96 = 8L - 48
4L - 8L = -48 - 96
-4L = -144
L = -144 / -4
L = 36
Therefore, the original length of the rectangle is 36 units.
Substituting this value back into the equation for the width:
Width = L - 12 = 36 - 12 = 24
Hence, the original length of the rectangle is 36 units, and the original width is 24 units.
To solve this problem, we can start by assigning variables to the length and width of the original rectangle. Let's say the length is L and the width is W.
We are given that the width of the rectangle is 12 units less than its length. This can be expressed as W = L - 12.
Next, we are told that if we add 30 units to both the length and width, we double the perimeter. The perimeter of a rectangle is given by the formula P = 2(L + W). So, if we add 30 units to both L and W, the new perimeter becomes 2(P + 30).
Now, let's put this information together:
1. W = L - 12
2. 2(P + 30) = 2(L + 30 + W + 30)
First, let's simplify equation 2:
2(P + 30) = 2(L + W + 60)
P + 30 = L + W + 60
P = L + W + 30
Since we know that the perimeter is twice its initial value, we can write:
P = 2(L + W)
L + W + 30 = 2(L + W)
Now, let's substitute W = L - 12 into the equation and solve for L:
L + (L - 12) + 30 = 2L + 2(L - 12)
2L + 18 = 4L - 24
2L - 4L = -24 - 18
-2L = -42
L = 21
Now that we have the length, we can substitute it back into the equation W = L - 12:
W = 21 - 12
W = 9
Therefore, the length of the original rectangle is 21 units and the width is 9 units.