rectangle problemThe length of a rectangle is 3 in. more than twice its width. If the perimeter of the rectangle is 18 in., what are the width of the rectangle?
First, we set up two equations based on the information you have given.
L = 2w + 3
2L + 2W = 18
Now, we use the value for L in the first equation and substitute it into the second equation.
2 (2W + 3) + 2W = 18
4W + 6 + 2W = 18
6W = 12
W = 2
Now we use this value for W to determine L.
L = 2*2 + 3
L = 4 + 3
L = 7
So, the length is 7 inches and the width is 2 inches.
The length of a rectangle is 3 inch more than twice its width.
Let the width of the rectangle be W. The length of the rectangle is 3 + 2*W
For a rectangle with sides W and L the perimeter is equal to 2*(W + L)
Substituting the values in the formula, the perimeter is:
2*(W + 3 + 2*W). This is equal to 18
2*(W + 3 + 2*W) = 18
W + 3 + 2*W = 9
3W + 3 = 9
W = 2
The width of the rectangle is 2 inches.
We'll put the length of the rectangle to be a inches and the width be b inches.
We know, from enunciation, that the length is 3 inches more than twice its width and we'll write the constraint mathematically:
a - 3 = 2b
We'll subtract 2b and add 3 both sides:
a - 2b = 3 (1)
The perimeter of the rectangle is 18 inches.
We'll write the perimeter of the rectangle:
P = 2(a+b)
18 = 2(a+b)
We'll divide by 2:
9 = a + b
We'll use the symmetric property:
a + b = 9 (2)
We'll add (1) + 2*(2):
a - 2b + 2a + 2b = 3 + 18
We'l eliminate and combine like terms:
3a = 21
We'll divide by 3:
a = 7 inches
7 + b = 9
b = 9 - 7
b = 2 inches
So, the length of the rectangle is of 7 inches and the width of the rectangle is of 2 inches.