The length of the rectangle is 4 more than twice its width and its perimeter is 70m. What are the dimensions of the rectangle?
let x = witdth
4x+2 = length
perimeter of rectangle = 2L + 2W
substitute:
34 = 2(4x+2) +2x
34 = 8x + 4 + 2x
34-4 = 10x
10x = 30
x = 3 - width
length = 4(3) + 2
= 12 + 2 = 14 :)
John, why isn't the length 2x + 4 ??
Good catch
John has his definition backwards, thus his answer is not correct
width --- x
length ---- 2x+4
2x + 2(2x+4) = 70
6x + 8 = 70
6x = 62
x = 31/3 or 10 1/3
width = 31/3
length = 2(31/3)+4 = 74/3 or 24/2/3
check: 2(31/3) + 2(74/3) = 70
To find the dimensions of the rectangle, we need to set up equations based on the given information and then solve them.
Let's assume the width of the rectangle is "w" meters.
According to the given information, the length of the rectangle is 4 more than twice the width. So, the length would be (2w + 4) meters.
The equation for the perimeter of a rectangle is: P = 2(l + w), where P is the perimeter, l is the length, and w is the width.
We know that the perimeter of the rectangle is 70m. Substituting the known values into the equation, we get:
70 = 2((2w + 4) + w)
Simplifying the equation, we have:
70 = 2(3w + 4)
70 = 6w + 8
Subtracting 8 from both sides, we get:
62 = 6w
Dividing both sides by 6, we have:
w = 10.33 (rounded to two decimal places)
Since the width of a rectangle cannot be in decimal, we need to round it down to the nearest whole number. Therefore, the width of the rectangle is 10 meters.
Now, substituting the value of width (w) into the equation for the length, we get:
Length = 2w + 4 = 2 * 10 + 4 = 20 + 4 = 24 meters.
So, the dimensions of the rectangle are:
Width = 10 meters
Length = 24 meters