The length of a rectangle is 5cm more than twice the width. The perimeter of the rectangle is 66 cm. Find the dimensions of the rectangle.

width ---- x

length ----2x+5

solve for x

2x + 2(x+5) = 66

The perimeter of a rectangle is 15 feet the height is three more than twice the length find the dimension of the rectangle

the length of a rectangle is 10 feet more than twice the width. The perimeter is 110 feet. Find the dimensions.

To find the dimensions of the rectangle, we can set up an equation based on the given information. Let's denote the width of the rectangle as "w" and the length as "l".

According to the problem, the length of the rectangle is 5cm more than twice the width. This can be written as:

l = 2w + 5

The perimeter of a rectangle is given by the formula:

Perimeter = 2(length + width)

In this case, the perimeter is given as 66 cm. So we can write:

66 = 2(l + w)

Substituting the value of "l" from the first equation, we have:

66 = 2((2w + 5) + w)

Now, simplify the equation:

66 = 2(3w + 5)
66 = 6w + 10
56 = 6w
w = 56/6
w = 9.33 cm

Now, we can substitute the value of "w" back into the first equation to find the length:

l = 2w + 5
l = 2(9.33) + 5
l = 18.67 + 5
l = 23.67 cm

Therefore, the dimensions of the rectangle are width = 9.33 cm and length = 23.67 cm.