Question

The perimeter of a rectangle is 66 feet. Find its dimensions assuming that its length is 9 feet less than twice its width.

Answer 1

The dimensions of the rectangle with a perimeter of 66 feet, where the length is 9 feet less than twice the width, are 19 feet in length and 14 feet in width.

Step-by-step explanation:

Solving for Rectangle Dimensions

We are given that the perimeter of a rectangle is 66 feet and that the length (L) is 9 feet less than twice the width (W). This can be written as L = 2W - 9. To find the dimensions, we use the formula for perimeter of a rectangle, which is P = 2(L + W). Substituting the given perimeter and the relationship between L and W into this formula, we get:

1. 66 = 2((2W - 9) + W)
2. 33 = 3W - 9
3. 42 = 3W
4. W = 14 feet
5. L = 2W - 9 = 2(14) - 9 = 28 - 9
6. L = 19 feet

Therefore, the dimensions of the rectangle are 19 feet (length) and 14 feet (width).

Answer 2

Length = 19 feet

Width = 14 feet

Step-by-step explanation:

The perimeter, P, of a rectangle is given by

`P=2(l+w)`

where l is the length and w is the width.

From the question, the length is 9 feet less than twice the width.

`l=2w-9`

Substitute this in the equation for the perimeter.

`P=2(2w-9+w)`

`P=2(3w-9)`

`66=2(3w-9)`

`33=3(w-3)`

`11=w-3`

`w=14`

Substitute for this in the equation for l.

`l=2w-9=2*14-9=28-9=19`

Hence, the length is 19 feet and the width is 14 feet.