Question

The perimeter of a rectangle is 140 feet. The length of the rectangle is triple the width increased by six. Find length and width.

Answer 1

Given:

The perimeter of a rectangle, P=140 ft.

Let l be the length and w be the width of the rectangle.

Since the length is triple the width increasesd by 6, the length of the rectangle can be expressed as,

`l=3w+6\text{ ----(1)}`

Now, the perimeter of the rectangle can be expressed as,

`\begin{gathered} P=2(l+w) \\ P=2(3w+6+w) \\ P=2(4w+6) \\ P=8w+12 \end{gathered}`

Substitute P=140 in the above equation and solve for w.

`\begin{gathered} 140=8w+12 \\ 8w=140-12 \\ 8w=128 \\ w=(128)/(8) \\ w=16\text{ ft} \end{gathered}`

Now, put w=16 in equation (1) and find l.

`\begin{gathered} l=3\cdot16+6 \\ =48+6 \\ =54\text{ ft} \end{gathered}`

Therefore, the length of the rectangle is 54 ft and width of the rectangle is 16 ft.