Question

a rectangle field is four times as long as it's wide if the length is decreased by 10 ft and the width is increased by 2 ft the perimeter will be 80 ft find the dimensions of the original field the original dimensions are blank feet long by blank feet wide

Answer 1

Let the width of the field = w

∵ The length is four times as the width

∵ The width = w

∴ The length = 4w

∵ The length is decreased by 10 feet

∴ The new length = 4w - 10

∵ The width is increased by 2 feet

∴ The new width = w + 2

The new perimeter is 80 feet

∵ The perimeter of the rectangle = 2(length + width)

`\therefore P=2(4w-10+w+2)`

Let us simplify it

`\begin{gathered} P=2(4w+w-10+2) \\ P=2(5w-8) \\ P=2(5w)-2(8) \\ P=10w-16 \end{gathered}`

Now equate it by 80

`10w-16=80`

Add 16 to both sides

`\begin{gathered} 10w-16+16=80+16 \\ 10w=96 \end{gathered}`

Divide both sides by 10

`\begin{gathered} (10w)/(10)=(96)/(10) \\ w=9.6 \end{gathered}`

The length is 4 times the width

`\begin{gathered} l=4(9.6) \\ l=38.4 \end{gathered}`

The length is 38.4 feet and the width is 9.6 feet