Question

Two rectangular fields with identical shapes and areas are to be fenced side by side. The total area enclosed is 8 mmmm2 and the perimeter for all the fencing is 16 miles. Find the possible dimensions of one of the fields (2 possible answers).

Answer 1

3 mi and
`(4)/(3)` mi, 1 mi and 4 mi

Explanation:

Length of one of rectangles = x

Breadth of one of the rectangles = y

Total enclosed area =
`8\ \text{mi}^2`

Fence available = 16 miles

Total area is given by

`2xy=8\\\Rightarrow y=(4)/(x)`

Perimeter of the fence

`2* 2x+3y=16\\\Rightarrow 4x+3y=16\\\Rightarrow 4x+3(4)/(x)=16\\\Rightarrow 4x+(12)/(x)=16\\\Rightarrow 4x^2-16x+12=0\\\Rightarrow x=(-\left(-16\right)\pm √(\left(-16\right)^2-4* \:4* \:12))/(2* \:4)\\\Rightarrow x=3,1`

If
`x=3`

`y=(4)/(x)=(4)/(3)`

If
`x=1`

`y=(4)/(x)=4`

The possible values of length and breadth are 3 mi and
`(4)/(3)` mi, 1 mi and 4 mi respectively.