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

Answer:

3 mi and $\dfrac{4}{3}$ mi, 1 mi and 4 mi

Step-by-step 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

$2xy=8\\\Rightarrow y=\dfrac{4}{x}$

Perimeter of the fence

$2\times 2x+3y=16\\\Rightarrow 4x+3y=16\\\Rightarrow 4x+3\dfrac{4}{x}=16\\\Rightarrow 4x+\dfrac{12}{x}=16\\\Rightarrow 4x^2-16x+12=0\\\Rightarrow x=\frac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\times \:4\times \:12}}{2\times \:4}\\\Rightarrow x=3,1$

If $x=3$

$y=\dfrac{4}{x}=\dfrac{4}{3}$

If $x=1$

$y=\dfrac{4}{x}=4$

The possible values of length and breadth are 3 mi and $\dfrac{4}{3}$ mi, 1 mi and 4 mi respectively.