How much

Mathematics

1 answer:

allsm [11]3 years ago

7 0

The farmer will need:

$\boxed{191\frac{11}{12}yd}$

In order to enclose the area shown in the figure below.

<h2>Explanation:</h2>

The diagram below shows the representation of this problem. Let:

$x: The \ length \ of \ the \ rectangular \ pastures \\ \\ y: The \ width \ of \ the \ rectangular \ pastures$

We know that:

$x=5\frac{5}{8}yd \\ \\ y=30\frac{2}{9}yd$

So the fencing the farmer needs can be calculated as the perimeter of the two adjacent rectangular pastures:

$P=2(x+y)+y \\ \\ P=2(50\frac{5}{8}+30\frac{2}{9})+30\frac{2}{9} \\ \\ P=100\frac{10}{8}+60\frac{4}{9}+30\frac{2}{9} \\ \\ P=100\frac{10}{8}+90\frac{6}{9} \\ \\ P=100\frac{5}{4}+90\frac{2}{3} \\ \\ P=190(\frac{15+8}{12}) \\ \\ P=190(\frac{23}{12}) \\ \\ \\ Expressing \ as \ a \ mixed \ fraction: \\ \\ P=190+1+\frac{11}{12} \\ \\ P=191+\frac{11}{12} \\ \\ \boxed{P=191\frac{11}{12}yd}$