Answer:
Step-by-step explanation:
Given
Total area of two plot [tex]A=336 ft^2[/tex]
Let l and b the length and width of plot
Perimeter for Fencing
[tex]P=3l+4b[/tex]
[tex]100=3l+4b-----1[/tex]
[tex]A=336=2lb[/tex]
[tex]lb=168 ft^2---2[/tex]
Put value of b in 1
[tex]4\times \frac{168}{l}+3l=100[/tex]
[tex]672+3l^2=100 l[/tex]
[tex]3l^2-100l+672=0[/tex]
[tex]l=\frac{100\pm \sqrt{100^2-4\times 3\times 672}}{2\times 3}[/tex]
[tex]l=\frac{144}{6},\frac{56}{6}[/tex]
so there can be two value of l i.e. [tex]24\ ft, 9.33\ ft[/tex]
for [tex]l=24\ ft,\ b=7\ ft[/tex]