Answer:
Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Step-by-step explanation:
Given the figure with dimensions. we have to find the area of given figure.
Area of figure=ar(1)+ar(2)+ar(3)
Area of region 1 = ar(ANGI)+ar(AIB)
[tex]=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha[/tex]
Area of region 2 = ar(DHBC)
[tex]=2000\times1500\\\\=3000000m^2=300ha[/tex]
Area of region 3 = ar(GFEH)
[tex](2000+1500)\times 1000\\\\=3500000m^2=350ha[/tex]
Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha
=987.5 ha
Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.
Let the fencing be done through x m downward from B which divides the two into equal area.
⇒ Area of upper part above fencing=Area of lower part below fencing
⇒[tex]ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m[/tex]
Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
