1) Diametrul unui semicerc este egal cu o treime din AG
d = AG/3 = 120/3 = 40 m
r=20 m
[tex]2)~A_{GHIJ}=GH\cdot HI=40^2=1600~m^2[/tex]
[tex]A_{semicerc}=\frac{\pi r^2}{2}=\frac{\pi20^2}{2}=\frac{400\pi}{2}=200\pi~m^2[/tex]
[tex]3\cdot A_{semicerc}=600\pi~m^2[/tex]
[tex]A_{AGKN}=AG\cdot AN=AG\cdot r=120\cdot20=2400~m^2[/tex]
[tex]A_{AGKN}-3\cdot A_{semicerc}=2400-600\pi=600(4-\pi)[/tex]
[tex]A_{hasurat}=1600+600(4-\pi)=200[8+3(4-\pi)]=200(8+12-3\pi)=200(20-3\pi)[/tex]
