L=64 m=AB=DC
l=48 m=AD=BC
EF=1,2 m
[tex]A_{ABCD}=3072\ m^2[/tex]
[tex]A_{ABC}=A_{ADC}=1536\ m^2[/tex]
In ΔADC aplicam Pitagora:
AC²=AD²+DC²
AC²=2304+4096
AC²=6400
AC=80 m
In ΔABC, BF inaltime⇒
[tex]BF=\frac{AB\times BC}{AC} =\frac{64\times 48}{80}= 38,4 m[/tex]
EF=1,2 m
BE=BF-EF=38,4-1,2=37,2 m
MNCA este trapez isoscel
Inaltime trapez=EF=1,2 m
Notam MN=x
[tex]A_{MNCA}=\frac{(b+B)\times h}{2} =\frac{(x+80)\times 1,2}{2} \\\\A_{BMN}=\frac{BE\times MN}{2} =\frac{37,2\times x}{2}[/tex]
Daca adunam cele doua arii:
[tex]A_{MNCA}+A_{BMN}=A_{ABC}\\\\\frac{(x+80)\times 1,2}{2}+\frac{37,2x}{2} = 1536[/tex]
0,6x+48+18,6x=1536
19,2x=1488
x=77,5 m
MN=PQ=77,5 m
- Daca vrem sa plantam arbusti din 2,5 m in 2,5 m vom avea:
77,5+77,5=155 m
155:2,5=62 arbusti
