ABCD-Trapez isocel
AB//CD
AB=24 cm
CD=18 cm
Concluzie
a) Aria trapezului ABCD?
b) AC=BD=?
c) d(C;AD)
Rezolvare
a) cos de 60 de grade=cateta.alaturata/ipotenuza
1/2=3/AD
AD=3*2/1
AD=6cm
sin 60 de grade=cateta opusa/ipotenuza
rad3/2=h/6
h=6*rad3/2
h=3rad3cm
De unde h(inaltimea in trapezul ABCD isocel
De unde rad(radical)
DO=EC=3rad 3cm
Aria trapez=B+b*h/2
Aria trapez=24+18*3rad3/2
Aria trapez=42*3rad3/2
Aria trapez=21*3rad3cm
Aria trapez=63rad3 cm patrati
b) AC=BD=diagonalele trapezului ACBD isocel
sin de 60 de grade=cateta opusa/ipotenuza
rad3/2=18/AC
AC=18*2/rad3
AC=36/rad3
Simplificam cu rad 3 si rezulta
AC=36*rad3/3
AC=12rad3cm
AC=BD=12rad3 cm
c) d(C;AD)=12rad3cm
