Răspuns:
2.<A=40° < C=30° <B =180°-70°=110°
<BCD=<A+<B=110°+40°=150°. b
3.AO mediană=>=BC/2
∆AOB echilateral (<BOA=60°)
AB=AO=4cm
pt arie ne mai trebuie AC
AC²=BC²-AB²=8²-4²=64-16=48
AC=4√3cm
Aria=AC×AB/2=4√3×4=16√3cm ²d)
4) dacă BD_l_peBC =>∆BCDdr cu
<D=30°și C=60°
BC=DC/2=18/2=9cm
Aria paralelogram=2×A∆BCD
=2×DB×DC/2=18√3×9=262√3cm²
5.R=5 r=3 R/r=5/3
SO/So=5/3
(So+r+ R)/So=5/3
(So+r+R)3=5 So
3(So+8)=5So
2So=24cm
So=12cm
SO=12+3+5=20cm c)
