1
MC linie mijlocie in triunghiul NAB
MC/AC=NC/NB
12/24=NC/(NC+15)
24NC=12NC+180
12NC=180
NC=15
A=15x12/2=90 cm²
2.
OM_I_AD
BD=24
OD=24/2=12
MD=√[12²-(6√3)²]=√(144-108)=√36=6
OM²=AM × MD
(6√3)²=AM×6
AM=108/6=18
AD=18+6=24
Aromb=4×[(24×6√3)/2]=288√3
3.
m(<BAC)=70⁰
m(<ABC)=80⁰ ⇒ m(<AOC)=80⁰ × 2=160⁰
m(<ABC)=80⁰ ⇒ m(<OBC)=80⁰ : 2=40⁰