Răspuns :
raza cercului circumscris Δ este AB·AC·BC÷4·Aria Δ
calculam i
Fie AD perpendiculara pe BC⇒BD=CD=BC÷2=36÷2=18
in ΔABD dr in D⇒cu teorema lui Pitagora AB²=BD²+AD²
30²=18²+AD²
900=324+AD²
AD²=900-324
AD²=576
AD=√576
AD=24
AriaΔ= b·i÷2=BC·AD÷2=36·24÷2=864÷2=432
R=30·30·36÷(4·432)=32400÷1728=18,75
calculam i
Fie AD perpendiculara pe BC⇒BD=CD=BC÷2=36÷2=18
in ΔABD dr in D⇒cu teorema lui Pitagora AB²=BD²+AD²
30²=18²+AD²
900=324+AD²
AD²=900-324
AD²=576
AD=√576
AD=24
AriaΔ= b·i÷2=BC·AD÷2=36·24÷2=864÷2=432
R=30·30·36÷(4·432)=32400÷1728=18,75