a) Se aplica formula a*b*sinC/2
A=6*10*1/2*1/2=15cm
b) se aplica formula lui Heron (https://ro.wikipedia.org/wiki/Formula_lui_Heron)
A=[tex] \sqrt{10*5*2*3} [/tex]=[tex] \sqrt{300} [/tex]=10[tex] \sqrt{3} [/tex]
c) este un triunghi echilateral deci aria = [tex] l^{2} \sqrt{3} /4[/tex]
A=9[tex] \sqrt{3} [/tex]
d) Aria triungiului dreptunghic = C1*C2/2
A=2[tex] \sqrt{2} [/tex]*3[tex] \sqrt{2} [/tex]/2=6
e)
Notam AC=BC=l
Aplicand teorema lui Pitagora rezulta:
2[tex] l^{2} [/tex]=384
l=8[tex] \sqrt{3} [/tex]
Aplicam aceeasi formula ca si la punctul precedent:
A=8*8*3/2=96
f) A=AB*AC*sin150/2=12*1/2*1/2=3
sin(A+B)=sinA cosB+ sinB cosA
sin150=sin(90+60)=sin90*cos60+sin60*cos90=1/2
