Răspuns :
C
mas <A = 60 A D B
CD = inaltimea aria = AB · CD / 2 = 8cm ·CD / 2 = 4·CD
aria = 20√3 = 4 ·CD ⇒ inaltimea CD = 20√3 : 4 = 5√3
Δ CDA drept cu mas<C = 30 atunci AC = 2·AD
AC² = AD² + CD² ⇒ ( 2AD)² = AD² + (5√3)²
4AD²- AD² = 25·3 ; 3AD² = 25·3 ; AD² =25 ; AD =√25 = 5
ATUNCI AC = 2·AD = 2·5 = 10
daca AD = 5 atunci DB = 8 - 5 = 3 cm
Δ CDB drept BC² = CD² + DB² = (5√3)² + 3² = 75 +9=84
BC = √84 = √4√21 = 2√21
perim = 10cm + 8cm + 2√21cm = 18 cm + 2√21 cm
mas <A = 60 A D B
CD = inaltimea aria = AB · CD / 2 = 8cm ·CD / 2 = 4·CD
aria = 20√3 = 4 ·CD ⇒ inaltimea CD = 20√3 : 4 = 5√3
Δ CDA drept cu mas<C = 30 atunci AC = 2·AD
AC² = AD² + CD² ⇒ ( 2AD)² = AD² + (5√3)²
4AD²- AD² = 25·3 ; 3AD² = 25·3 ; AD² =25 ; AD =√25 = 5
ATUNCI AC = 2·AD = 2·5 = 10
daca AD = 5 atunci DB = 8 - 5 = 3 cm
Δ CDB drept BC² = CD² + DB² = (5√3)² + 3² = 75 +9=84
BC = √84 = √4√21 = 2√21
perim = 10cm + 8cm + 2√21cm = 18 cm + 2√21 cm
[tex]AB=\boxed{c=8}\ ,BC=a,\ AC=b\\
A_{[ABC]}=\frac{bc\sin A}{2}\Leftrightarrow\ 20\sqrt3=\frac{b\cdot 8\sin60\textdegree}{2}\\
\Leftrightarrow 40\sqrt3=b\cdot 8 \frac{\sqrt3}{2}\Leftrightarrow\boxed{ b=10}\\
a^2=b^2+c^2-2bc\cdot\cos60\textdegree\\
a^2=100+64-2\cdot8\cdot10\cdot\frac{1}{2}\\
a^2=164-80=84\\
a=\sqrt{84}=2\sqrt{21}\\
\boxed{a=2\sqrt{21}}\\
P=a+b+c=2\sqrt{21}+10+8=18+2\sqrt{21}
[/tex]