a)
[tex]\it \mathcal{A} = \sqrt{p(p-a)(p-b)(p-c)}
\\\;\\
p=\dfrac{a+b+c}{2} = \dfrac{10+11+9}{2} =\dfrac{30}{2} = 15[/tex]
[tex]\it p-a = 15-9 = 6 \\\;\\
p-b = 15-11=4
\\\;\\
p-c=15-10=\ 5[/tex]
[tex]\it \mathcal{A} = \sqrt{15\cdot6\cdot4\cdot5} = \sqrt{3\cdot5\cdot3\cdot2\cdot4\cdot5}= \sqrt{9\cdot25\cdot4\cdot2}=
\\\;\\
=3\cdot5\cdot2\sqrt2 =30\sqrt2\ cm^2[/tex]
b)
d (C; AB) reprezintă lungimea înălțimii duse din C pe AB.
Aria(ABC) = (1/2) AB·d ⇒ d = 2·Aria/AB = 2·30√2/10 = 6√2 cm.
c)
Aria(ABC) = (1/2)AB·BC· sinB ⇒ sinB= 2·Aria/(AB·BC) = 2·30√2/(10·9) =2√2/3
