Question

triunghiul ABC este dreptunghic in A daca ab=12 cm si m unghiului b este=60 de grade calculati aria si perimetrul triunghiului

Answer 1

=> c =90-60 =30 grade
BC =2*AB =2*12 =24 cm (  T. <) de 30 grade )
AC^2 =BC^2 - AB^2
AC^2 =24^2-12^2 =576 -144 =432
AC =20.78 cm

P =12+24+20.78=56.78 cm

AB^2 =BD*BC
144 =BD*24
BD =144:24 =6 cm (BD = inaltime)

A =24*6/2 =42 cm^2

Answer 2

[tex]Daca~ \Delta~ABC~ este~ dreptunghic ~in ~A ~si~ m(B)=60 =\ \textgreater \ masura \\\\ ~unghiului~C=30 \°=\ \textgreater \ AB= \frac{BC}{2}\\\\ AB=12~cm=\ \textgreater \ BC=2*AB=\ \textgreater \ BC=2*12=\ \textgreater \ BC=24~cm.\\\\ AC^2+AB^2=BC^2\\\\ AC^2=BC^2-AB^2\\\\ AC^2=24^2-12^2\\\\ AC^2=576-144\\\\ AC^2=432\\\\ AC= \sqrt{432}\\\\ AC=12 \sqrt{3}\\\\\\ P_{_{\Delta~ABC}} =AB+BC+AC\\\\ P=12+24+12 \sqrt{3} \\\\ \boxed{\bold{P=36+12 \sqrt{3} =12*(3+ \sqrt{3})}} \\\\\\ Ducem~AD \perp BC=\ \textgreater \ prin~teorema~catetei=\ \textgreater \ \\\\ AB^2=BC*BD\\\\ BD= \frac{AB^2}{BC}\\\\ BD= \frac{12^2}{24}\\\\[/tex]
[tex]BD= \frac{144}{24}=6~cm\\\\ CD=BC-DB\\\\ CD= 24-6\\\\ CD=18 ~cm\\\\ =\ \textgreater \ ~teorema~inaltimii~=\ \textgreater \ \\\\ AD^2=CD*BD\\\\ AD^2=6*18\\\\ AD^2=108\\\\ AD= \sqrt{108}\\\\ AD= 6 \sqrt{3} \\\\\\ Aria_{\Delta~ABC}= \frac{AD*BC}{2} = \frac{6 \sqrt{2}*24 }{2}= 72 \sqrt{2}\\\\\\ ~\boxed{\bold{Aria_{\Delta~ABC}=72 \sqrt{2} cm^2}}[/tex]