Question

va Rog rapid Dau multe puncte​

Answer 1

Răspuns:

Explicație pas cu pas:

in ΔABD   AD⊥BC   ⇒sin∡BAD=BD/AB=sin 30°=1/2

⇒BD=3√3

daca ∡BAD=30°  ⇒∡ABC=30°  ⇒BC=2AB=12√3 cm

DC=BC-BD=12√3-2√3=10√3 cm

T Inaltimii

AD²=BD·DC=2√3·10√3=60    ⇒AD=2√15 cm

Aabc=AD·BC/2=2√15·12√3/2=12·3·√5=36√5 cm²

Aabd=BD·AD/2     Aabc=AD·BC/2

Aabd/Aabc=BD·AD/AD·BC=BD/BC=3√3/12√3=1/4

Answer 2

$\it \widehat{ABD}=60^o\ (complementul\ lui\ 30^o\ \hat\imath n\ \Delta DAB)\\ \\ \widehat{BCA}=30^o\ (complementul\ lui\ 60^o\ \hat\imath n\ \Delta ABC)\\ \\ T.\angle\ 30^o\ \hat\imath n\ \Delta DAB \Rightarrow BD=\dfrac{6\sqrt3}{2}=3\sqrt3\\ \\ T.\ Pitagora\ \hat\imat n\ \Delta DAB\ \Rightarrow AD=9\ cm\\ \\ T.\angle\ 30^o\ \hat\imath n\ \Delta ADC \Rightarrow AC=18cm$

$\it \mathcal{A}_{ABC}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AB\cdot AC}{2}=\dfrac{6\sqrt3\cdot18}{2}=54\sqrt3cm^2$