Explicație pas cu pas:
ΔABC, ∢B = 30°, ∢C = 45°
[tex]\frac{AC}{ \sin(B) } = \frac{AB}{ \sin(C) } < = > \frac{18 \sqrt{2} }{ \sin(30) } = \frac{AB}{ \sin(45) } \\ \frac{18 \sqrt{2} }{ \frac{1}{2}} = \frac{AB}{ \frac{ \sqrt{2} }{2} } = > AB = 36 \: cm [/tex]
în ΔABD dreptunghic: ∢ABD = 30°
→ AD = AB÷2 => AD = 18 cm
→ T.P.: BD² = AB² - AD² = 36² - 18² = 972
[tex]=> BD = 18\sqrt{3}\: cm[/tex]
în ΔADC dreptunghic: ∢ACD = 45°
→ ΔADC dreptunghic isoscel
=> AD = DC = 18 cm
[tex]BC = BD + DC = 18\sqrt{3} + 18 \\ => BC = 18(\sqrt{3}+1)\: cm[/tex]
a)
[tex]Aria_{(ABC)} = \frac{AD \cdot BC}{2} = \frac{18 \cdot 18(\sqrt{3}+1)}{2} \\ = 162(\sqrt{3}+1) \: {cm}^{2}[/tex]
b)
[tex]Perimetrul_{(ABC)} = AB + BC + AC \\ = 36 + 18(\sqrt{3}+1) + 18\sqrt{2} = 18(3 + \sqrt{3} + \sqrt{2})\: cm [/tex]
c)
[tex]h_{A} = AD = 18\: cm[/tex]
[tex]Aria_{(ABC)} = \frac{h_{B} \cdot AC}{2} = \frac{h_{C} \cdot AB}{2} \\ [/tex]
[tex]\frac{h_{B}\cdot AC}{2} = 162(\sqrt{3}+1) \\ h_{B} = \frac{2 \cdot 162(\sqrt{3}+1)}{18 \sqrt{2} } = > h_{B} = 9\sqrt{2}(\sqrt{3}+1) \: cm [/tex]
[tex]\frac{h_{C} \cdot AB}{2} = 162(\sqrt{3}+1) \\ h_{C} = \frac{2 \cdot 162(\sqrt{3}+1)}{36} = > h_{C} = 9(\sqrt{3}+1) \: cm [/tex]
