Question

Repede va rogggg dau coroana ce vreți voii

Answer 1

2.

a)

$\it a=|1-\sqrt3|+|2-\sqrt3|=-1+\sqrt3+2-\sqrt3=-1+2=1\\ \\ (a-2)^{2023}=(1-2)^{2023}=(-1)^{2023}=-1$

b)

$\it \sqrt{121}=11\\ \\ m_g=\sqrt{11\cdot99}=\sqrt{11\cdot11\cdot9}=\sqrt{11^2\cdot3^2}=11\cdot3=33$

3.

a)

AD -înălțime în ΔABC- isoscel (AB=AC) ⇒ [AD - bisectoare ⇒

∡(DAB) = ∡(DAC) = 120° : 2 = 60° ⇒∡B = ∡C = 30°, complement pt. 60°

Cu teorema ∡30° ⇒ AB = AC= 2· AD = 2√3 cm.

Cu teorema lui Pitagora ⇒ BD=CD = 3 cm ⇒ BC = 3 + 3 = 6cm

$\it \mathcal{P}_{ABC} =2\sqrt3+2\sqrt3+6=4\sqrt3+6\ cm$

b)

 $\it\mathcal{A}_{ABC}=\dfrac{BC\cdot AD}{2}=\dfrac{6\cdot\sqrt3}{2}=3\cdor\sqrt3 < 3\cdot1,74 \Rightarrow \mathcal{A}_{ABC} < 5,12\approx 5cm^2$

c)

$\it \mathcal{A}_{ABC}=\dfrac{AC\cdot d(B, \ AC)}{2} \Rightarrow 3\sqrt3=\dfrac{2\sqrt3\cdot d(B,\ AC)}{2} \Rightarrow\\ \\ \\ \Rightarrow 3\sqrt3=\sqrt3\cdot d(B,\ AC)\Big|_{:\sqrt3} \Rightarrow d(B,\ AC)=3\ cm$