Question

Ajutați-mă dau coroana!...........
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Answer 1

Răspuns:

$\red{a)\,\, \mathcal{P_{ABC}} = 96\,cm \,\,\, \,\, \mathcal{A_{ABC}}=384 \, cm^2}$

$\red{ b )\,\,d \mathcal{\,(O,AB)} = 16\,cm \,\,\, d \mathcal{\,(O,AC)} = 12\,cm}$

Explicație pas cu pas:

$BO=CO=r \Rightarrow BC=2\,r=d=40\,cm$

$\triangle ABC: \sphericalangle A=90^{\circ} \xrightarrow {T.P.} AC^2=BC^2-AB^2$

$AC=\sqrt{40^2-24^2} =\sqrt{(40-24)(40+24)}$

$AC = \sqrt{16 \cdot 64 } = 4 \cdot 8 = 32\, cm$

$\,$

$\red{{\mathcal P}_{ABC}} =AB+AC+BC = 40+24+32=\red{96\,cm}$

$\red{{\mathcal A}_{ABC} }= \dfrac{c_1 \cdot c_2}{2}=\dfrac{AB \cdot AC}{2}=\dfrac{24 \cdot 32}{2}=\red{384\,cm^2}$

${\mathcal Fie}: d\,(O,AB)=OM, M \in AB\\\\{\mathcal Fie} : d\,(O,AC)=ON, N \in AC$

$\left.\begin{aligned} AC \perp AB \,\, \\ OM \perp AB \end{aligned} \right\} \Rightarrow AC \parallel OM \Rightarrow \triangle BMO \sim \triangle BAC$

$\dfrac{BM}{AB}=\dfrac{MO}{AC}=\dfrac{BO}{BC}=k \, , \,\, \dfrac{BO}{BC}=\dfrac{20}{40}^{(20}=\dfrac{1}{2}$

$\dfrac{MO}{AC}=k \Rightarrow \dfrac{MO}{32}=\dfrac{1}{2} \Rightarrow \red{MO=\dfrac{32}{2}=16}$

${analog}\, , \triangle CNO \sim \triangle CAB \Rightarrow \dfrac{CN}{AC}=\dfrac{NO}{AB}=\dfrac{CO}{BC}$

$\dfrac{CO}{BC}=\dfrac{BO}{BC}=\dfrac{1}{2} \Rightarrow \dfrac{NO}{AB}=\dfrac{NO}{24}=\dfrac{1}{2} \Rightarrow \red{ NO=12}$