Question

Problema 18, va rog din suflet!!!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Fie a - muchia cubului.

BD = A'B = A'D = a√2 cm (diagonale ale fețelor cubului).

Deci, ΔA'BD - echilateral .

$\it \mathcal{A}_{A'BD}=\dfrac{(a\sqrt2)^2\sqrt3}{4} \Rightarrow 18\sqrt3=\dfrac{\ 2a^2\sqrt3\ ^{(2}}{4} \Rightarrow 18\sqrt3=\dfrac{a^2\sqrt3}{2} \Rightarrow \\ \\ \\ \Rightarrow 36\sqrt3=a^2\sqrt3 \Rightarrow a^2=36 \Rightarrow a=6\ cm$

ABC'D' - dreptunghi,  AB=6cm,  AD'=6√2 cm .

$\it \mathcal{A}_{ABC'D'}=AD'\cdot AB=6\sqrt2\cdot6=36\sqrt2\ cm^2$

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AC' = BD' = 6√3 cm (diagonale ale cubului) .

Fie AC' ∩ BD'={Q}, AQ = BQ = 3√3 cm .

$\it \mathcal{A}_{ABQ}=\dfrac{\mathcal{A}_{ABC'D'}}{4}=\dfrac{36\sqrt2}{4}=9\sqrt2cm^2\\ \\ \\ Dar,\ \mathcal{A}_{ABQ}=\dfrac{AQ\cdot BQ\cdot sin(\widehat{AC',\ BD'})}{2} \Rightarrow 9\sqrt2=\dfrac{3\sqrt3\cdot3\sqrt3\cdot sin(\widehat{AC',\ BD'})}{2} \Rightarrow\\ \\ \\ \Rightarrow sin(\widehat{AC',\ BD'})=\dfrac{18\sqrt2\ ^{(9}}{27}=\dfrac{2\sqrt2}{3}$