ΔABC=dreptunghic isoscel
⇒AC²=BC²+AB²
⇒AC²=6²+6²
⇒AC²=36+36
⇒AC²=72
⇒AC=√72
⇒AC=6√2 cm
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Observatie!AC,B'C si B'A sunt diagonalele cubului⇒sunt congruente⇒ΔACB'=echilateral.
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⇒A=[tex] \frac{l^2 \sqrt{3} }{4} [/tex]
⇒A=[tex] \frac{(6 \sqrt{2} )^2 \sqrt{3} }{4} [/tex]
⇒A=[tex] \frac{72 \sqrt{3} }{4} [/tex]
⇒A=18√3cm²
Desenul este in atasament,spor!
