BC||A’D’⇒∠(BC,D’O)=∠(A’D’,D’O)=∠A’D’O=60 °
A’O=D’O ⇒ △A’OD’ isoscel ⇒ ∠A’D’O=∠D’A’O=60°⇒△A’OD’ echilateral
Deci A’O = D’O=4 cm
Calculam AO:
[tex] AO=\dfrac{l\sqrt{2}}{2}=\dfrac{4\sqrt{2}}{2} =2\sqrt{2} \ cm [/tex]
Deci putem afla AA’ (inaltimea) folosind Teorema lui Pitagora in △AA’O:
[tex] (AA^{\prime} )^2 = A^{\prime}O^2 -AO^2 \\ h^2 = 4^2 -(2\sqrt{2})^2 \\ h^2 =16-4\cdot2 \\ h^2 =16-8=8 \\ \implies \tt h=\sqrt{8} =2\sqrt{2} \ cm [/tex]
