Question

Aria totală a unui tetredru regulat este de 6√3 m².Aflați înălțimea tetredrului.

Answer 1

   
Tetraedrul regulat are 4 fete triunghiuri echilaterale.
Aria unei vete este:

[tex]\displaystyle\\ A_f = \frac{At}{4}= \frac{6 \sqrt{3}}{4}= \boxed{\frac{3 \sqrt{3}}{2}m^2}[/tex]

Calculam latura unei fete (muchia piramidei), inaltimea unei fete (apotema piramidei) si apotema bazei.


[tex]\displaystyle\\ A_f=\frac{l^2\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}\\\\ l^2=\frac{3\sqrt{3}\times4}{2\sqrt{3}}=\frac{3\times4}{2}=\frac{12}{2}=6\\\\ l =\boxed{\sqrt{6}~m}~~\text{(latura)}\\\\ h_t=a_p=\frac{l\sqrt{3}}{2}=\frac{\sqrt{6}\times\sqrt{3}}{2}=\frac{ \sqrt{18}}{2}=\boxed{\frac{3\sqrt{2}}{2} m}~~\text{(apotema piramidei)}\\\\ a_b=\frac{h_t}{3}=\frac{a_p}{3}=\frac{\frac{3\sqrt{2}}{2}}{3} =\frac{3\sqrt{2}}{2\times3}}=\boxed{\frac{\sqrt{2}}{2}m}}[/tex]


Acum aflam inaltimea piramidei din triunghiul dreptunghic format din:
         inaltimea piramidei, apotema piramidei si apotema bazei
folosind teorema lui Pitagora.


[tex]\displaystyle\\ h = \sqrt{a_p^2 - a_b^2} = \sqrt{\left(\frac{3\sqrt{2}}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2} = \\ \\ \\ \\ = \sqrt{\frac{9 \times 2}{4} - \frac{2}{4}} =\sqrt{\frac{18}{4} - \frac{2}{4}} =\sqrt{\frac{18-2}{4} } = \\ \\ \\ =\sqrt{\frac{16}{4} } = \sqrt{4} =\boxed{\boxed{2~m}} ~~\text{(inaltimea piramidei)}[/tex]