Question

Fie ABCD tetraedru regulat de muchie a.Demonstrati ca volumul este de a³ radical din 2 supra 12.​

Answer 1

Răspuns:

Explicație pas cu pas:

$V=\frac{A_{b}\cdot h}{3}$

$A_{b}=\frac{a^{2}\sqrt{3} }{4}$

$a_{b}=\frac{a\sqrt{3} }{2}$

AO inaltime

OB=$\frac{2}{3} \cdot a_{b}=\frac{2}{3} \cdot \frac{a\sqrt{3} }{2}=\frac{a\sqrt{3} }{3}$

Aplicam Pitagora in ΔAOB

$a^{2} =\frac{3a^{2} }{9} +h^{2}$

$h^{2}=a^{2} -\frac{3a^{2} }{9} =\frac{6a^{2} }{9} \\h=\frac{a\sqrt{6} }{3}$

$V=\frac{A_{b}\cdot h}{3}=\frac{\frac{a^{2}\sqrt{3} }{4}\cdot\frac{a\sqrt{6} }{3} }{3} =\frac{3a^{3}\sqrt{2} }{36} =\frac{a^{3}\sqrt{2} }{12}$