Question

1.Completați tabelul,știind că elementele sunt măsurate în centimetri,iar trunchiul este un trunchi de piramidă triunghiulară regulată.
Vreau seriozitate maximă!​

Answer 1

• Avem urmatoarele formule pentru trunchiul de piramida triunghiulara regulata:

$A_b=\frac{l^2\sqrt{3} }{4} \\\\A_B=\frac{L^2\sqrt{3} }{4} \\\\\\A_l=\frac{(P_b+P_B)\times a_t_r}{2} \\\\A_t=A_l+A_b+A_B\\\\V=\frac{h}{3}(A_B+A_b+\sqrt{A_B\times A_b} )$

$h_b=\frac{l\sqrt{3} }{2} \\\\h_B=\frac{L\sqrt{3} }{2}$

$R_b=\frac{2}{3} \times \frac{l\sqrt{3} }{2} =\frac{l\sqrt{3} }{3} \\\\R_B=\frac{L\sqrt{3} }{3}$

$a_b=\frac{l\sqrt{3} }{6} \\\\\\a_B=\frac{L\sqrt{3} }{6}$

$m^2=(\frac{L}{2} -\frac{l}{2})^2+a_{tr}^2$

$m^2=h^2+(R_B-R_b)^2$

$a_{tr}^2=h^2+(a_B-a_b)^2$

a)

L=18 cm

l=12 cm

m=6 cm

$R_b=\frac{12\sqrt{3} }{3} =4\sqrt{3}\ cm$

$R_B=\frac{18\sqrt{3} }{3} =6\sqrt{3}\ cm$

$a_b=\frac{12\sqrt{3} }{6} =2\sqrt{3} \ cm$

$a_B=\frac{18\sqrt{3} }{6} =3\sqrt{3} \ cm$

6²=h²+(6√3-4√3)²

h²=36-12=24

h=2√6 cm

$P_b=12\times 3=36\ cm\\\\P_B=18\times 3=54 \ cm$

$A_b=\frac{144\sqrt{3} }{4} =36\sqrt{3}\ cm^2\\\\ A_B=\frac{324\sqrt{3} }{4} =81\sqrt{3} \ cm^2$

$a_{tr}^2=24+\sqrt{3} ^2\\\\a_{tr}=3\sqrt{3}\ cm$

$A_l=\frac{(54+36)\times 3\sqrt{3} }{2} =135\sqrt{3} \ cm^2$

$A_t=135\sqrt{3} +36\sqrt{3} +81\sqrt{3} =252\sqrt{3} \ cm^2$

$V=\frac{2\sqrt{6} }{3} (36\sqrt{3} +81\sqrt{3}+\sqrt{8748} )=\frac{2\sqrt{6} }{3} (117\sqrt{3} +54\sqrt{3} )=\frac{2\sqrt{6} }{3}\times 171\sqrt{3} =114\sqrt{18} =342\sqrt{2} \ cm^3$

b)

L=16√3 cm

l=8√3 cm

h=4√3 cm

$R_B=\frac{16\times3 }{3} =16\ cm\\\\R_b=\frac{24}{3} =8\ cm$

$a_B=\frac{48}{6} =8\ cm\\\\a_b=\frac{24}{6} =4\ cm$

m²=48+(16-8)²

m²=48+64=112

m=4√7 cm

$a_{tr}^2=48+16=64\\\\a_{tr}=8\ cm$

$P_b=24\sqrt{3} \ cm\\\\P_B=48\sqrt{3} \ cm$

$A_l=\frac{(24\sqrt{3}+48\sqrt{3} )\times 8 }{2} =288\sqrt{3} \ cm^2$

$A_b=\frac{192\sqrt{3} }{4} =48\sqrt{3}\ cm^2$

$A_B=\frac{768\sqrt{3} }{4} =192\sqrt{3}\ cm^2$

$A_t=288\sqrt{3} +48\sqrt{3} +192\sqrt{3} =528\sqrt{3} \ cm^2$

$V=\frac{4\sqrt{3} }{3} (48\sqrt{3} +192\sqrt{3} +\sqrt{27648} )\\\\V=\frac{4\sqrt{3} }{3} (48\sqrt{3}+192\sqrt{3}+96\sqrt{3})=448\sqrt{3}\ cm^3$