Question

un trunchi de piramidă triunghiulară regulată are laturile bazelor de 3 cm respectiv 9 cm iar apotema piramidei din care provine trunchiul de 6 cm calculează aria laterală a trunchiului de piramidă volumul trunchiului de piramidă și volumul piramidei din care provine trunchiul​

Answer 1

Folosim urmatoarele formule:

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

$P_b=3l\\\\P_B=3L$

$V=\frac{h}{3}(A_B+A_b+\sqrt{A_B\times A_b})$

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

Cunoastem:

l=3 cm

L=9 cm

apotema piramida (SE)=6 cm

a)

$P_b=3\times 3=9\ cm\\\\P_B=3\times 9=27\ cm$

Notam trunchiul conform desen atasat

Calculam apotemele bazelor, OE si O'E'

$OE=\frac{9\sqrt{3} }{6} =\frac{3\sqrt{3} }{2} \\\\O'E'=\frac{3\sqrt{3} }{6} =\frac{\sqrt{3} }{2}$

In ΔSOE avem OE║O'E'⇒

$\frac{SE'}{SE} =\frac{SO'}{SO} =\frac{O'E'}{OE}$

Luam primul si ultimul raport, unde SE=6 cm

$\frac{SE'}{6} =\frac{\frac{\sqrt{3} }{2} }{\frac{3\sqrt{3} }{2} } \\\\\frac{SE'}{6}=\frac{1}{3}$

SE'=2 cm

SE=SE'+EE'

EE'=6-2=4 cm (apotema trunchi)

$A_l=\frac{(9+27)\times 4}{2} =72\ cm^2$

b)

$A_b=\frac{9\sqrt{3} }{4}\ cm^2$

$A_B=\frac{81\sqrt{3} }{4}$

Stim de la punctul anterior ca:

$\frac{SE'}{SE} =\frac{SO'}{SO} =\frac{O'E'}{OE}\\\\\frac{SO'}{SO} =\frac{1}{3}$

• Luam trapezul O'E'EO si aplicam Pitagora (Suma catetelor la patrat este egala cu ipotenuza la patrat) in triunghiul E'FE, unde E'F⊥EO

FO=E'O'=$\frac{\sqrt{3} }{2}$

$EF=EO-EF=\frac{3\sqrt{3} }{2} -\frac{\sqrt{3} }{2} =\sqrt{3}$

EE'²=E'F²+EF²

16=E'F²+3

E'F=√13 cm=O'O

$\frac{SO'}{SO} =\frac{1}{3}$

Facem proportie derivata si obtinem:

$\frac{SO-SO'}{SO} =\frac{3-1}{3} \\\\\frac{\sqrt{13} }{SO} =\frac{2}{3} \\\\SO=\frac{3\sqrt{13} }{2}$

Calculam volum trunchi de piramida

$V=\frac{\sqrt{13} }{3}( \frac{9\sqrt{3} }{4} +\frac{81\sqrt{3} }{4}+\sqrt{\frac{81\sqrt{3} }{4}\times \frac{9\sqrt{3} }{4} }=\frac{\sqrt{13} }{4} (\frac{90\sqrt{3} }{4} +\frac{27\sqrt{3} }{4})= \frac{\sqrt{13} }{4}\times \frac{117\sqrt{3} }{4} =\frac{117\sqrt{39} }{16} \ cm^3$

Calculam volum piramida

$V=\frac{A_B\times SO}{3}$

$V=\frac{\frac{81\sqrt{3} }{4}\times \frac{3\sqrt{13} }{2} }{3}$

$V=\frac{81\sqrt{39} }{8} \ cm^3$