Question

Copiați și completați tabelul de mai jos, unde L, h, m, ap, Al, At și V reprezintă respectiv: latura bazei înălțimea, muchia laterală, apotema, aria laterală, aria totală și volumul piramidei triunghiulare regulate.​

Answer 1

Aplicam urmatoarele formule:

$P_b=3l$

$A_l=\frac{P_b\times a_p}{2}$

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

$A_t=A_l+A_b$

$V=\frac{A_b\times h}{3}$

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

$a_p^2=h^2+a_b^2$

$m^2=\frac{l^2}{3} +h^2$

$m^2=\frac{l^2}{4}+a_p^2$

a)

l=6

h=3

P=18

$A_b=\frac{36\sqrt{3} }{4} =9\sqrt{3} \ cm^2$

$a_b=\frac{6\sqrt{3} }{6} =\sqrt{3}$

$a_p^2=9+3=12\\\\a_p=2\sqrt{3}$

$A_l=\frac{18\times 2\sqrt{3} }{2} =18\sqrt{3}$

$A_t=9\sqrt{3} +18\sqrt{3} } =27\sqrt{3}$

$m^2=12+9=21\\m=\sqrt{21}$

$V=\frac{9\sqrt{3}\times 3 }{3} =9\sqrt{3}$

b)

l=2

m=√13

$13=\frac{4}{3}+h^2\\\\ h=\frac{\sqrt{35} }{\sqrt{3} } =\frac{\sqrt{105} }{3}$

$a_b=\frac{\sqrt{3} }{3}$

$a_p^2=\frac{105}{9}+\frac{3}{9} =\frac{108}{9}=12\\\\ a_p=2\sqrt{3}$

$P_b=6$

$A_b=\frac{4\sqrt{3} }{4}=\sqrt{3}$

$A_l=\frac{6\times 2\sqrt{3} }{2} =6\sqrt{3}$

$A_t=7\sqrt{3}$

$V=\frac{\sqrt{3} \times \sqrt{105} }{9} =\frac{\sqrt{35} }{3}$

c)

m=13

$a_p=12$

$169=\frac{l^2}{4}+144 \\\\l^2=100\\\\l=10$

$P_b=30$

$A_b=\frac{100\sqrt{3} }{4}=25\sqrt{3}$

$A_l=\frac{30 \times 12}{2}=180$

$A_t=25\sqrt{3} +180$

$a_b=\frac{10\sqrt{3} }{6} =\frac{5\sqrt{3} }{3}$

$144=h^2+\frac{75}{9} \\\\h^2=\frac{1221}{9} \\\\h=\frac{\sqrt{1221} }{3}$

$V=\frac{25\sqrt{3}\times \frac{\sqrt{1221} }{3} }{3}=\frac{25\sqrt{407} }{3}$

d)

l=2√3

Al=9√3

$P_b=6\sqrt{3}$

$9\sqrt{3} =\frac{6\sqrt{3} \times a_p}{2}$

$a_p=3$

$a_b=\frac{6}{6} =1$

$A_b=\frac{12\sqrt{3} }{4}=3\sqrt{3}$

$A_t=12\sqrt{3}$

$9=1+h^2\\\\h=2\sqrt{2}$

$V=\frac{3\sqrt{3}\times 2\sqrt{2} }{3} =2\sqrt{6}$

$m^2=4+8\\\\m=2\sqrt{3}$

e)

Al=75√3

At=100√3

$A_b=100\sqrt{3}-75\sqrt{3} =25\sqrt{3} =\frac{l^2\sqrt{3} }{4} \\\\l=10$

$P_b=30\\\\75\sqrt{3} =\frac{30\times a_p}{2} \\\\a_p=5\sqrt{3}$

$m^2=25+75=100\\\\m=10$

$100=\frac{100}{3}+h^2\\\\ 3h^2=200\\\\h=\frac{10\sqrt{2} }{\sqrt{3} }=\frac{10\sqrt{6} }{3}$

$V=\frac{25\sqrt{3}\times \frac{10\sqrt{6} }{3} }{3}=\frac{250\sqrt{2} }{3}$

f)

h=4√6

V=144√2

$144\sqrt{2} =\frac{A_b\times 4\sqrt{6} }{3} \\\\A_b=\frac{108}{\sqrt{3} }=36\sqrt{3}$

$36\sqrt{3} =\frac{l^2\sqrt{3} }{4} \\\\l=12$

$m^2=48+ 96=144\\\\m=12$

$144=36+a_p^2\\\\a_p=6\sqrt{3}$

$P_b=36$

$A_l=\frac{36\times 6\sqrt{3} }{2} =108\sqrt{3}$

$A_t=144\sqrt{3}$