Question

in tabelul următor am notat cu l, h, ap, R, ab, At, Al si V latura bazei unei piramide patrulatere regulate, înălțimea piramidei, apoteoza piramidei, muchia laterala a piramidei, raza cercului circumscris bazei, apotema bazei, aria totala și respectiv volumul piramidei. Completați tabelul, știind ca dimensiunile sunt măsurate in centimetri.
Va rog, ma asculta mâine la tema și nu am înțeles!

Answer 1

$A_b=l^2\\\\P_b=4l\\\\R=\frac{l\sqrt{2} }{2} \\\\a_b=\frac{l}{2}\\\\ A_l=\frac{P_b\times a_p}{2} \\\\A_t=A_l+A_b\\\\V=\frac{A_b\times h}{3}$

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

$m^2=R^2+h^2$

a)

l=24 cm

h=9 cm

$R=\frac{24\sqrt{2} }{2} =12\sqrt{2} \ cm$

$m^2=288+81=369\\\\m=3\sqrt{41}\ cm$

$a_b=12\ cm$

$a_p^2=144+81=225\\\\a_p=15\ cm$

$P_b=4\times24=96\ cm$

$A_b=24^2=576\ cm^2$

$A_l=\frac{96\times 15}{2} =720\ cm^2\\\\\\A_t=720+576=1296\ cm^2$

$V=\frac{576\times 9}{3} =1728\ cm^3$

b)

h=12 cm

$a_p=15\ cm$

$a_b=225-144=81\\\\a_b=9\ cm$

l=18 cm

$R=9\sqrt{2} \ cm$

$m^2=162+144=306\\\\m=3\sqrt{34} \ cm$

$P_b=72\ cm$

$A_b=324\ cm^2$

$A_l=\frac{72\times 15}{2} =540\ cm^2\\\\A_t=540+324=864\ cm^2$

$V=\frac{324\times 12 }{3} =1296\ cm^3$

c)

$a_b=9\ cm$

$A_l=540\ cm^2$

l=18 cm

$R=9\sqrt{2} \ cm$

$540=\frac{P_b\times a_p}{2}$

$P_b=72\ cm$

$1080=72\times a_p$

$a_p=15\ cm$

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

$h^2=225-81=144\\\\h=12\ cm$

$A_b=324\ cm^2$

$A_t=324+540=864\ cm^2$

$V=\frac{324\times 12}{3} =1296\ cm^3$

d)

h=12 cm

V=576 cm³

$576=\frac{A_b\times 12}{3}$

$A_b=144\ cm^2$

l=12 cm

$a_b=6\ cm$

R=6√2 cm

$a_p^2=144+36\\\\a_p=6\sqrt{5} \ cm$

$P_b=48\ cm$

$A_l=\frac{48\times 6\sqrt{5} }{2}= 144\sqrt{5} \ cm^2$

$A_t=144\sqrt{5} +144$

e)

$A_l=960\ cm^2$

$A_t=1536\ cm^2$

$A_b=576\ cm^2$

l=24 cm

R=12√2 cm

$m^2=288+256\\\\m^2=544\\\\m=4\sqrt{34} \ cm$

$a_b=12\ cm$

$P_b=96\ cm$

$960=\frac{96\times a_p}{2} \\\\a_p=20\ cm$

$400=144+h^2\\\\h^2=256\\\\h=16\ cm$

$V=\frac{576\times 16}{3} =3072\ cm^3$

f)

m=18 cm

R=9√2 cm

$9\sqrt{2} =\frac{l\sqrt{2} }{2} \\\\l=18\ cm$

$a_b=9\ cm$

$324=162+h^2$

h=9√2 cm

$162+81=a_p^2\\\\a_p=9\sqrt{3} \ cm$

$P_b=72\ cm$

$A_l=\frac{72\times 9\sqrt{3} }{2} =324\sqrt{3} \ cm^2$

$A_b=324\ cm$

$A_t=324+324\sqrt{3}$

$V=\frac{324\times 9\sqrt{2} }{3} =972\sqrt{2} \ cm^3$