Oblicz pole powierzchni całkowitej stożka o promieniu podstawy r, tworzącej l i wysokości h , jeżeli :
a)r=9 cm, l=41cm
b)r=9 cm, l=15cm
c)r=15cm, h=20cm
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Pp=πr²
Pb=πrl
Pc=Pp+Pb
a]
r=9cm
l=41cm
Pp=π×9²=81π
Pb=π×9×41=369π
Pc=81π+369π=450πcm²
b]
r=9cm
l=15cm
Pp=π×9²=81π
Pb=π×9×15=135π
Pc=81π+135π=216πcm²
c]
r=15cm
h=20cm
l=√[15²+20²]=√625=25cm
Pp=π×15²=225π
Pb=π×15×25=375π
Pc=225π+375π=600πcm²
P=πr (r+l)
a) P=π*9*(9+41)=9π*50=450π cm²
b) P=π*9* (9+15)=9π*24=216π cm²
c)
15²+20²=l²
l²=225+400
l²=625
l=25
P=π*15*(15+25)=15π*40=600π cm²