Question

oblicz pole powierzchni całkowiej i objętość stożka o wysokości 9 cm i obwodzie 24 pi cm

Answer 1

obwod ( l ) = 24 pi cm

l= 2 pi r

24 pi = 2pi r |:2pi

12 = r

 

Pp= pir^2 = 144Pi cm^2

 

l=?

l^2=12^2+9^2=144+81=225

l=15

 

Pc= pir(r+l)=12pi(12+15)=12pi*27= 324Pi Cm^2

V= 1/3 * Pp * H= 1/3* 9 * 144Pi=3*144Pi=432Pi Cm^3

 

Answer 2

Pc = pp + pb

Pc = πr^2 + πrl

 

L = 2πr

 

r = L / 2π

r = 24πcm / 2π

r = 12 cm

 

l^2 = r^2 + h^2

 

l^2 = 144 + 81

 

l^2 = 225

 

l = 15 cm

 

Pc = 12^2π cm + 15*12π 

Pc = 144π + 180π 

Pc = 322π cm2

 

v = 1/3 Pp * h

V = 1/3 * 144π cm2 * 9cm

V = 432π cm3