r = 4
Pp = 0,5 Pb
więc
π r2 = 0,5 π*r*l / : π*r
r = 0,5 l
2 r = l
zatem h2 + r2 = l2
h2 + r2 = (2 r)2
h2 = 4 r2 − r2 = 3 r2
h = √ 3 r2 = √3*r = 4√3
−−−−−−−−−−−−−−−−−−−−
Objętość stożka
| 1 | 1 | 1 | 64 | |||||
V = | Pp*h = | π r2 *h = | π*42 * 4√3 = | √3π | ||||
| 3 | 3 | 3 | 3 |
r = 4
Pp = 0,5 Pb
więc
π r2 = 0,5 π*r*l / : π*r
r = 0,5 l
2 r = l
zatem h2 + r2 = l2
h2 + r2 = (2 r)2
h2 = 4 r2 − r2 = 3 r2
h = √ 3 r2 = √3*r = 4√3
−−−−−−−−−−−−−−−−−−−−
Objętość stożka
| 1 | 1 | 1 | 64 | |||||
V = | Pp*h = | π r2 *h = | π*42 * 4√3 = | √3π | ||||
| 3 | 3 | 3 | 3 |