| a√3 | 4√3 | |||
R=2/3h lub R= | stąd R = | |||
| 3 | 3 |
| a√3 | 2√3 | |||
R= 1/3h lub R= | stąd R= | |||
| 6 | 3 |
| √3 | √3 | |||
h = a | = 4* | = 2 √3 | ||
| 2 | 2 |
| 1 | 2 | 4 | 4 | |||||
r = | h = | √3 ⇒ r2 = | * 3 = | |||||
| 3 | 3 | 9 | 3 |
| 2 | 4 | 16 | 16 | |||||
R = | h = | √3 ⇒ R2 = | *3 = | |||||
| 3 | 3 | 9 | 3 |
| 16 | 4 | |||
Pp = PR − Pr = π R2 − π r2 = π *( | − | ) = π*4 = 4π | ||
| 3 | 3 |
| √3 | √3 | |||
h = a | = 4* | = 2 √3 | ||
| 2 | 2 |
| 1 | 2 | 4 | 4 | |||||
r = | h = | √3 ⇒ r2 = | * 3 = | |||||
| 3 | 3 | 9 | 3 |
| 2 | 4 | 16 | 16 | |||||
R = | h = | √3 ⇒ R2 = | *3 = | |||||
| 3 | 3 | 9 | 3 |
| 16 | 4 | |||
Pp = PR − Pr = π R2 − π r2 = π *( | − | ) = π*4 = 4π | ||
| 3 | 3 |
Liczymy pola kół
obie odpowiedzi są dobre
sprawdzałam w odpowiedziach i faktycznie ma być
4π cm2