| 3 | |
R. Znajdź pole powierzchni całkowitej stożka ściętego. | |
| 2 |

| 3 | ||
r1 = | R | |
| 4 |
| 7 | ||
x = | R | |
| 12 |
| 3 | 7 | 4 | ||||
r2 = r1 + x = | R + | R = | R | |||
| 4 | 12 | 3 |
| 25 | ||
l = r1 + r2 = | R | |
| 12 |
AB=2r1
| 1 | 3 | 3 | ||||
r2= | * | R= | R | |||
| 2 | 2 | 4 |
| 3 | ||
|PC|= | R | |
| 2 |
| 3 | 3 | |||
AD=BC=OB+PC=2* | R+e= | R+e | ||
| 4 | 2 |
| 3 | ||
e2+(2R)2=( | R+e)2 | |
| 2 |
| 9 | ||
e2+4R2= | R2+3eR+e2 | |
| 4 |
| 7 | ||
e= | R | |
| 12 |
| 3 | 7 | |||
r1= | R+ | R | ||
| 4 | 12 |
| 4 | ||
r1= | R | |
| 3 |