| π | π | |||
i | . Oblicz objętość tego ostrosłupa. | |||
| 6 | 3 |
1) Wyobraź sobie, że ostrosłup jest "wpisany" w graniastosłup prosty.
a*b=1
| π | ||
α= | ||
| 3 |
| π | ||
γ= | ||
| 6 |
| 1 | 1 | |||
V= | *a*b*H= | H | ||
| 3 | 3 |
| H | ||
tg60= | ⇔H=b*√3= | |
| b |
| H | ||
tg30= | ||
| a |
| √3 | ||
H=a* | ||
| 3 |
| √3 | ||
b√3=a* | ||
| 3 |
| 1 | ||
b= | a | |
| 3 |
| 1 | ||
a*b=1⇔a* | a=1 | |
| 3 |
| √3 | ||
H=√3* | =1 | |
| 3 |
| 1 | ||
6) V= | ||
| 3 |
A ja wolę trójkąty "ekierki"
a=3k√3, b=k√3, H=3k, k>0
| 1 | ||
P=ab=1 m2 ⇒ k√3*3k√3=1⇒ k= | to H=3k=1 m | |
| 3 |
| 1 | ||
V= | m3 | |
| 3 |