Suma długości środkowych trójkąta.
1) WΔABC:
a2=b2+c2−2*b*c*cosα⇔32=22+42−2*2*4*cosα
| 11 | ||
cosα= | ||
| 16 |
| 7 | ||
cosβ= | ||
| 8 |
| 11 | ||
sb2= 12+42−2*1*4* | ||
| 16 |
| 23 | ||
sb2= | ||
| 2 |
| √46 | ||
sb= | ||
| 2 |
| 11 | ||
sc2=22+22−2*2*2* | ||
| 16 |
| 5 | ||
sc2= | ||
| 2 |
| √10 | ||
sc= | ||
| 2 |
| 7 | ||
sa2=42+1.52−2*4*1.5* | ||
| 8 |
| √31 | ||
sa= | ||
| 2 |
| √31+√46+√10 | ||
s= | ||
| 2 |
https://pl.wikipedia.org/wiki/%C5%9Arodkowa_tr%C3%B3jk%C4%85ta
| 1 | ||
sa= | √2b2+2c2−a2 | |
| 2 |
| 1 | √31 | |||
sa= | √8+32−9= | |||
| 2 | 2 |