3 lip 07:54
jc:
(12+2):(3+4) = 2:1
3 lip 08:22
Dek: Nie wiem skad te x−sy , możesz wyjaśnić?
3 lip 08:50
Saizou :
ΔABC ≡ Δ EFG (bkb)
| | 4x | |
ΔGHI ~ΔABC w skali |
| = 4, zatem |
| | x | |
HI = 4b
HG = 4a
| | 2x | |
ΔIJA ~ΔABC w skali |
| = 2, zatem |
| | x | |
IJ = 2a
AJ = 2b
FG + GH = JA + AB
b + 4a = 2b+a
a = b
| JA+a | | 3a | | 1 | |
| = |
| = |
| |
| HI + IJ | | 6a | | 2 | |
3 lip 09:16
Saizou :
poprawka
...
b =3a,
| | 2b+a | | 7a | | 1 | |
zatem szukany stosunek to |
| = |
| = |
| |
| | 4b+2a | | 14a | | 2 | |
3 lip 09:25
Dek: Dzięki
3 lip 09:45
