bjr
on applique le théorème de Pythagore au triangle abc rectangle en A
AB = 1
AC = 1
BC² = BA² + AC²
BC² = 1² + 1²
BC² = 2
BC = √2
si on continue
• triangle BCD ( CD = 1)
BD² = BC² + CD²
BD² = 2 + 1²
BD² = 3
BD = √3
• triangle BDE (DE = 1)
BE² = BD² + DE²
BE² = 3 + 1²
BE² = 4
BE = √4 (= 2)
• triangle BEF (le dernier triangle)
BF² = BE² + EF²
BF² = 4 + 1²
BF² = 5
BF = √5
observe les mesures des segments
BC = √2
BD = √3
BE = √4
BF = √5
