Réponse :
l= BF
l1=OF
l2 = FK
1) A milieu de BF
→AB=AF=3m
→l = AB+AF = 3+3=6m
2) l1 =OF
l1 = BF-BO= 6-5 =1
→l1 =1m
3a) l2= FK
OK rayon du cercle de centre O=AO=2m
OK² =OF²+(l2)²
OK² = (l1)²-(l2)²
2² = 1²-(l2)²
(l2)² = 2²-1²=3
l2 = √3 ≈1,72
→l2=1,72m
b) L =FD=l2+KD=1,72+6 =7,72
→L = 7,72m
4a) A1 = L*BF = 7,73*6 = 46,38m²
b) A2=(l2*l1)/2 = 0,865m²
Ap = 7,72+0,865+8,38 = 16,96 = 17m² arrondi à l'unité
→Ap = 17m²
