A
3cm 3cm
T₁ T₂
9 cm
B T₃ C
cercul inscris are punctele de tangenta T₁ , T₂ , T₃
atunci , stim AT₁ = AT₂ = 3 cm
BT₁ = BT₃ = 9 cm
CT₂ = T₃C = x
catetele AB = 12cm si AC = 3 +x
ipotenuza BC = 9 + x
Δ ABC drept : BC² = AB² + AC²
( 9 + x)² = 12² + ( 3 +x )²
81 + 18x + x² =144 + 9 + 6x+ x²
18x - 6x = 144 + 9 - 81
12x = 72
x = 72 : 12 = 6
AB = 12cm ; AC = 9cm si BC = 9cm + 6cm = 15cm