Question

Problema 8
Punctula a

Answer 1

notam tangenta AB
O₁A=10
O₂B=5
ducem BA_I_ O₁A
DA=10-5=5
teorema lui Pitagora in triunghiul ADB
AB=√(15²+5²)=√(225+25)=√250=5√10

Answer 2

Fie  VT1  tangenta la   cele  2   cercuri.VT1∩C(O2,5)=T2
VT1_l_O1T1 (raza  e   perpendiculara pe   tangenta  in punctul de contact.).
Analog  VT2_l_O2T1=>
O1T1//O2T2=>
ΔVO1T1~ΔVO2T2 (Teorema  fundamentala asemanarii)=>
VO2/VO1=O2T2/O1T1=5/10=1/2
Vo2/VO1=1/2=> O2T2=linie mijlocie=>
VO1=2*O1O2=2*15=30cm
APlici  Teorema  Pitagora in  ΔVO1T1
VT1=√(VO1²-O1T1²)=√(30²-10²)=√800=2√2*10cm