Geometry Practice Test 1

Chord AB of the outer circle intersects inner circle at point C and R.
OP = 3 cm

Radius of the outer circle = r1 = 13 cm
Radius of the inner circle = 7 cm
In right triangle OPB,
(OP)2+(PB)2 = (OB)2
(3)2+(PB)2 = (13)2
PB2 = 169 − 9 = 160
PB = √160 = 4√10 cm
Also ΔOPC = (OC)2 = (OP)2+(PC)2
(7)2 = (3)2+(PC)2
49 - 9 = PC2
PC = √40 = 2√10 cm
AP = PB = 4√10 cm
CP = PD = 2√10 cm
AC = AP − CP
AC = 4√10−2√10
AC = 2√10 cm