math problems

圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/circle3.jpg?t=1184797244


a. ∠OPA = 90° (tangent ⊥ radius)
∠OPA = ∠ACS = 90°
∠OAP = ∠SAC (tangent property)

So, ΔAPO ~ ΔACS (AA)

b. Let r cm be the radius of the circle

AB = √(AC2 + BC2)
AB = √(242 + 182)
AB = 30 cm

Let BP be x cm
So, BQ = x cm (tangent property)
QC = (18 - x) cm
RC = QC = (18 - x) cm (tangent property)
AR = 24 – (18 - x) = (6 + x) cm
AP = AR = (6 + x) cm (tangent property)

AP + BP = AB

So, 6 + x + x = 30
x = 12

Draw OQ, OR,

OQ = OR (radius)
∠QOR = ∠OQC = ∠QCR = ∠ORC = 90° (tangent ⊥ radius)

So, OQCR is a square.

So, OQ = QC

i.e. r = 18 – x = 18 – 12 = 6

So, the radius of the circle is 6 cm.


c. AP = 6 + 12 = 18 cm
OP = r cm = 6 cm

Proved in (a), we have

AP / AC = OP / CS (corr. sides of ~ Δs)

i.e. 18 / 24 = 6 / CS
CS = 8 cm