急! Properties of Circles 12q6

19) TF = TG and TE = TH (Tangent properties)

Hence CE = HF

Extend AB and DC so that they meet at S, then SB = SC and SA = SD (Tangent Prop)

Hence AB = DC

20a) ∠OAQ = ∠OAR = 45 (Tangent properties)

OQ = OA sin 45

OA = 3√2 cm = 4.24 cm

b) OP + OA = 3√2 + 3 + 7.24 cm

21a) Diameter of the semi-circle = 10 cm, hence OC = 2 cm

b) Let r be the radius of Q, then

∠QCO = 90 (Radius perp. tangent)

OQ = √(r2 + 22) = √(r2 + 4)

OQ + QP = 10 (radius of semi-circle)

√(r2 + 4) + r = 10

(r2 + 4) = (10 - r)2

r2 + 4 = r2 - 20r + 100

20r = 96

r = 4.8 cm

22a) Drop a perp. from B to AP and let the foot be C, then BCPQ is a rectangle.

Also let r be the radius of circle with centre B, then:

BQ = r, AC = 9 - r, AB = 9 + r

Using Pyth. thm:

(9 + r)2 = (9 - r)2 + 122

r2 + 18r + 81 = r2 - 18r + 81 + 144

36r = 144

r = 4 cm

b) (4 + 9) x 12/2 = 216 cm2