急! Properties of Circles 12q10

19.
Let BP = x cm and CR = y cm
19.
The two tangents from a point to the circle are equal in length.
AP = AR = 15 cm
BQ = BP = x cm
CQ = CR = y cm

The perimeter of ΔABC
= AB + BC + AC
= (AP - BP) + (BQ + CQ) + (AR - CR)
= (15 - x) + (x + y) + (15 - y) cm
= 30 cm


20.
(a)
Draw the common tangent HRK, and join AR and RB.
∠ARH = 90° (tangent⊥radius at point ofcontact)
∠BRH = 90° (same as above)

∠ARB = ∠ARH + ∠BRH
∠ARB = 90° + 90°
∠ARB = 180°
ARB is a st. line. (adj. ∠s supplementary)

(b)
∠PAR = 2∠PQS (∠ at centre = 2∠ at circum.)
∠PAR = 2 x 40°
∠PAR = 80°

∠APT = 90°(tangent⊥radius at point ofcontact)
∠BTP = 90° (same as above)

In quad. ABPT :
∠TBR + ∠PAR + ∠APT + ∠BTP = 360° (∠ sum of quad.)
∠TBR + 80° + 90° + 90° = 360°
∠TBR = 100°

∠TBR = 2∠TSQ (∠ at centre = 2∠ at circum.)
∠TSQ = 100° x (1/2)
∠TSQ = 50°


21.
(a)
Let ∠RLN = 5k°
Then ∠NPT = 4k°

∠RQN = ∠RLN (∠s in same segment)
∠RQN = 5k°

∠PQN = ∠NPT (∠s in same segment)
∠PQN = 4k°

∠RQN + ∠PQN = 180° (adj. ∠s on st. line)
5k° + 4k° = 180°
k = 20
5k° = 100°
∠RLN = 100°

(b)
∠NPT = 4k°
∠NPT = 4 x 20°
∠NPT = 80°