中4maths 7C 4,6,7,8

(4) By Pyth. theorem, O'A = √(42 + 32) = 5
(6) By tangent properties, we have:
OR is perp. to RS
OP is perp. to PQ
MQ is perp. to PQ
MS is perp. to RS
Also let the point of intersection between PQ and RS be T, then PORT and SMQT are both cyclic quad.
Hence x = y = 60° (Ext. angle of cyclic. quad)
(7) Join PD, then PD and QE are perpendicular to DE by tangent properties and hence PD//QE.
So, ∠DPF = ∠FQE (Alt. ∠s, PD//QE)
∠PDF = ∠QEF (proved)
PD = QE (Given)
△PDF and △QEF are congurent (ASA)
So DF = FE = 8 cm
PF = FQ = (1/2)√(62 + 82) = 5 cm
(8a) Let r be the radius of the smaller circle.
Also, join AB and BQ, then BQ = r and AB = 9 + r
Draw a perpendicular from B to AP and let the foot be C, then BC = PQ = 12 cm and AC = 9 - r
By Pyth theorem, AB2 = AC2 + BC2
(9 + r)2 = (9 - r)2 + 122
81 + 18r + r2 = 81 - 18r + r2 + 144
36r = 144
r = 4 cm
(b) ABQP is in fact a trapezium and therefore its area is:
(1/2) × (9 + 4) × 12 = 78 cm2.

PF = FQ = (1/2)√(62 + 82) = 5 cm
This is wrong, the correct solution is
PF = FQ = √(6^2 + 8^2) = 10 cm and
PQ = 2 PF = 20 cm

4. OO' = 4-1 = 3cm
O'A = (4^2 - 3^2)^1/2 = 5 cm


6. angle ORS = angle OPQ = 90 degree
y = 60 degree
angle MQP = angle MSR = 90 degree
x = 60