properties of circle(2題）（10分）

23.
(a)
PQ = RS (given)
arc PQ = arc RS (equal chords cut equal arcs)
arc PR + arc RQ = arc RQ + arc QS
Hence, arc PR = arc QS = 7 cm

Let C cm be the circumference of the circle.

Length of arc QR :
C x (48/360) = C - (12 + 7 + 7)
13C/15 = 26
C = 30

∠POR
= 360° x (7/30)
= 84°

(b)
arc RQ
= 30 - (12 + 7 + 7) cm
= 4 cm


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26.
Join DA.

BC = CD (given)
ΔCBD is an isoscelestriangle with ∠CBD = ∠CDB
Let ∠CBD = ∠CDB = a

In ΔCBD: ∠BCD + ∠CBD + ∠CDB = 180° (∠ sum of Δ)
∠BCD + a + a = 180°
∠BCD = 180° - 2a

∠TAD= ∠BCD(ext. ∠ of concyclic quad.)
∠TAD= 180° -2a

∠ADB= 90° (∠ in semi-circle)
∠TDA+ ∠ADB+ ∠CDB= 180°
∠TDA+ 90° + a = 180°
∠TDA= 90° - a

In ΔTDA: ∠BTC + ∠TAD + ∠TDA = 180° (∠ sum of Δ)
18° + (180° - 2a) + (90° - a) = 180°
3a = 108°
a = 36°
∠CBD= 36°

2012-10-07 15:16:09 補充：
1 樓 26 題的 「A+angle dcb+18度=180度(angle sum of triangle)」是錯誤的，並沒有任何一個 triangle 有此 angle sum。

23.RS=PQ (given)
arc rs=arc pq (equal chords,equal arc)
arc pq=arc rq+arc pr
arc rs=arc rq+arc qs
arc qs=arc pr
arc qs=7 (given)
arc qs=arc pr=7
arc ps=12 (given)
angle roq=48degree (given)
angle rop+angle roq+angle soq+angle pos=360度(angles at a pt.)
angle pos=312度-2 x angle rop
angle pos/12=angle roq/7 (arc prop. to angle at centre)
(312度-2 x angle rop)/12=angle roq/7
7x312度-14 x angle rop=12 x angle rop
2184度=26 x angle rop
angle rop=84度=angle por



rq/48度=7/84度
rq=4cm
26聼日做埋比你

2012-10-05 22:59:05 補充：
26.let angle cbd=angle cdb=A
2A+angle dcb=180度(angle sum of triangle)
angle dcb=180度-2A
A+angle dcb+18度=180度(angle sum of triangle)
180度-2A+ A+18度=180度
180度-A+18度=180度
A=18度







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