Circle (DSE Maths)
2014-08-21 1:19 am
O is the centre of the circle. AOB is the diameter of the circle, and it intersects the chord CD at M. If CM= 8 cm, MD= 6cm and OM=MA, find the radius of the circle.
回答 (3)
2014-08-21 3:15 am
✔ 最佳答案
The answer is as follows :圖片參考：https://s.yimg.com/lo/api/res/1.2/0b.2s4YFYSC26ktM5OVVhQ--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/https://farm4.staticflickr.com/3869/14955791576_1eab0048b4_o.png
https://farm4.staticflickr.com/3869/14955791576_1eab0048b4_o.png
2014-08-22 03:03:03 補充：
If you do not know intersecting chords, use the following 2 paragraphs to replace the 2nd paragraph.
Join AD and BC.
∠AMD = ∠CMB (vert. opp. ∠s)
∠BAD = ∠DCB (∠s in same segment)
∠DAB = ∠BCD (∠s in same segment)
ΔAMD ~ ΔCMB (AAA)
AM / CM = MD / MB
(r/2) / 8 = 6 / (3r/2)
3r²/4 = 48
r² = 64
r = 8
參考: pingshek, pingshek
2014-08-21 6:04 am
Let OC=OD=r (radii)
∠MCO=∠MDO=θ (base ∠s,isos△)
Using cos law,
MC²+OC²-2(MC)(OC)cos∠MCO=OM²
8²+r²-16rcosθ=r²/4 ... 1
MD²+OD²-2(MC)(OD)cos∠MDO=OM²
6²+r²-12rcosθ=r²/4 ... 2
Combine 1&2,
8²+r²-16rcosθ=6²+r²-12rcosθ
cosθ=7/r ... 3
Sub 3 into 1
8²+r²-16r(7/r)=r²/4
3r²/4=48
r=8
∠MCO=∠MDO=θ (base ∠s,isos△)
Using cos law,
MC²+OC²-2(MC)(OC)cos∠MCO=OM²
8²+r²-16rcosθ=r²/4 ... 1
MD²+OD²-2(MC)(OD)cos∠MDO=OM²
6²+r²-12rcosθ=r²/4 ... 2
Combine 1&2,
8²+r²-16rcosθ=6²+r²-12rcosθ
cosθ=7/r ... 3
Sub 3 into 1
8²+r²-16r(7/r)=r²/4
3r²/4=48
r=8
2014-08-21 5:36 am
pingshek 答得好好！
╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\\╯
同學如果不明白 Intersecting chord 的話，可以考慮
△MAD ~ △MCB
╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\\╯
同學如果不明白 Intersecting chord 的話，可以考慮
△MAD ~ △MCB
收錄日期: 2021-05-01 10:01:49
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