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標題:

maths 03 MC Q16,51,52,53

發問:

plz help me to solve Maths CE2003 mc Q 16, 51, 52 , 53 thanks~

最佳解答:

(16) 圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Apr09/Crazygeom1.jpg As shown in the diagram above, let r be the radius of the circumscribing circle and then, we have: r sin 22.5 = 3 r = 7.84 cm Then each isos. triangle has an area = 0.5 x 7.84 x 7.84 sin 45 = 21.73 cm2. So total area is 21.73 x 8 = 174 cm2. (51) Join AD and BC, we have: ∠XDA = ∠XBC and ∠XAD = ∠XCB (Ext. ∠s of cyclic. quad) Therefore △XDA ~ △XBC (AAA) Now we have: XD/XB = XA/XC 5/10 = 6/XC XC = 12 CD = 7 (52) Join AC, then ∠BCD = ∠CAD and ∠BAD = ∠ACD (∠s in alt. segment) Since ∠CAD + ∠ACD = 80 (∠s sum of triangle) ∠BCD + ∠BAD = 80 Finally, ∠ABC + ∠BCD + ∠BAD = 100 ∠ABC = 20 (53) From the figure, we have: △HDG ~ △HAF and △EBF ~ △ECG So, HD/HA = HG/HF HG/HF = 2/5 --> HG/(HG + GF) = 2/5 And EB/EC = EF/EG EF/EG = 3/7 --> EF/(EF + FG) = 3/7 Now, HG/(HG + GF) = 2/5 GF/HG + 1 = 5/2 GF = 3HG/2 EF/(EF + FG) = 3/7 FG/EF + 1 = 7/3 FG = 4EF/3 So, 3HG/2 = 4EF/3 9HG = 8EF EF : GH = 9 : 8

其他解答:

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