05 mc ce maths Q.29.Q43－百科全書

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05 mc ce maths Q.29.Q43

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05 mc ce maths Q.29.Q43 解釋比我聽

最佳解答:

http://www.yy1.edu.hk/~yy1-mat/math05_mc.pdf 這裹有齊05年MC的做法

其他解答:

Q29 ∵△OFA~△ODC(AAA) OFFA1 ∴----- = ------- = ------ (corr. sides,~△s) ODDC5 ∴OF:OD=1:5 ∴OF:FD=1:4 ∵OA:AC=OF:FD(intercept theorem) =1:4 ∴Let OA=k and AC=4k. ∵AB:BC=2:3 ∴Let AB=2m and BC=3m ∵AB+BC=AC 2m+3m=4k 4 ∴m=-----k 5 8 AB=2m= -------k 5 8 OA:AB=k:----------k 5 OA:AB=5:8 Q43 Let AB=x and the perpendicular height from CD of ABDC be h. (3+2) X perpendicular height from CD of CDFE Area of CDFE=--------------------------------------------------------------------=5 2 perpendicular height from CD of CDFE=2 (x+3)h Area of ABDC=----------=16 2 32 h= --------- ..........................(*) x+3 Height of ABFE =perpendicular height from CD of CDFE+perpendicular height from CD of ABDC=h+2 (x+2)(h+2) Area of ABFE=----------------- =16+5 2 (x+2)(h+2)=42 Sub. (*) into the above equation, 32 (x+2)( ------------- +2)=42 x+3 32+2x+6 (x+2)(----------------)=42 x+3 (x+2)(38+2x)=42(x+3) 2x2+38x+4x+76=42x+126 2x2+42x+76=42x+126 2x2=50 x2=25 x=5 AB=5cm