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[Maths] 1996 Maths MC - 27,46,47,52

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1996 Maths MC - 27,46,47,52

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27. A
shortest distance can be found by considering the height and diameter of the cylinder as the radius of a circle. Since angle CAB = 40 degree
therefore, required length = 2(pi)(r)(1/4) = 2(4)(1/4) (pi) = 2pi
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27)
AC=4pi/2=2pi
BC is the shortest length
So
BC=[(2pi)^2+4^2]^(1/2)=(4pi^2+16)^(1/2)=2(pi^2+4)^(1/2)
ANS: B

46)
加入點G及H使CH垂直BD及DG垂直AC
Area of triangle CDE/Area of triangle BCE
=(1/2)(DE)(CH)/(1/2)(BE)(CH)=1/2
So
DE/BE=1/2
BE=2DE
triangle CDE is similar to triangle ABE(AAA)

Because BE=2DE, so AE=2CE

SO
Area of triangle CDE/Area of triangle ADE=1/2

Area of triangle CDE: Area of triangle ABE=1^2:2^2=1:4

So
Area of triangle CDE/Area of trapezium ABCD
=1/(1+2+4+2)
=1/9
ANS: B

47)By the concept of sin cos tan
tan35=y/x
tan(35+theta)=2y/x=2(y/x)=2tan35
tan(35+theta)=1.4004
35+theta=54
theta=54-35=19
ANS: B

52)
Join CG to make AB and CG parallel
Because BC=CD and CG is parallel BF, so DG=GF

Because angle AGE=angle EAF, angle GEC=angle AEF and CE=EA
so triangle CEG is equal to triangle AEF(ASA)

SO
GE=EF=(1/2)GF=(1/2)DG
SO
DE:EF
=(DG+1/2DG):1/2DG
=(3/2)DG1/2)DG
=3:1
ANS: C

[ 本帖最後由 021044 於 2010-3-25 09:55 AM 編輯 ]

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