F3 Maths Problems

2015-04-11 6:08 pm

1) Figure 4 shpws two perpendicular lines AB and PR intersect at Q(3,a), with A and R lying on the x-axis and P lying on the y-axis. The coordinates of A and B are (-1,0) and (5,3) respectively.
(a) Find the value of a.
(b) FInd the equation of PQ.
(c) Find the area of Triangle AQR.
FIgure4:http://postimg.org/image/xdzn6umvn/

2) The mean test score of a class of n students is x. If a new student who scored 91is added to the class,the mean score become 71.
(a) Show that xn-71n+20=0
(b) If the mean score becomes 68 when another new student who scored 5 is added, find the values of n and x.

3)Figure 5 shows triangle ABC and three parallel lines such that AB//DE//FG.
(a) Show that triagnle CDE ~triangle CAH.
(b) Given CH is a median of triangle ABC, show that DE:FG=CE:CF.
Figure5:http://postimg.org/image/bialng2wd/

回答 (1)

pui

2015-04-11 9:58 pm

✔ 最佳答案

1a
斜率AQ=斜率AB
(a-0)/(3+1)=(3-0)/(5+1)
a/4=1/2
a=2

1b
the equation of PQ:
(斜率AB)(斜率PR)=-1
[(3-0)/(5+1)][(y-2)/(x-3)]=-1
0.5(y-2)/(x-3)=-1
y-2=-2x+6
y=-2x+8

1c
R的坐標=(x,0)
0=-2x+8
x=4
面積AQR=0.5(4+1)(3)
面積AQR=7.5

2a
n位同學的成績總和=nx
新成績的mean=71
(nx+91)/(n+1)=71
nx+91=71n+71
nx-71n-20=0

2b
新成績的mean=68
(nx+5)/(n+2)=68
nx+5=68=136
nx-68n-131=0 i
nx-71n-20=0 ii
ii-i
-3n+111=0
n=37
代n=37入ii
37x-71(37)-20=0
x=71.54054054

3a
角DCE=角ACH(同頂角)
角CDE=角CAE(同位角，DE//AB)
角CED=角CHA(同位角，DE//AB)
triagnle CDE ~triangle CAH (A.A.A)

3b
角DCE=角FCG(CH平均角ACB)
DE/CE=FG/CF
DE/FG=CE/CF
DE:FG=CE:CF

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