F.2 Maths question

1.
πr²×θ÷360°= 36 cm²-----------------------(1)
2πr×θ÷360°= 12 cm-------------------------(2)
(1)÷(2): r÷2 = 3 cm
r = 6 cm

sub r=6cm into (2), we have
2π(6 cm)×θ÷360°= 12 cm
θ = 360°÷π

2.
level in cylinder A=20cm
∴x=20

level in cylinder B=5cm
∴level in cylinder B=x÷4

2007-05-27 22:42:25 補充：
1.180°= π rad.∴θ = 2π÷π = 2∴如果:θ is measured in degree, θ = 360°÷πθ is measured in radian, θ = 2

1) 2pi*r* θ/360=12- - - - - 1
r^2*pi* θ/360=36- - - - -2
according 1&2,
6pi*r=r^2*pi
r=6
substitute in 1
2(6)*pi* θ/360=12
θ=360/pi
2) base area A:B
= 1:4
height of all liquid in A:B
40cm:10cm
if the level of the liquid in A=xcm
B=(40-x)/4

1) πr^2 x θ/360 = 36 ... (1)
2πr x θ/360 = 12 ... (2)
θ = 2160/πr ... (3)
sub (3) into (1):
πr^2 x (2160/πr)/360 = 36
r x 2160 = 12960
r = 6 cm
sub r = 6 into (3):
θ = 2160/6π
θ = 360/π
θ = 115 (3 sig.fig.)

2) the volume of measuring cylinder A:
π5^2 x 20 = 500π
the volume of measuring cylinder B:
π10^2 x 5 = 500π
Let y be the level of the liquid in cylinder B.
25πx = 100πy
x = 4y
y = x/4

πr^2*θ/360=36 ..............1
2πr*θ/360=12 ................2
from1
πr^2*θ/360=36
θ=12960/πr^2 ................3
sub 3 into 2
2πr*12960/πr^2=4320
25920=4320r

r=6 ............................4
sub 4 into 2
2π6*θ/360=12
θ=360/π
θ=115(cor to 3 sig fig)


2.
20=x
5=x/4

Area = pi(r^2)*(θ/360) = 36
=> r = sqrt[36(360)/pi(θ)]

arc length AB = 2pi (r)*(θ/360) = 12
=> r = 12(360)/(2piθ)

兩條式兩個unknown搞掂~~~~

θ = 360/pi
r = 6

1)
L = rθ
A = 1/2 r^2 θ = 1/2 r L

36 = 1/2 r (12)
r = 6 cm

2)
V = pi r^2 h

pi (5)^2 x = pi (10)^2 h
h = x/4

唔知你有無打錯野,個20cm同10cm用唔著

2007-05-27 21:45:45 補充：
計少個θ12 = 6θθ = 2 radian