maths~ hkcee

a i)
Arc CD in cm :
2 * π * (56 + 24) * (x/360) = 30π
160x/360 = 30
x = 30 * (360/160)
x = 67.5

a ii)
Area of ABCD
= (Area of sector DOC) - (Area of sector AOB)
= [π(56 + 24)² - π(56)²] * (67.5/360) cm²
= 612π cm²


b i)
Due to the similarity,
(Area of ABCD) : (Area of EFGH) = BC² : FG²
(612π cm²) : (Areaof EFGH) = 24² : 18²
Area of EFGH = 612π * (18/24)² = 344.25π cm²

b ii)
Arc GH : Arc CD = FG : CD
Arc GH : 30π cm = 18cm : 24 cm
Arc GH = 30π * (18/24)= 22.5 cm

Circumference of the base of the frustum :
2πr = 22.5cm
r = 22.5/(2π) cm
r = 11.25/π cm
(or 3.581 cm)

因輸入的問題 以下 V代表3.14........圓周率

ai)
r=80cm
(2Vr)(x/2V)=30V
80x=30V
x=3V/8

aii)
area= (Vr^2)(x/2V) - (VR^2)(x/2V)
=3/24 (6400V - 3136V)
=408V cm^2

Bi)
let A be the area of EFGH
A/408V = (18/24)^2
A=229.5V cm^2

bii)
18/24=GH/30V
GH=22.5V
Let O be centre of the circle of EFGH
(R+18)(3V/8)=22.5V
R=42cm

Let W be the radius of circle EF in figure c
42/60=W/r
W=0.7r

so V(r)(60) - V(0.7r)(42) = 229.5V
r=7.5cm