F5 MATH* ratio & area questions

Q1.
Let C1 be the circle passing through A and B, C2 be the one passing through A and C and C3 be the one passing through B and C.
Let O1, O2, O3 be the centre of C1, C2 and C3 respectively.
Join O1 to A and B, O2 to A and C, O3 to B and C.
O1A=O2A=O2C=O3C=O3B=O1B= 1cm.
This forms a regular hexagon.
∴∠AO1B = ∠AO2C=∠BO3C= (6-2)180/6 = 120 (∠sum of polygon)
Area of sector AO1B = Area of sector AO2C= Area of sector BO3C
=π(1)^2 X 120/360 = π/3
The total areas of the 3 sectors overlaps with each other. The overlapped area is equal to the shaded area. You may try to draw the diagram to see if it is true.
Therefore,
(π/3) X 3 = Area of the regular haxagon + overlapped area
π = 6 X 0.5(1)(1)sin60 + shaded area （property of eq.△）＜－ Area of six eq.△ of which each angle equals to 60
Overlapped area = π-3sin60 = π - (3root3 )/2
[If the shaded regoin is the overlapped area, this is the answer]
If the shaded regoin is the non-overlapped area, the answer is
π(1)^2 x3 - [π - (3root3 )/2]
=2π + (3root3 )/2
But your answer with root 2 should be wrong I think.

Q2.
EB //DC
∠EBF=∠CDF (alt.∠s, EB //DC)
∠FEB=∠FCD(alt.∠s, EB //DC)
∠BFE=∠DFC(vert. opp. ∠s)
∴△EFB~△CFD（AAA）
Area of△ADE :Area of△CDB
=1/2 (AD)(DE) : 1/2(BC)(CD)
Since AD=BC and CD = 4DE
Area of△ADE :Area of△CDB
=1:4

Area of△CFB :Area of△CFD
=BF:FD
=EB:CD (corr. sides, ~△s)
=3:4

Area of△CFD : Area of△CDB
=Area of△CFD :Area of△CFB +Area of△CFD
=4:3+4
=4:7

Area of△ADE : Area of△CDB = 1:4
Area of△CFD : Area of△CDB = 4:7
Area(△ADE) / Area (△CFD)
=[Area(△ADE)/Area(△CBD)] / Area(△CFD)/Area(△CBD)
=(1/4) / (4/7)
= 7/16

2007-04-19 02:33:49 補充：
一滴step都唔跳嫁喇…希望你睇得明啦=]

2007-04-19 02:38:17 補充：
會考喇…加油=] 我都要會考簡單到AAA既野都要寫足proof 唔係一定扣分…順帶一題…如果你識砌下野d overlapped region…其實佢地可以擺落1個圓形入面圍住個邊… 淨返既位就係一個regular hexagon…同我上面所講果個係全等既…

Q1
AB = AC = BC = sqrt [1^2 + 1^2 - 2 * 1 * 1 cos(120 deg)] = sqrt(3) --- [1] [2]

The area of Equilateral Triangle ABC is 1/2 * sqrt(3) * sqrt(3) * sin(60 deg)
= 3 * sqrt(3) / 4

Area of segment of each AB, AC and BC is 1/3 * pi * 1^2 - 1/2 * 1 * 1 * sin(120 deg)
= 1/3 * pi - sqrt(3) / 4

The area of the shaded part = Area of segment of AB, AC and BC - The area of Equilateral ABC
= 3 (1/3 * pi - sqrt(3) / 4) - 3 * sqrt(3) / 4
= pi - 3 * sqrt(3) / 4 - 3 * sqrt(3) / 4
= pi - 3 * sqrt(3) / 2

Remark:
[1] draw a line joints each center of circle and pass through A, B and C forms regular hexagon, internal angles of it is 120 deg

[2] Cosine Law

Q2
Triangle BEF ~ Triangle DCF (AAA)
BE : CD = 3 : (3 + 1) = 3 : 4
Area of EBF : Area of CDF = 3^2 : 4^2 = 9 : 16

Area of EBF : Area of CBF = 3 : 4 (since heights are same)

Area of CBE : Area of CDF = (4 + 3) * 3 : 16 = 21 : 16

Area of CBE : Area of ADE = 3 : 1 (since heights are same)

Therefore Area of ADE : Area of CDF = 21 / 3 : 16 = 7 : 16
or Area of ADE / Area of CDF = 7 / 16