F2 Maths急求solution!! 20點!!!

let OF = r,
ie, OC = OB = OF = r (radii)
hence
OA² = OB² + OC² = 2r² (pyth. theorem)
OA = √2r
AF = OA + OF = √2r + r = (1+√2)r

shaded area = area of sector - area of inscribe circle
(2√2-1)π = π(AF)²/4 - π(OF)²
2√2-1 = (1+√2)²r² /4 - r²
2√2-1 = {[(3+2√2)/4] - 1}r²
2√2-1 = [(2√2-1)/4]*r²
hence r = 2
hence OC = r cm = 2cm ##

2012-06-04 16:19:56 補充：
001 答案錯在由尾數上去第五行
RHS唔記得乘以4

2012-06-04 16:23:15 補充：
更正 :

第四行 "OA² = OB² + OC² = 2r² (pyth. theorem)"
應為:

OA²
= OB² + AB² (pyth. theorem)
= OB² + OC² (prop. of square)
= 2r²

Answer: 1

Solution:

shaded area = (2√2 - 1)π
area of sector ADE - area of circle = (2√2 - 1)π
1/4 x (AF)^2 x π - (OC)^2 x π = (2√2 - 1)π
1/4 x (AO + OF)^2 x π - (OC)^2 x π = (2√2 - 1)π
1/4 x (√[(OB)^2 + (AB)^2] + OF)^2 x π - (OC)^2 x π = (2√2 - 1)π
1/4 x (√[(OC)^2 + (OC)^2] + OC)^2 x π - (OC)^2 x π = (2√2 - 1)π
1/4 x (√2(OC) + OC)^2 x π - (OC)^2 x π = (2√2 - 1)π
1/4 x (√2 + 1)^2 x (OC)^2 x π - (OC)^2 x π = (2√2 - 1)π
(√2 + 1)^2 x (OC)^2 - 4 x (OC)^2 = 2√2 - 1
[(√2 + 1)^2 - 4] (OC)^2 = 2√2 - 1
(2√2 - 1) (OC)^2 = 2√2 - 1
(OC)^2 = 1
OC = 1