Maths 06 ce mc question

) Join OB , then ㄥBOA = 60° since OA/OB = 1/2 = cos60°,
ㄥCOB = 90 - 60 = 30°
OABC = 扇形 OBC + △OBA
= π (2^2) (30/360) + (1/2) (2)(1)sin60°
= π/3 + √3 / 2 cm^2

2)Let r be the radius of the in-circle , then the in-centre is (r , r)

Since △ABO is an right-angle△,

2r = OA + OB - AB

2r = 6 + 6 - √(6^2 + 6^2)

2r = 12 - 6√2

r = 6 - 3√2

the in-centre is ( 6 - 3√2 , 6 - 3√2)