skalevar wrote:If the radius of the circle above (see attachment) is equal to the chord AB, then what is the value of x?


I don't know where to start here. Is there some relationship between chords and radii that I have somehow missed?

A. 25
B. 30
C. 40
D. 45
E. 50

OA is B from Ready for Gmat.com


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its very simple... this is how i solve it...
O is centre of the circle,
because AB=r thus the triangle AOC is equilateral triangle thus angle AOB=ABO^=OAB^ so OAC^=10
(C is the intersection between AC and BC )
because triangle AOC is isosceles triangle thus OAC^=OCA^=10 so angle AOC=160
so angle BOC=160-AOB=160-60=100
100+OBC^+10+x=180
or OBC^+x=70
futhermore OBC^=x+10
thus > 2x+10=70 so x=30

Thanks - it makes sense now. although it took me almost 10 min to figure it out, even with the answer!