jjaacc Posted May 9, 2006 Share Posted May 9, 2006 Could someone help with this question please. I know my basic geom but I guess there is some formula I don't know of? Thanks C Quote Link to comment Share on other sites More sharing options...
Cloudnineast Posted May 9, 2006 Share Posted May 9, 2006 There is a formula n/360 = Arc ABC/Circumference(pi d) Im not sure if that's what needs to be used here. But since its eq triangle n would be the measure of the angles that faces the arc. Quote Link to comment Share on other sites More sharing options...
GMAT-HELP Posted May 9, 2006 Share Posted May 9, 2006 Length of the arc ABC = 24 (given) Let O be the centre of the circle..OA = OB = OC = radius arc ABC = arc AB+ arc BC arc OAB = 120 degrees => length AB = 120/360(2*pi*r) arc OBC = 120 degrees => length BC = 120/360(2*pi*r) summing up, we get 2/3(2*pi*r) = 24 => 2r = 36/pi = 11.46 (Ans C) Note: For the geometry, draw a line from O to the midpoint of AB, u'll get a 30-60-90 triangle and u can compute arc OAB : 120 degrees Quote Link to comment Share on other sites More sharing options...
meb2183 Posted May 10, 2006 Share Posted May 10, 2006 should be simpler really... you know that the length of an arc is directly proportional to the inscribed angle creating it. further we know that the triangle is equilateral so all angles are the same and thus the three arcs will have the same length (12). from here you just plug into the pi*d = circum (36=pi*d ~ 11) Quote Link to comment Share on other sites More sharing options...
jjaacc Posted May 14, 2006 Author Share Posted May 14, 2006 Thanks gmathelp & meb....very good explanations Quote Link to comment Share on other sites More sharing options...
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