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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Geometry Problem ##### This topic has 1 expert reply and 3 member replies ABCD is a circle and circles are drawn with AO, CO, DO and OB as diameters. Areas E and F are shaded. E/F is equal to a. 1 b. 1/2 c. 1/π d. π/4 e. π/2 ### GMAT/MBA Expert GMAT Instructor Joined 21 Jan 2009 Posted: 3650 messages Followed by: 82 members Upvotes: 267 GMAT Score: 760 vikrantr93 wrote: ABCD is a circle and circles are drawn with AO, CO, DO and OB as diameters. Areas E and F are shaded. E/F is equal to a. 1 b. 1/2 c. 1/π d. π/4 e. π/2 Good question! Things to remember here are the areas of sectors and segments of a circle. Let’s take AD = BC = 2 r so that OA = OB = OC = OD = r and hence each included common chord (not named) is r √2/2. Area of each of the 8 segments = area of sector - area of triangle = ¼ X π X r^2 - ½ X r √2/2 X r √2/2 = ¼ X r^2 (π - 1) Hence total shaded F = 8 X ¼ X r^2 (π - 1) = 2 r^2 (π - 1). Area of big circle = π X (2 r) ^2 = 4 π r^2 Area inside the big circle EXCEPT the shaded E = area of 4 small circles - total shaded area F = 4 X π r^2 - 2 r^2 (π - 1) = 2 π r^2 + 2 r^2 = 2 r^2 (π + 1) Hence, total shaded area E = 4 π r^2 - (2 π r^2 + 2 r^2) = 2 π r^2 - 2 r^2 = 2 r^2 (π - 1). Hence E/F is equal to [2 r^2 (π - 1)]/ [2 r^2 (π - 1)] = 1 A _________________ The mind is everything. What you think you become. –Lord Buddha Sanjeev K Saxena Quantitative Instructor The Princeton Review - Manya Abroad Lucknow-226001 www.manyagroup.com Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Legendary Member Joined 17 May 2011 Posted: 1448 messages Followed by: 53 members Upvotes: 375 vikrantr93 wrote: ABCD is a circle and circles are drawn with AO, CO, DO and OB as diameters. Areas E and F are shaded. E/F is equal to a. 1 b. 1/2 c. 1/π d. π/4 e. π/2 Hi, I dont know if some of the unshaded regions marked as F are mistyped as G. The question states E and F are shaded regions. so, I am taking for granted that the 4 inner shaded regions are F and the four outer shaded regions are E Let the radius of smaller circles be r. So, the radius of outer circle is 2r Adding the the areas of 4 smaller circles gives 4.πr^2 = unshaded region + 2.F (as each shaded F is repeated twice) So, 4.πr^2 = unshaded region + 2F --eqn(1) The area of outer circle is π(2r)^2 = 4πr^2 = unshaded region+F+E --eqn(2) From eqns(1) & (@) we get 4.πr^2 = unshaded region + 2F = unshaded region+F+E => E = F. So, E/F = 1 Hence, A Cheers! Legendary Member Joined 04 Apr 2011 Posted: 1309 messages Followed by: 123 members Upvotes: 310 Test Date: 13th Oct Target GMAT Score: 750+ GMAT Score: 750 Isn't there a problem with the question?? Legendary Member Joined 17 May 2011 Posted: 1448 messages Followed by: 53 members Upvotes: 375 cans wrote: Isn't there a problem with the question?? Hi, I think some of the unshaded regions are mistakenly marked 'F' instead of 'G'. But, in the ques, it is mentioned that E and F are marked regions. So, I have assumed that in my post. Moreover, the options seem to be having 'n'. In fact, it is π(Pie). Cheers! • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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