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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 ## Right triangle problem tagged by: ##### This topic has 8 member replies For right triangle RST, RS and RT are legs and RT+RS=12. ST is hypotenuse with length 10. What is the area of the right triangle? A) 24 B) 11 C) 15 D) 2*sqrt(30) E) 12 Newbie | Next Rank: 10 Posts Joined 09 Aug 2009 Posted: 3 messages Is it B Senior | Next Rank: 100 Posts Joined 28 Jan 2010 Posted: 97 messages Upvotes: 9 mattocks wrote: For right triangle RST, RS and RT are legs and RT+RS=12. ST is hypotenuse with length 10. What is the area of the right triangle? A) 24 B) 11 C) 15 D) 2*sqrt(30) E) 12 RS + ST = 12 using pythagoras theorem 100 = x^2 + (x-12)^2 solving we get X as 6+sqrt(14) and 6-sqrt(14) Area = 1/2(6+sqrt(14) and 6-sqrt(14)) = 11 Hence B Master | Next Rank: 500 Posts Joined 05 Jul 2010 Posted: 324 messages Followed by: 3 members Upvotes: 70 Target GMAT Score: 720+ Area of triangle = 1/2 x RS x RT Now, ST^2 = RS^2 + RT^2 RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS) Thus, ST^2 = (12)^2 - 2xRTxRS => 2RTxRS = 100 - 144 => RT x RS = 22 => 1/2 x RT x RS = 11 This, answer is (B) In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle) Junior | Next Rank: 30 Posts Joined 07 Mar 2010 Posted: 11 messages A circle is circumscribed about an equilateral triangle of side 'a'. The diameter of the circle is? A. a*root3 B. a/root3 C. 2a/root3 D. 2*root3(3a) E. 3root3(a) Senior | Next Rank: 100 Posts Joined 22 Jun 2010 Posted: 36 messages Upvotes: 1 GMAT Score: 760 mattocks wrote: A circle is circumscribed about an equilateral triangle of side 'a'. The diameter of the circle is? A. a*root3 B. a/root3 C. 2a/root3 D. 2*root3(3a) E. 3root3(a) Option (C). Circumradius = (2/3 * length of altitude). We can remember this as a rule. Junior | Next Rank: 30 Posts Joined 11 Jul 2010 Posted: 12 messages Upvotes: 2 Target GMAT Score: 730+ GMAT Score: 630 Hey there, my answer is C. Diameter of circumscribed circle of any triangle is abc/(2*area of triangle), where a, b, c are the sides of the triangle. 1. We know that all the sides are equal and length = a 2. To find the area of the triangle, use the 30 60 90 triangle rule, whereby the height will be (root3*a)/2 Therefore, the area of the triangle = 1/2 base* height = 1/2*a*(root3*a)/2) = (a*a*root3)/4 3. Diameter= (a*a*a)/((a*a*root3)/4) = 2a/root3 [/u] Senior | Next Rank: 100 Posts Joined 10 May 2010 Posted: 51 messages Upvotes: 4 Target GMAT Score: 750 kmittal82 wrote: Area of triangle = 1/2 x RS x RT Now, ST^2 = RS^2 + RT^2 RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS) Thus, ST^2 = (12)^2 - 2xRTxRS => 2RTxRS = 100 - 144 => RT x RS = 22 => 1/2 x RT x RS = 11 This, answer is (B) In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle) Now that is an interesting approach and the takeaway is really helpful, thanks Legendary Member Joined 29 Aug 2009 Posted: 758 messages Followed by: 2 members Upvotes: 67 kmittal82 wrote: Area of triangle = 1/2 x RS x RT Now, ST^2 = RS^2 + RT^2 RS^2 + RT^2 = (RS+RT)^2 - (2xRTxRS) Thus, ST^2 = (12)^2 - 2xRTxRS => 2RTxRS = 100 - 144 => RT x RS = 22 => 1/2 x RT x RS = 11 This, answer is (B) In other words, (hypotenuse)^2 = (sum of sides)^2 - (4x area of triangle) Nice solution without any approximation.. _________________ I am on a break !! • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • 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 • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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