• NEW! FREE Beat The GMAT Quizzes
Hundreds of Questions Highly Detailed Reporting Expert Explanations
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
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 ## MGMAT Question Bank Geometry: 20 ##### This topic has 1 expert reply and 6 member replies ## MGMAT Question Bank Geometry: 20 Hi all, I came across question number 20 in the geometry question bank section and would like to see if my analysis is correct. Below is the question: What is the length of segment BC? (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30. In the diagram, there is a triangle ABC, with side AB=5 and side AC=13. Side BC is not labeled. None of the 3 angles is labeled. The answer is supposed to be A. I wonder why statement (2) alone isn't enough to solve this question. I envision that the height of the triangle can be calculated using area=30 and a base of 13. With that, the triangle can be split into two smaller right triangles, one with hypotenuse AB=5, and other one with hypotenuse BC. x^2 + y^2 = z^2 can then be used to find the unknown side of triangle that has hypotenuse AB. With that, BC can then be eventually calculated. This method alone does not assume that angle ABC is 90 deg. Is there something wrong with this method? With this, I answered D for this question. I uploaded the image below, hopefully everyone can see it. Thanks everyone for looking into this. Legendary Member Joined 20 Jun 2007 Posted: 1153 messages Followed by: 2 members Upvotes: 146 Target GMAT Score: V50 I agree with you. Answer should be D. Master | Next Rank: 500 Posts Joined 11 Jun 2008 Posted: 418 messages Upvotes: 65 x^2 + y^2 = z^2 only holds true for right-angled triangles. You can't assume this for all triangles. In statement 2, the information for angle ABC is not given. Without this information, you dont know if it is a right angled or not. Therefore, only statement 1 is sufficent. OA is A. Legendary Member Joined 20 Jun 2007 Posted: 1153 messages Followed by: 2 members Upvotes: 146 Target GMAT Score: V50 bluementor wrote: x^2 + y^2 = z^2 only holds true for right-angled triangles. You can't assume this for all triangles. In statement 2, the information for angle ABC is not given. Without this information, you dont know if it is a right angled or not. Therefore, only statement 1 is sufficent. OA is A. Thats true, but without any assumptions, if we draw a perpendicular from angle ABC to the base AC, you get 90 degrees angle. The perpendicular is the height of the triangle and then you can use the pythagoreous theorem, since area is given. Any reasoning why this is wrong. Thanks. Legendary Member Joined 07 Jul 2008 Posted: 829 messages Followed by: 3 members Upvotes: 84 Target GMAT Score: 700+ bluementor wrote: x^2 + y^2 = z^2 only holds true for right-angled triangles. You can't assume this for all triangles. In statement 2, the information for angle ABC is not given. Without this information, you dont know if it is a right angled or not. Therefore, only statement 1 is sufficent. OA is A. i go with A as well.. Master | Next Rank: 500 Posts Joined 26 Feb 2008 Posted: 294 messages Followed by: 1 members Upvotes: 13 Target GMAT Score: 750 I feel the ans should be "A" here. From stem 2: Area of triangle is 30 The altitute drop can be either from angle B or from angle C as well.... If we try with altitude that drops from angle B then we get the value of BC as 12 which is same as from stem 1 value. But when we try with altitude droping from angle C ... area=1/2*AB*CD=30=>CD=30*2/AB=30*2/5=12 So AD should be 5 here (applying Pythagoras theorm) but this contradicts as AB itself is 5 so AD has to be less than 5. To get a definite value of BC it has to satisfy with respect to all perceived altitues droping from any of the three angles. Amit Attachments This post contains an attachment. You must be logged in to download/view this file. Please login or register as a user. ### GMAT/MBA Expert GMAT Instructor Joined 02 Jun 2008 Posted: 2525 messages Followed by: 352 members Upvotes: 1090 GMAT Score: 780 Using the second statement, there are only two possible values of BC. One of these values is 12, and one is quite a bit larger. The diagram is misleading; the only way to find the second possible triangle is to let the angle at A be greater than 90 degrees. Look at the problem this way: put A at the origin on the xy plane, and C at (13,0). Draw a circle of radius 5 around the origin: B must be somewhere on this circle, since it is 5 away from A. Now, if the co-ordinates of point B are (g, h), then h is clearly the height of the triangle. If we assume B is somewhere above the x-axis, there are only two possible points on the circle that give the correct height- one in the first quadrant (x > 0) which will make BC = 12, and one in the second quadrant (where x is negative), which will make angle A much bigger than 90 degrees. In each case we get a different value for the length of BC. _________________ If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Junior | Next Rank: 30 Posts Joined 26 Dec 2007 Posted: 15 messages Hi Ian, Thanks for your explanation. I understand my mistake now. I drew out what you described. At least with this drawing that I can see that I only considered the first (top) case, where I assumed the height of the triangle would split AC into two segments, one having a length of x and the other having the length of 13-x. In the second (bottom) case, the two lengths of the two triangles formed are x and 13. It is clear that both triangles ABC can have area 30, since they both have the same base and same height. Thanks for your explanation. • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• FREE GMAT Exam
Know how you'd score today for \$0

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 Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

### Top First Responders*

1 Ian Stewart 41 first replies
2 Brent@GMATPrepNow 40 first replies
3 Scott@TargetTestPrep 39 first replies
4 Jay@ManhattanReview 32 first replies
5 GMATGuruNY 26 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

### Most Active Experts

1 Scott@TargetTestPrep

Target Test Prep

159 posts
2 Max@Math Revolution

Math Revolution

92 posts
3 Brent@GMATPrepNow

GMAT Prep Now Teacher

60 posts
4 Ian Stewart

GMATiX Teacher

50 posts
5 GMATGuruNY

The Princeton Review Teacher

37 posts
See More Top Beat The GMAT Experts