MGMAT Question Bank Geometry: 20

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andy9923: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Dec 26, 2007 3:46 pm

by andy9923 » Sat Sep 06, 2008 11:02 pm

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.

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Source: Beat The GMAT — Data Sufficiency | Sat Sep 06, 2008 11:02 pm

parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Sun Sep 07, 2008 5:28 am

I agree with you. Answer should be D.

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bluementor: Master | Next Rank: 500 Posts; Posts: 418; Joined: Wed Jun 11, 2008 5:29 am; Thanked: 65 times

by bluementor » Sun Sep 07, 2008 5:36 am

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.

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Sun Sep 07, 2008 5:42 am

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.

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Sun Sep 07, 2008 6:35 am

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..

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amitansu: Master | Next Rank: 500 Posts; Posts: 294; Joined: Tue Feb 26, 2008 9:05 pmFollowed by: 1 member; Thanked: 13 times

by amitansu » Sun Sep 07, 2008 7:01 am

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

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sun Sep 07, 2008 8:56 am

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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andy9923: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Dec 26, 2007 3:46 pm

by andy9923 » Sun Sep 07, 2008 9:26 am

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.

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