BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

700+ BTG problem (Triangle inscribed)

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Mon Nov 01, 2010 5:17 pm

700+ BTG problem (Triangle inscribed)

by johnconnor99 » Fri Nov 12, 2010 10:04 am
Hi,

A question from BTG practice questions.
[spoiler]
The answer to this question is given as C. But using the second option alone, we can say as radius is always perpendicular to the diameter (The line joining the center on BC and A which is nothing but radius) the area is 1/2*base*height; which transpires to 1/2 * 12*6 giving 36. Please correct me if I'm wrong somewhere.
[/spoiler]

Image
Top
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Fri Nov 12, 2010 12:28 pm
Using only Statement 2, you can be certain that BC is a diameter. You do not, however, have any information about the location of point A. It might be that point A is extremely close on the circle to point B, for example, in which case the area of the triangle will be very small, or A could be halfway between B and C, and the area could be as you've calculated. So we need Statement 1.

I think you've assumed that AO, where O is the center of the circle, will be perpendicular to BC. That's not usually going to be true - in fact that's only true if AB = AC, which is why we need Statement 1.
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
Top
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Mon Nov 01, 2010 5:17 pm

by johnconnor99 » Sat Nov 13, 2010 1:08 am
Ian Stewart wrote:Using only Statement 2, you can be certain that BC is a diameter. You do not, however, have any information about the location of point A. It might be that point A is extremely close on the circle to point B, for example, in which case the area of the triangle will be very small, or A could be halfway between B and C, and the area could be as you've calculated. So we need Statement 1.

I think you've assumed that AO, where O is the center of the circle, will be perpendicular to BC. That's not usually going to be true - in fact that's only true if AB = AC, which is why we need Statement 1.


Hi Ian,

Thanks a lot for the quick reply!! Yeah..That was the assumption I made because of the fact that, the chords emanating from a diameter always subtends a rtangle and LA is that right angle. Please correct me if I'm wrong.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Sat Nov 13, 2010 2:03 am
johnconnor99 wrote: Hi Ian,

Thanks a lot for the quick reply!! Yeah..That was the assumption I made because of the fact that, the chords emanating from a diameter always subtends a rtangle and LA is that right angle. Please correct me if I'm wrong.
Yes, if BC is a diameter, then ABC is a right triangle here, with the right angle at A. But if you take BC as the base of your triangle, then you don't really care if angle A is a right angle. Instead, you need a right angle where you connect your height to BC. I think you've assumed that OA, where O is the center of the circle, is at 90 degrees to BC, and that is almost never going to be true (if you draw points B and A very close together, you can see that); OA will only be perpendicular to BC if the triangle is isosceles. If the triangle is not isosceles, then the point on BC that you would join to A to make a height will not be the center of the circle.

OA is of course a radius, so its length is 6. If you can be certain that OA is perpendicular to BC, then OA is the height of the triangle, and the area is indeed 6*12/2 = 36, but you need both statements to be sure that OA is a height.
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
Top
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Mon Nov 01, 2010 5:17 pm

by johnconnor99 » Sat Nov 13, 2010 8:07 am
Ian Stewart wrote:
johnconnor99 wrote: Hi Ian,

Thanks a lot for the quick reply!! Yeah..That was the assumption I made because of the fact that, the chords emanating from a diameter always subtends a rtangle and LA is that right angle. Please correct me if I'm wrong.
Yes, if BC is a diameter, then ABC is a right triangle here, with the right angle at A. But if you take BC as the base of your triangle, then you don't really care if angle A is a right angle. Instead, you need a right angle where you connect your height to BC. I think you've assumed that OA, where O is the center of the circle, is at 90 degrees to BC, and that is almost never going to be true (if you draw points B and A very close together, you can see that); OA will only be perpendicular to BC if the triangle is isosceles. If the triangle is not isosceles, then the point on BC that you would join to A to make a height will not be the center of the circle.

OA is of course a radius, so its length is 6. If you can be certain that OA is perpendicular to BC, then OA is the height of the triangle, and the area is indeed 6*12/2 = 36, but you need both statements to be sure that OA is a height.

Wow!! That cleared things up...I assumed each and everything you thought I assumed :) Great Explanation!!
Top
Post new topic Post Reply
• Page 1 of 1