| View previous topic :: View next topic |
| Author |
Message |
chatekar
Rising GMAT Star
Joined: 13 Jul 2007
Posts: 62
Thanks given: 0
Thanked 0 times in 0 posts
|
 Posted: Sat Aug 04, 2007 9:56 am Post subject: a tough geometry question |
 |
|
Is there a standard method to solve this problem?
What is the greatest possible area of a triangular region with one vertex at the centre of the circle of radius 1 and other two vertices on the circle
1. sqrt(3)/4
2. ½
3. pi/4
4. 1
5. sqrt(2)
Thanks |
|
| Back to top |
|
 |
|
|
devesh
Rising GMAT Star
Joined: 18 Jun 2007
Posts: 57
Thanks given: 0
Thanked 0 times in 0 posts
|
| Posted: Sat Aug 04, 2007 10:54 am Post subject: |
|
|
| is it 1/2. |
|
| Back to top |
|
|
chatekar
Rising GMAT Star
Joined: 13 Jul 2007
Posts: 62
Thanks given: 0
Thanked 0 times in 0 posts
|
| Posted: Sat Aug 04, 2007 12:33 pm Post subject: |
|
|
Thats correct.
Could you please explain how to solve that?
thanks |
|
| Back to top |
|
|
jayhawk2001
Moderator
Joined: 28 Jan 2007
Posts: 789
Thanks given: 0
Thanked 11 times in 11 posts
Location: Silicon valley, California
|
| Posted: Sat Aug 04, 2007 3:01 pm Post subject: Re: a tough geometry question |
|
|
| chatekar wrote: | Is there a standard method to solve this problem?
What is the greatest possible area of a triangular region with one vertex at the centre of the circle of radius 1 and other two vertices on the circle
1. sqrt(3)/4
2. ½
3. pi/4
4. 1
5. sqrt(2)
Thanks |
The area of a triangle given 2 sides a and b is
1/2 * ab * sin (angle between a and b). The max of this value is when
sin (angle) is 1. Hence 1/2 * 1 * 1 = 1/2
I think for purposes of GMAT, you can just remember that the largest
area of a triangle given 2 sides a & b = 1/2 * a * b |
|
| Back to top |
|
|
|
