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What is the greatest possible area of a triangular region

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What is the greatest possible area of a triangular region [#permalink] New post 26 Dec 2007, 19:05
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What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

A) sqrt (3) / 4

B) 1/2

C) PI / 4

D) 1

E) SQRT (2)
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 [#permalink] New post 26 Dec 2007, 20:15
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I'm thinking B 1/2

The area of a triangle is .5(b*h). Since we know that one vertex is at the center of the circle and the other vertices are on the circle we can see that the base and height will both be 1. so .5(1*1) = 1/2. Anyway you shift around the two other points, the longest base and tallest height you can find will be 1 and 1.
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 [#permalink] New post 27 Dec 2007, 13:01
I think the answer is A.

Diagram it out. The Radius, or base of triangle, is 1. I think you were mislead to beleieve H is 1 as well.

Break the equilateral tri up into two right triangles, base=1/2, hyp=1, solve for H.

H = sqrt(3/4)

Now, B*H=1*sqrt(3/4)= sqrt(3/4)
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 [#permalink] New post 27 Dec 2007, 16:53
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.
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 [#permalink] New post 27 Dec 2007, 17:18
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option
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 [#permalink] New post 27 Dec 2007, 18:51
eschn3am wrote:
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option


Quite right.
OA is 1/2.
This one is from GMATPrep

Normally, I would be tempted to go for an equilateral triangle for maximum area.
However, in such cases, I will try to go for derivatives based method (first derivative of Area function = 0)

(cant think of conventional method to tackle this problem)
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 [#permalink] New post 27 Dec 2007, 21:07
eschn3am wrote:
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option


Good point :)
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 [#permalink] New post 27 Dec 2007, 22:15
eschn3am wrote:
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option


Not sure if you can have a right triangle with the given constraints. I pick sqrt(3)/4.

What's the OA jimjohn?
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 [#permalink] New post 28 Dec 2007, 03:55
Image

triangle AEB or CEB. both are right triangles with one point at the vertex and the other two on the circle. if the radius = 1 then the area of this triangle would be 1*1*1/2 = 1/2
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 [#permalink] New post 28 Dec 2007, 07:44
GK_Gmat wrote:
eschn3am wrote:
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option


Not sure if you can have a right triangle with the given constraints. I pick sqrt(3)/4.

What's the OA jimjohn?


I think he/she's right. You can have a right triangle and the area is 1/2 so I guess the is the right answer here. The reason why I wasn't sure if this was a gmat problem was: how do we know that this is really the maximum area?
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 [#permalink] New post 28 Dec 2007, 08:12
eschn3am wrote:
Image

triangle AEB or CEB. both are right triangles with one point at the vertex and the other two on the circle. if the radius = 1 then the area of this triangle would be 1*1*1/2 = 1/2



Some thing missing here.. the orignal stem goes like this..

Code:
What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?


One vertex is at the centre of the circle

The other 2 on the circle, by which the base can never be a diameter. But your diagram shows otherwise. Is that true? Just a quick clarification.
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 [#permalink] New post 28 Dec 2007, 08:17
Quote:
The other 2 on the circle, by which the base can never be a diameter. But your diagram shows otherwise. Is that true? Just a quick clarification.


This was just a diagram I pulled off of Google images. You'll notice that in my caption I refer to triangle AEB or CEB. While the picture shows a triangle with the diameter as a base, it's just a combination of two smaller triangles that follow the question stem exactly (and those are the triangles I'm referring to).
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 [#permalink] New post 28 Dec 2007, 08:27
eschn3am wrote:
Quote:
The other 2 on the circle, by which the base can never be a diameter. But your diagram shows otherwise. Is that true? Just a quick clarification.


This was just a diagram I pulled off of Google images. You'll notice that in my caption I refer to triangle AEB or CEB. While the picture shows a triangle with the diameter as a base, it's just a combination of two smaller triangles that follow the question stem exactly (and those are the triangles I'm referring to).


now it makes sense...Thank you..
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Re: PS Triangle Circle [#permalink] New post 01 Jan 2008, 11:51
I went completely different...It says triangular region

So..in order for it to still be a triangle is has to be less than 180 degress. I said (179/360) * 1pie = 1/2 pie.

Could also look at it as...A triangle has maximum area when all sides are equal. In this case, since one vertex is in the center of the triangle, we know the best we can do is a right triangle. Hence the question is now...What is the maximum area of a right triangle with a center of the circle as a vertex and a radius of 1. Answer is 1/2.
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Re: [#permalink] New post 01 Jan 2008, 12:16
parsifal wrote:
eschn3am wrote:
Hayabusa wrote:
If we assume that the greatest possible are would be the one of a equil. triangle, the answer is A.

I don't believe this is a GMAT problem.


but sqrt(3)/4 is less than 1/2 and you can easily have a triangle with the area 1/2 (a right triangle with legs of 1). if they're looking for the greatest possible area and we know 1/2 is possible then sqrt(3)/4 shouldn't even be considered as an option


Quite right.
OA is 1/2.
This one is from GMATPrep

Normally, I would be tempted to go for an equilateral triangle for maximum area.
However, in such cases, I will try to go for derivatives based method (first derivative of Area function = 0)

(cant think of conventional method to tackle this problem)


Here it goes-
For a triangle ABC with sides a, b & c, area is given by - 1/2 ab sin C = 1/2 bc sinA = 1/2 ca sinB

Area of a triangle (A) with sides 1, 1 and included angle say x is given by
A = 1/2 sin x.
Then (A)max= 1/2 when x = 90 i.e. rt. angle.

ANS B
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Re: PS Triangle Circle [#permalink] New post 03 Mar 2008, 10:19
jimjohn wrote:
What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

A) sqrt (3) / 4

B) 1/2

C) PI / 4

D) 1

E) SQRT (2)


A = (BH)/2

Since one of the vertices is at the origin of the circle, the radius is given as 1, in turn the base must be 1 and the height must be 1. no matter how u stretch the triangle, the base and height are in tandem with the radius. the only side that will change is the largest side, the hypotenuse, which is irrelevant to the calc of the area of a triangle.

(1*1)/2 = 1/2

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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: PS Triangle Circle   [#permalink] 03 Mar 2008, 10:19
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