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What is the greatest possible area of a triangular region [#permalink]
26 Dec 2007, 19:05
00:00
A
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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?
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.
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
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)
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
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.
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
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?
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.
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).
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).
Re: PS Triangle Circle [#permalink]
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.
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.
Re: PS Triangle Circle [#permalink]
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
gmatclubot
Re: PS Triangle Circle
[#permalink]
03 Mar 2008, 10:19