Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

What is the greatest possible area of a triangular region [#permalink]

Show Tags

26 Dec 2007, 20:05

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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).

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.

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, 11:19