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

15 Jul 2008, 09:32

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? a) sqrt(3)/4 b) 1/2 c) PI/4 d) 1 e) sqrt(2)

I think calculus will be involved to know for sure? \(B^2+H^2=1 --> H=sqrt(1-B^2)\) Now solve for: \(f(B)=1/2(2B)sqrt(1-B^2)\)

Take the derivative, set it equal to zero to find local max. Should give you the answer for sure.

Edit: maybe there is an easier way? I used intuition when solving this problem and the answer was A naturally.... maybe someone else can find a solution where fancy math needs not be involved.
_________________

guys!!!! why complicating matters with all these... the concept is simple... area of a triangle is A=1/2 *b*h. for A(max), we should maximise either b or h or both. but b is constant as it is the radius of the circle. so h is max when it is perpendicular to b. so A max= 1/2 *1*1= 1/2

guys!!!! why complicating matters with all these... the concept is simple... area of a triangle is A=1/2 *b*h. for A(max), we should maximise either b or h or both. but b is constant as it is the radius of the circle. so h is max when it is perpendicular to b. so A max= 1/2 *1*1= 1/2

Sorry but I don't think it is "obvious" that h is max when it is perpendicular to b (or else explain why: I don't get your "so" here )

the thick line is the height. as the point is moved along the circumference, the height of the circle increases, till the second point is perpendicular to he radius. this is where the height is max. at any other point on the circumference, height< height corresponding to the radius. is the explanation ok? or any ambiguity?? i shall explain if reqd

the thick line is the height. as the point is moved along the circumference, the height of the circle increases, till the second point is perpendicular to he radius. this is where the height is max. at any other point on the circumference, height< height corresponding to the radius. is the explanation ok? or any ambiguity?? i shall explain if reqd

+1

nicely explained
_________________

----------------------------------------------------------- 'It's not the ride, it's the rider'

The question says one vertex is on the centre and the other two on the circumference. How will a right triangle have one vertex on the centre and have max area. Please explain.

the thick line is the height. as the point is moved along the circumference, the height of the circle increases, till the second point is perpendicular to he radius. this is where the height is max. at any other point on the circumference, height< height corresponding to the radius. is the explanation ok? or any ambiguity?? i shall explain if reqd

the thick line is the height. as the point is moved along the circumference, the height of the circle increases, till the second point is perpendicular to he radius. this is where the height is max. at any other point on the circumference, height< height corresponding to the radius. is the explanation ok? or any ambiguity?? i shall explain if reqd

guys!!!! why complicating matters with all these... the concept is simple... area of a triangle is A=1/2 *b*h. for A(max), we should maximise either b or h or both. but b is constant as it is the radius of the circle. so h is max when it is perpendicular to b. so A max= 1/2 *1*1= 1/2

Thanks arjtryarjtry. This is a very good explanation ..

Put the circle in the xy-plane. Put the centre at (0,0). Put one of the vertices of the triangle at (1,0). The base is 1. If the other point is (g, h), the height of the triangle will be |h|. The largest possible value of |h| is clearly 1.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

gmatclubot

Re: Greatest possible area
[#permalink]
16 Jul 2008, 10:56