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

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15 Jul 2008, 08:32

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.
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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
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----------------------------------------------------------- '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.
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Re: Greatest possible area
[#permalink]
16 Jul 2008, 09:56