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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 12 Mar 2008, 12:07
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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?

sqrt3 /4
1 / 2
π / 4
1
sqrt2
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Re: gprep geo [#permalink] New post 12 Mar 2008, 19:31
I think it is B

I don't know if this is an overkill, but this is what came to my mind...

If you drop a perpendicular line to secant from the center of the circle (which becomes height of the traingle), this height is given by SinX, where X is the angle at each side of secant (this is an isosceles triangle).

Now the length of secant is given by 2sqrt(1-SinX^2) = 2CosX (because SinX^2+CosX^2=1)

Now the area of the traingle = (1/2)*SinX*2CosX = SinX*CosX

The product of Sin and Cos are maximum when X=45, so the area becomes 1/2 (in other words this is a right angle triangle)
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Re: gprep geo [#permalink] New post 12 Mar 2008, 22:05
sreehari wrote:
I think it is B

I don't know if this is an overkill, but this is what came to my mind...

If you drop a perpendicular line to secant from the center of the circle (which becomes height of the traingle), this height is given by SinX, where X is the angle at each side of secant (this is an isosceles triangle).

Now the length of secant is given by 2sqrt(1-SinX^2) = 2CosX (because SinX^2+CosX^2=1)

Now the area of the traingle = (1/2)*SinX*2CosX = SinX*CosX

The product of Sin and Cos are maximum when X=45, so the area becomes 1/2 (in other words this is a right angle triangle)


so we can say that the area of a right angle triangle is maximized when its angles are 45°, can't we?
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Re: gprep geo [#permalink] New post 13 Mar 2008, 01:38
B is the correct answer.
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Re: gprep geo [#permalink] New post 13 Mar 2008, 02:13
Is it not C pi/4

my approach is => area of the circle Pi r^2 => Pi

if the triangle has one vertex at the center of the circle, let us say the angle at the vertex is 90, thus 1/4 of the circle, the area would not be pi/4?
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Re: gprep geo [#permalink] New post 13 Mar 2008, 06:39
marcodonzelli wrote:
so we can say that the area of a right angle triangle is maximized when its angles are 45°, can't we?


Yes - Product of Sin and Cos is a bell curve peaking at 45. to give you precise numbers..

Sin0*Cos0=0
Sin30*Cos30=sqrt(3)/4
Sin45*Cos45=1/2
Sin60*Cos60=sqrt(3)/4
Sin90*Cos90=0
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Re: gprep geo [#permalink] New post 13 Mar 2008, 06:43
prasannar wrote:
Is it not C pi/4

my approach is => area of the circle Pi r^2 => Pi

if the triangle has one vertex at the center of the circle, let us say the angle at the vertex is 90, thus 1/4 of the circle, the area would not be pi/4?


Prasanna - you are talking about sectors, not triangle. areas of both segments are not propotional. But you were right on the angle though - right angled triangle has the maximum area.
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Re: gprep geo [#permalink] New post 13 Mar 2008, 07:24
marcodonzelli 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?

sqrt3 /4
1 / 2
π / 4
1
sqrt2


Guys,
Can someone draw the diagram? I cannot visualise this question. Thks
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Re: gprep geo [#permalink] New post 13 Mar 2008, 08:29
Hope the below diagram helps...

OA = OB = 1
OC = SinX
AC = CosX => AB = 2CosX
Area = (1/2)*AB*OC = (1/2)*SinX*2CosX = SinX*CosX
The max value of SinX*CosX is 1/2 at Angle OAC=45 (or Angle AOB=90)
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Re: gprep geo [#permalink] New post 13 Mar 2008, 09:12
Answer is D

AS AREA OF TRIANGLE IS BASED ON TWO VARIABLES BASE OF TRIANGLE AND HEIGHT. TO GET MAX AREA EITHER MAXIMISE BASE OR HEIGHT.
MAXIMISE BASE : BASE IS SLIGHTLY LESS THAN DIAMETER OF CIRCLE AND HEIGHT IS SLIGHTLY MORE THAN ZERO ,
Area = 1/2 x base x height = 1/2 x ( nearly 2) x ( height negligible as little more than zero)=1

MAXIMISE HEIGHT : HEIGHT NEARLY EQUALS RADIUS AND BASE NEGLIGIBLE OR LITTLE MORE THAN ZERO
Area = 1/2 x negligible base x 1=1/2

AREA =1/2

THIRD CASE : EQUILATERAL TRIANGLE WITH ALL SIDE EQUAL RADIUS IN A CIRCLE
AREA= 1/2 x 1/2 x .865 = .2

SO MAXIMUM AREA =1 ANSWER D
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Re: gprep geo [#permalink] New post 13 Mar 2008, 09:41
Let's start with the left circle. If x = 45deg, this is an isosceles triangle with the angle at the centre point = 90deg.
Given that the radius = 1 and it is the hypotenuse, the base AC and height OC must be 1 / sqrt 2. The area of this AOB triangle = 1/2 * 1 / sqrt2 * 1/sqrt 2 * 2 = 1/2 = 0.5

For the right circle, if x = 60deg, this is an equilateral triangle with angle at the centre point = 60deg.
Again, given that the radius is 1, and it is the hypotenuse, the base AC = 1 / 2 and height OC = sqrt 3 / 2. The area of this AOB triangle = 1/2 * 1/2 * sqrt 3 / 2 * 2 = sqrt 3 / 4 = ~ 1.7 / 4 = ~ 0.4.

Therefore the answer is B

I solve the above using these rules:
For 45-45-90 triangle (isosceles triangle)
> 45deg : 45 deg : 90deg has a ratio of 1 : 1 : sqrt 2

For 30-60-90 triangle (equilateral triangle)
> 30deg : 60 deg : 90deg has a ratio of 1 : sqrt3 : 2

This takes away the need to calculate sin and cosin etc. You dont have that luxury in GMAT :-D
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Re: gprep geo [#permalink] New post 13 Mar 2008, 11:19
neeraj.kaushal wrote:
Area = 1/2 x base x height = 1/2 x ( nearly 2) x ( height negligible as little more than zero)=1


You got to be more precise than this - you are assuming that the "negligible height" is 1 but actually it gets closer to zero (because radius is 1). So it brings down the total area of the triangle.

If you draw a graph, it looks like a bell curve. So you need to find the peak. So taking the mid point which should be halfway between 0 and 180 degrees, you get 90 hence right angle triangle. Another way of looking at the problem!
Re: gprep geo   [#permalink] 13 Mar 2008, 11:19
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