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

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12 Mar 2008, 13:07

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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?

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)

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?

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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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.

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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Jimmy Low, Frankfurt, Germany Blog: http://mytrainmaster.wordpress.com GMAT Malaysia: http://gmatmalaysia.blogspot.com

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)

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

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

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Jimmy Low, Frankfurt, Germany Blog: http://mytrainmaster.wordpress.com GMAT Malaysia: http://gmatmalaysia.blogspot.com

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!