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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 01 Nov 2006, 10:08
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A
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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?

A) sqrt3/4
B) 1/2
C) pi/4
D) 1
E) sqrt2
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 [#permalink] New post 01 Nov 2006, 10:28
I would go for B.

A line from the center of the circle to a point on the circle will have its length equal to its radius. Therefore, you have 2 equal sides and the triangle formed will be a 60:60:90 triangle with sides 1:1:sqrt2

To get the area of the triangle, you need to get the hieght of the triangle, since we already have the base (sqrt2)

Split the triangle in half, and we get a 30:60:90 triange with sides
(sqrt2)/2:??:1. Solve and you get 1/sqrt2 for the other side.

so area will be 1/2 (sqrt2) (1/sqrt2) = 1/2
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 [#permalink] New post 01 Nov 2006, 11:34
Hermione wrote:
To get the area of the triangle, you need to get the hieght of the triangle, since we already have the base (sqrt2)



to get the area in this case you need only the two sides, as the triangle is right

so it's 1*1*1/2 = 1/2

B.
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 [#permalink] New post 01 Nov 2006, 11:46
The triangle so formed will be right angled trianle and the area will be 1/2
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 [#permalink] New post 01 Nov 2006, 11:51
yogeshsheth wrote:
The triangle so formed will be right angled trianle and the area will be 1/2



i'm not sure i got your point - why can it be only right?

is there a rule that right triangles are the largest area-wise?
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 [#permalink] New post 01 Nov 2006, 12:19
Well, it's not so easy to prove without a blackboard. The triangle will be an isoceles triangle with sides being the two radii and base being the secant. Call the radius R and drop a perpendicular bisector from the center of the circle to the secant. The bisector length is a, the base is 2*b so the area of the triangle is a*b. Now call the angle between the secant and the radii x. a*b = R^2*cos(x)*sin(x) = R^2/2*sin(2x) which is maximized for any R with x= Pi/4. So now draw the picture and see that the triangle must be right because the angle of the two isoceles legs is Pi/2 (cuz the other two angles are each Pi/4).
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 [#permalink] New post 01 Nov 2006, 13:57
A right triangle would be the one with the largest area (easier to see if you draw diagrams). 1/2 would be the best choice.
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  [#permalink] 01 Nov 2006, 13:57
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