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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 16 Jun 2006, 17:10
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What is the greatest possible area of a triangular region with one vertex at the center of the circle of a radius of 1 and the other two vertices on the circle?

a. sqrt(3)/4

b. 1/2

c. pi/4

d. 1

e. sqrt(2)
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Re: PS: Geometry - Triangles [#permalink] New post 16 Jun 2006, 20:19
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.
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Re: PS: Geometry - Triangles [#permalink] New post 16 Jun 2006, 20:38
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.


so i can assume this is a general rule? how could i prove this?
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Re: PS: Geometry - Triangles [#permalink] New post 16 Jun 2006, 21:17
mrmikec wrote:
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.


so i can assume this is a general rule? how could i prove this?


i guess yes. it should be therom which we can verify with an example.

suppose a an equilatereal triangle with side s = 6.
area = 9 sqrt(3)

now lets change a side, the base, of the original triangle and make the triangle issoceles.
new base, b = 8
now area = 8 sqrt (5).

now lets decrease onle a side, the base, of original triangle and make the triangle issoceles.
new base, b = 4
now area = 8 sqrt(2)

so from the example, the largest area for a given range of sides of a triangle is of equilateral triagle.
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Re: PS: Geometry - Triangles [#permalink] New post 18 Jun 2006, 12:32
MA wrote:
mrmikec wrote:
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.


so i can assume this is a general rule? how could i prove this?


i guess yes. it should be therom which we can verify with an example.

suppose a an equilatereal triangle with side s = 6.
area = 9 sqrt(3)

now lets change a side, the base, of the original triangle and make the triangle issoceles.
new base, b = 8
now area = 8 sqrt (5).

now lets decrease onle a side, the base, of original triangle and make the triangle issoceles.
new base, b = 4
now area = 8 sqrt(2)

so from the example, the largest area for a given range of sides of a triangle is of equilateral triagle.


The OA is B. Can you please explain??? Thanks.
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 [#permalink] New post 18 Jun 2006, 15:24
If the triangle is 90-45-45 then the sides would be sqrt2 - 1 - 1. The height for this would be 1/2sqrt(2).

A=1/2hb ==> 1/2*1/2sqrt(2)*sqrt(2) ==> 1/2.

This value is is greater than 1/4sqrt(3). Thus, a 90-45-45 triangle is the greatest in value with respect to the area of it.

Can anyone else confirm this?

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 [#permalink] New post 19 Jun 2006, 04:09
Area = 1/2*Base*height

So in order to maximize the area we should attempt to choose the highest possible values for base and height.
In a right triangle, Base=1 and Height=1.

Height will be maximum when it is equal to radius. Imagine a rt. traingle with vertex at center o.

You could go with a higher base, as close to the length of the diameter as possible but only lesser however (imagine a line almost perpendicular cutting the circle into half, but not quite perpendicular). In this case the height will be small.

Best case if 1 and 1 so ans is 1/2
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 [#permalink] New post 19 Jun 2006, 05:47
If we know that the two sides (a and b) are always going to be 1, then we could perhaps use this area formula: S=1/2absinO, where O is the angle formed by the two sides. Then we can see that sin90=1, which gives us the greatest value. In that case S=1/2.
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 [#permalink] New post 19 Jun 2006, 06:19
It`s even simpler than that. With the center of the circle serving as the vertice, then the largest inscribed triangle must be 90-45-45. The two shorter sides, also the length and width, are given as 1.

A=1/2b*h

A=1/2(1)*(1)

A=1/2

(B)
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 [#permalink] New post 19 Jun 2006, 08:09
Here is another approach,

Imagine the largest rectangle in a circle - it would be a square. So if we fit in a square to the circle of radius 1, then it's sides would be root(2), and therefore area is 2.

Area of square = 2 (maximizes area), and the triangle in question is exactly (1/4) the area of the square, so the area of triangle is

2/4 = 1/2
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  [#permalink] 19 Jun 2006, 08:09
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