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by tanviet » Tue Feb 02, 2010 12:19 am
what is greatest area of a triangular which have one vertex on the center of circle with radius of 1 and other vertexes on the circle

oa will be latter. this is from GMATPrep software
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by sanju09 » Tue Feb 02, 2010 12:26 am
duongthang wrote:what is greatest area of a triangular which have one vertex on the center of circle with radius of 1 and other vertexes on the circle

oa will be latter. this is from GMATPrep software
We have an isosceles triangle with its each leg 1. Its area in these strict conditions is the maximum, only when the two legs include a right angle. So, the maximum area is ½ *1*1 = [spoiler]½[/spoiler].
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by ajith » Tue Feb 02, 2010 12:52 am
duongthang wrote:what is greatest area of a triangular which have one vertex on the center of circle with radius of 1 and other vertexes on the circle

oa will be latter. this is from GMATPrep software
Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4

in this case a = 1

so we should maximize b*sqrt(4-b^2)

Now when the angle is 60 b = 1

the area = Sqrt(3)/4

when the angle is 90
b= sqrt(2)

the area = sqrt(2)*sqrt(2)/4 =2/4

when the angle is 120

b =root(3)

area = root(3)/4

So, the maximum area occurs when the angle is 90 degrees and the area is 1/2
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by tanviet » Tue Feb 02, 2010 3:30 am
ajith wrote:
duongthang wrote:what is greatest area of a triangular which have one vertex on the center of circle with radius of 1 and other vertexes on the circle

oa will be latter. this is from GMATPrep software
Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4

in this case a = 1

so we should maximize b*sqrt(4-b^2)

Now when the angle is 60 b = 1

the area = Sqrt(3)/4

when the angle is 90
b= sqrt(2)

the area = sqrt(2)*sqrt(2)/4 =2/4

when the angle is 120

b =root(3)

area = root(3)/4

how do you have the formular for area of the isocelles, pls, help

So, the maximum area occurs when the angle is 90 degrees and the area is 1/2
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by tanviet » Tue Feb 02, 2010 3:31 am
ajith wrote:
duongthang wrote:what is greatest area of a triangular which have one vertex on the center of circle with radius of 1 and other vertexes on the circle

oa will be latter. this is from GMATPrep software
Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4

in this case a = 1

so we should maximize b*sqrt(4-b^2)

Now when the angle is 60 b = 1

the area = Sqrt(3)/4

when the angle is 90
b= sqrt(2)

the area = sqrt(2)*sqrt(2)/4 =2/4

when the angle is 120

b =root(3)

area = root(3)/4

So, the maximum area occurs when the angle is 90 degrees and the area is 1/2
how do you know the formular of area of isoceles
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by ajith » Tue Feb 02, 2010 3:38 am
duongthang wrote:
how do you know the formular of area of isoceles
It is a derived formula from the general area formula for a triangle
b*sqrt(4a^2-b^2)/4

where b is the side with unequal length and a is the length of the other sides
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