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].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
hard one from GMATPrep
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sanju09
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ajith
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Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4duongthang 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
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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tanviet
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ajith wrote:Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4duongthang 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
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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tanviet
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how do you know the formular of area of isocelesajith wrote:Area of an isosceles triangle is b*sqrt(4a^2-b^2)/4duongthang 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
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
-
ajith
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It is a derived formula from the general area formula for a triangleduongthang wrote:
how do you know the formular of area of isoceles
b*sqrt(4a^2-b^2)/4
where b is the side with unequal length and a is the length of the other sides
Always borrow money from a pessimist, he doesn't expect to be paid back.
