OA after a few reply. Please provide explanations.
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. sqrt(3) / 4
B. 1/2
C. pi / 4
D. 1
E sqrt(2)
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GMATprep - triangle inside circle
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jayhawk2001
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Hi,
For greatest possible area, triangle should be an isosceles right angled triangle.
With one vertex at centre of the circle n other two on the circle.
Area will be=1/2*B*H
Base and Height are 1 (radius of the circle)
So, area should be 1/2*1*1 = 1/2
So, it should be B
Amit
For greatest possible area, triangle should be an isosceles right angled triangle.
With one vertex at centre of the circle n other two on the circle.
Area will be=1/2*B*H
Base and Height are 1 (radius of the circle)
So, area should be 1/2*1*1 = 1/2
So, it should be B
Amit
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jayhawk2001
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Good call. OA is B.
An isosceles right angled triangle is the isosceles triangle with the
greatest area.
An isosceles right angled triangle is the isosceles triangle with the
greatest area.
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simplythebest
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I would Still go with Option A.
An Equilateral triangle will have Maximum area.
Jay, I hope we can draw an Equi triangle with the above mentioned constraints in the Problem.
An Equilateral triangle will have Maximum area.
Jay, I hope we can draw an Equi triangle with the above mentioned constraints in the Problem.
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f2001290
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I got a formula from Trigonometry to solve this:
Area of a Triangle = (1/2)ab sin(x)
where a,b are the sides of the triangle and x is the included angle between these two sides.
We have maximum area when x is 90, because sin(90) is 1.
whereas sin(60) is sqrt(3)/2
Area of a Triangle = (1/2)ab sin(x)
where a,b are the sides of the triangle and x is the included angle between these two sides.
We have maximum area when x is 90, because sin(90) is 1.
whereas sin(60) is sqrt(3)/2
