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Interesting GMATFix Problem-42

by arora007 » Mon Oct 04, 2010 4:19 am
The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
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by sumit.sinha » Mon Oct 04, 2010 4:32 am
arora007 wrote:The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
IMO C
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by diebeatsthegmat » Mon Oct 04, 2010 4:55 pm
arora007 wrote:The circle with center O bisects the side AB and points A and C lie on the circle such that AC is the diameter. IF BC=4, what is the maximum possible area of triangle ABC?
A)4
B)4root3
C)8
D)8root3
E)12
i just guess the answer,,,,
is the answer C???
i dont understand the world bisect much but i consider that O is midpoint of diametter AC and OB is perpendicular AC and AB=AB then we will get the maximum =1/2 AB*BC
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by Rahul@gurome » Mon Oct 04, 2010 7:30 pm
Solution:
O is the centre of circle and AC is the diameter.
Or O lies on AC as its midpoint.
This circle bisects AB.
Let it bisect AB at P.
So P is the midpoint of AB.
Join P to O.
Consider the triangle ABC.
Using the midpoint theorem which states that the straight line joining the mid-points of two sides of a triangle is parallel to and equal to half the third side, we get that PO is half of BC.
Since BC is 4, PO is 2.
PO is also the radius of the circle.
Also AC is the diameter.
So AC is 2*2 = 4.
So area of triangle ABC is ½ * BC * AC * sin (angle ACB) = ½ * 4 * 4 * sin (angle ACB).
Now the maximum value of sin (angle ACB) is 1.
So the maximum value of the area of triangle ABC is ½ * 4 * 4 = 8.
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