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greatest possible area of triangle.....

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Source: — Problem Solving |
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by liferocks » Sun May 02, 2010 6:04 pm
let the triangle be AOB..o is center

hence area of the triangle is 1/2*OA*OB*Sin(AOB)=1/2*Sin(AOB)

now 0<=|Sin(AOB)|<=1
hence max area of the triangle is 1/2*1=1/2
option B
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by ansumania » Mon May 03, 2010 7:05 am
hey thanks man for the reply......
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by Ashish8 » Wed May 05, 2010 10:48 am
No point in using Trig functions. For easy math lets imagine the triangle is a 45 45 90

Since the radius is 1 The base and height will be 1, hypotenuse will be sqrt(2), but we don't need it for this problem

A = 1/2(B)(H)

A = 1/2(1)(1) = 1/2
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by tanviet » Thu May 06, 2010 3:04 am
I found a good way

sin(AOB) is out of scope of GMAT TEST, we do different way


draw the fugure. O is center, A, B on the circle, OA perpendicular to BX at the point X

the area=1/2 0A*BX(the side in the trangular OBX)

in the trangular OBX, BX is max when BX=OB=1

so 1/2
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