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Stmt1 . We know only one co-ordinate. we don't know altitude, base or one side of triangle. Hence insufficient.

Stmt2: We know altitude when draw from C to opposite base is 6. Also this altitude will bisect the angle ACB making each angle 30.
So cos30 = altitude/CA {imagine a line from C to AB}
\(\sqrt{3}/2=6/CA\)

\(CA= 12/\sqrt{3}.\) We know one side of EQUILATERAL triangle.

\(Apply formula Area of equilateral triangle= s^2 * \sqrt{3}/4\\
= (12/\sqrt{3})^2* \sqrt{3}/4\\
= 12\sqrt{3}\)
Hence sufficient.

OA B.
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jamifahad

So sin30 = altitude/CA {imagine a line from C to AB}

I know it doesn't change the answer. But, is it sin{30} or cos{30}?
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oh yes cos30. i was timing my solution. got mixed up. thanks.
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a the distance from origin to A is not known.Hence side of the equilateral triangle cannot be calculated. Not sufficient.

b C has x coordinate = 6 = altitude of equilateral triangle
y coordinate = 3* 3^(1/2) also does not account for the distance from origin to A.

however altitude a = 6 means side = 6*2/3^(1/2) as in a triangle with angle 60 deg sides are in the ratio 3^(1/2)(perpendicular):1 (base):2(hypotenuse)

thus area can be calculated.

B
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Thanks guys, indeed the OA is B.
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Yup B is sufficient

With point C there is only one possible equilateral triangle. Hence we can find it area
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(1)
Let Co-ord of A = (0,y)

5root(3) - y = Side

Insufficient

(2)

A median drawn from C will bisect AB

and there will be 30-60-90 triangle

The length of perpendicular is known

so other sides can be found, and this the area can be found

Sufficient

Answer - B



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