Given: A Rhombus is said to be mystic quadrilateral if its diagonals are in the ratio 1:root3.
Asked: Is the quadrilateral ABCD a mystic quadrilateral?
1) Angle ABC = 60 deg
Case 1: Quadrilateral ABCD is NOT a rhombus - > It is also not a mystic quadrilateral.
Case 2: Quadrilateral ABCD is a rhombus.
Let the side of the rhombus be x
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\(DE = x/2\sqrt{3}\); BE = x + x/2 = 3x/2
\(BD = \sqrt{(DE^2 + BE^2)} = \sqrt{(3x^2/4 + 9x^2/4)}= x/2\sqrt{(3+9)}= x\sqrt{3}\)
\(AF = x\sqrt{3}\); FC = x/2
\(AC = \sqrt{(AF^2 + FC^2)} =\sqrt{(3x^2/4 + x^2/4)} = x/2\sqrt{(3+1)}= x\)
\(\frac{BD}{AC} = \sqrt{3}\)
If ABCD is a rhombus, it is a mystic quadrilateral
Since it is not provided whether ABCD is a rhombus.
NOT SUFFICIENT
2) Diagonals are perpendicular and (diag1/diag2)^2= 3
Diag1/diag2 = root3/1
A quadrilateral may or may not be a rhombus if diagonals are perpendicular. It becomes a parallelogram if diagonals are perpendicular and are also bisecting each other.
NOT SUFFICIENT
(1) + (2)
1) Angle ABC = 60 deg
2) Diagonals are perpendicular and (diag1/diag2)^2= 3
Since it is still not provided whether ABCD is a rhombus.
NOT SUFFICIENT
IMO E
Hi
GMATBustersPlease check what is the flaw in the solution.