In the figure, ABCDEF is a regular hexagon with side 1cm. What is the

BCD is a 120°,30°,30° triangle (given property of regular hexagon)
Draw a line CY from C to BD such that angle C divide into 2 equal angles of 60°
Such that BCD is divided equally into two parts of 30°,60°,90° triangle.
Of side X,√3x,2x
Where 2x = 1.
X = 1/2 = CY
Half BD = √3/2
BD = 2*√3/2 = √3

Statement 1: AF=FX = 1
No other information to drive height or sides of triangle.
(Insufficient)

Statement 2: Angle AFX = angle XFE
(No determining factor to drive value of triangle height or side)
(Insufficient)

Combining both: Extend CY to Z such that triangle BDC is similar to triangle BZD
BZ = BC
YZ = CY
And BY common.
Also BZCD is a rhombus.
BZAF and DZFE area also rhombus.
Height of triangle BDC = FX+ZF+ZY
=> 1+1+.5 = 2.5
And base = √3

Area = 1/2*2.5*√3 = 1.25√3
(Sufficient)

IMO C

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