maths~~~~~~~~

6∠BAF = 180(6 - 2)度 (多邊形內角和)
∠BAF = 120度
∠BAH = 120度 – 90度 = 30度
相似地,∠AFG = 30度
AG = xtan30度
= x/√3
AH = x/cos30度
= 2x/√3
y = AH – AG = x/√3
x = √3y

Because ABCDEF is a regular hexagon, so angles abc bcd cde def efa fab are 120 degrees. Angles gaf lfe ked jdc icb hba are 90 degrees, that means angles gab hbc icd jde kef lfa are 30 degrees. Let gf=1, x=cos 30 degrees x 1 = sqrt(3)/2; ag=sin 30 degrees x 1 = 0.5. Because triangles agf and bha are congurent, that means ah=1, y=ah-ag=1-0.5=0.5.

y=0.5,x=sqrt(3)/2
x/y=[sqrt(3)/2]/0.5
x/y=sqrt(3)
x=sqrt(3).y

∠ BAF = 18*4/6=120 => ∠ BAG=30

AG/AF=tan30=>AG=x/√3

The area of triangle AFG=x^2/(2√3)

The area of ABCDEF=(3/2)x^2(√3)

So, the area of GHIJKL=(3/2)x^2(√3)-6*x^2/(2√3)
=x^2[3√3/2-√3]
=√3x^2/2

Let (3/2)y^2(√3)=√3x^2/2
3y^2=x^2
y=x/√3

Angle of BAF = [(6 - 2) x 180] / 6 = 120

Angle of BAH = 120 - 90 = 30

Angle of BHA = 180 - 90 - 30 = 60

x = AF = AB

AB / BH = tan (angle of BHA)

x / BH = tan 60

x = BH * (sqrt 3)

x = (AH - y)*(sqrt 3)

Because AB / AH = sin 60, AH = 2x / (sqrt 3)

x = (2x / (sqrt 3) - y)*(sqrt 3)

x = 2x - y*(sqrt 3)

- x = -y*(sqrt 3)

x = y*(sqrt 3)