Mathematics Question and Solution
Let AC be 2 units.
a = √(2²+2²)
a = √(8) units.
a = 2√(2) units.
Let the side length of the inscribed regular hexagon be b.
c² = 2b²-2b²cos120
c = √(3)b units.
d = 0.5b units.
e = √(2) units.
It implies;
e+d+c = a
√(2)+½(b)+√(3)b = 2√(2)
½(b)+√(3)b = √(2)
b+2√(3)b = 2√(2)
b = 2√(2)/(2√(3)+1) units.
b = (4√(6)-2√(2))/11 units.
Again, b is the side length of the inscribed regular hexagon.
c = √(3)b
c = √(3)(4√(6)-2√(2))/11
c = (12√(2)-2√(6))/11 units
(((12√(2)-2√(6))/11)/sin90) = (f/sin45)
f = 2(6-√(3))/11 units.
Therefore;
Total Rend Length is;
4(2(6-√(3))/11)+6((4√(6)-2√(2))/11)
= (48-8√(3)+24√(6)-12√(2))/11 units.
Perimeter of Largest Square is;
4*2
= 8 units.
It implies;
Total Red Length ÷ Largest Square Perimeter as a single fraction is;
((48-8√(3)+24√(6)-12√(2))/11)/8
= (12-2√(3)+6√(6)-3√(2))/22
= 0.86319073429
