Mathematics Question and Solution
a² = 2√(6)²
a² = 12
a = √(12)
a = 2√(3) units.
b = ½(a)
b = ½(2√(3))
b = √(3) units.
2c² = √(3)²
2c² = 3
c = √(3/2) units.
c =½√(6) units.
d = 2√(6)-½√(6)
d = ½(4√(6)-√(6))
d = ½(3√(6)) units.
d is the side length of the inscribed square.
Observing similar triangles side length ratios.
½√(6) - e
2√(6) - ½(3√(6))
Cross Multiply.
¼(18) = 2√(6)e
9 = 4√(6)e
e = ⅜√(6) units.
tanf = 2√(6)÷½√(6)
f = atan(4)°
g = f-45
g = (atan(4)-45)°
h = 180-45-(atan(4)-45)
h = (90+atan(¼))°
tan(atan(4)-45) = j/√(3)
j = ⅗√(3) units.
j = 1.0392304845 units.
Therefore, Area Red is;
2(area triangle with height √(3) units and base ⅗√(3) units - Area triangle with height ⅜√(6) units and base √(3)sin45 units)
= 2(0.5*⅗√(3)*√(3)-0.5*⅜√(6)*√(3)sin45)
= 2((9/10)-(9/16))
= (9/5)-(9/8)
= (72-45)/40
= (27/40) square units.
