Mathematics Question and Solution
Calculating Area Blue.
Let x be the side length of the inscribed regular pentagon.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the inscribed regular pentagon.
b² = 2x²-2x²cos108
b = 1.61803398875x units.
c²+(0.5x))² = (1.61803398875x)²
c² = 2.61803398875x²-0.25x²
c = √(2.36803398875x²)
c = 1.53884176859x units.
c is the height of the inscribed regular pentagon and also the height of the ascribed isosceles trapezoid.
d = 120-90
d = 30°
cos30 = e/4
e = 4cos30
e = 2√(3) units.
e = 3.46410161514 units.
e is the height of the inscribed regular pentagon and also the height of the ascribed isosceles trapezoid.
Calculating x.
It implies;
c = e
1.53884176859x =
x = 3.46410161514/1.53884176859
x = 2.25110968902 units.
Again, x is the side length of the inscribed regular pentagon.
sin30 = f/4
f = 2 units.
g = 2f+4
g = 2(2)+4
g = 8 units.
h = ½(4-x)
h = 0.5(4-2.25110968902)
h = 0.87444515549 units.
j = ½(180-a)
j = ½(180-108)
j = ½(72)
j = 36°
It implies, area blue is;
Area trapezoid with parallel lengths 4 units and 0.87444515549 units, and height 3.46410161514 units - Area triangle with height 4 units and base 2.25110968902sin36 units - Area triangle with height 2.25110968902 units and base 2.25110968902sin108 units.
0.5(4+0.87444515549)*3.46410161514-0.5*4*2.25110968902sin36-0.5*2.25110968902*2.25110968902sin108
= 8.44278666802-2.646338153-2.40973699063
= 3.38671152439 square units.
