Mathematics Question and Solution
Let BF = CF be a.
It implies;
BC = 2a
BC is the side length of the regular pentagon.
Calculating a.
b = ⅕(180(5-2))-90
b = 108-90
b = 18°
b is angle BFG
Notice.
108° is the single interior angle of the regular pentagon.
c+108+108+54 = 360
c = 360-108-108-54
c = 370-270
c = 90°
c is angle BFG.
Therefore;
tan18 = d/a
d = (a)tan18
d = 0.3249196962a units.
d is FG.
It implies;
0.5*0.3249196962a*a = Red Area.
0.1624598481a² = √(25-10√(5))
a² = √(25-10√(5))/(0.1624598481)
a² = 10
a = √(10) units.
And;
BC = 2a
BC = 2√(10) units.
And;
d = 0.3249196962a
d = 0.3249196962*√(10)
d = 1.0274862966 units.
d is FG.
e = 108-½(180-108)
e = 108-36
e = 72°
e is angle ECF.
tan72 = f/√(10)
f = 9.7324898947 units.
f is EF.
2√(10)+GH+1.0274862966 = 9.7324898947
GH = 9.7324898947-2√(10)-1.0274862966
GH = 2.3804482777 units.
cos18 = √(10)/g
g = 3.3250155022 units.
g is BG.
tanh = √(10)/1.0274862966
h = 72°
h is angle BGF.
j = 180-h
j = 180-72
j = 108°
j is angle BFG.
k = j-90
k = 108-90
k = 18°
cos18 = l/2.3804482777
l = 2.2639408462 units.
l is BI.
sin18 = m/2.3804482777
m = 0.735598972 units.
n = m+BG
n = 0.735598972+3.3250155022
n = 4.0606144742 units.
n is HI
Therefore, Area Trapezoid (Quadrilateral BGHI) Blue is;
½(BG+HI)*BI
= 0.5(3.3250155022+4.0606144742)*2.2639408462
= 0.5*2.2639408462(7.3856299764)
= 8.3603146892 square units.
