Mathematics Question and Solution
Let the side length of the regular pentagon be 1 unit.
a² = 2-2cos108
a = 1.6180339887 units.
cos18 = b/1
b = 0.9510565163 units.
Area ascribed regular pentagon is;
½*sin108+½(1+1.6180339887)*0.9510565163
= 0.4755282581+1.2449491424
= 1.7204774005 square units.
Calculating the inscribed yellow triangle area.
c = ⅓*½sin108
= 0.1585094194 square units.
d = ⅓(a)
d = ⅓(1.6180339887)
d = 0.5393446629 units.
e² = 0.5393446629²+1-2*0.5393446629cos36
e = 0.6466951903 units.
(0.6466951903/sin36) = (1/sinf)
f = 65.3546280909°
g = 180-65.3546280909
g = 114.6453719091°
h = 180-65.3546280909-72
h = 42.6453719091°
(0.5393446629/sin42.6453719091) = (j/sin72)
j = 0.757163917 units.
k = 0.5*0.757163917*0.5393446629sin114.6453719091
k = 0.1855860607 square units.
l = c+k
l = 0.1585094194+0.1855860607
l = 0.3440954801 square units.
I is the inscribed yellow triangle area.
Therefore,
The fraction of the shaded regular pentagon is;
l÷(area ascribed regular pentagon)
= 0.3440954801÷1.7204774005
= 0.2
= ⅕
