Mathematics Question and Solution
Let r be 1 unit.
It implies;
Radius, R of the ascribed semi circle is;
R = 2 units.
2² = a²+(½)²
4-¼ = a²
a = √(15/4)
a = ½(√(15)) units.
1² = b²+(½)²
b² = (3/4)
b = ½(√(3)) units.
c = a-b
c = ½(√(15))-½(√(3))
c = ½(√(15)-√(3)) units.
tand = 0.5√(3)/(½)
d = atan(√(3))
d = 60°
tan60 = (½(√(15)-√(3)))/e
√(3) = (½(√(15)-√(3)))/e
e = (½(√(15)-√(3)))/√(3)
e = ⅙(3√(5)-3)
e = ½(√(5)-1) units.
Area Green is;
½*½(√(5)-1)*½(√(15)-√(3))
= ⅛(√(75)-√(15)-√(15)+√(3))
= ⅛(5√(3)-2√(15)+√(3))
= ⅛(6√(3)-2√(15))
= ¼(3√(3)-√(15)) square units.
Calculating Area Purple.
cos60 = 0.5(√(5)-1)/f
f = √(5)-1 units.
g = f+1
g = √(5)-1+1
g = √(5) units.
h² = √(5)²+(½(√(5)-1))²-2*√(5)*½(√(5)-1)cos60
h = 2 units.
or
h² = (½)²+(0.5√(15))²
h = 2 units.
h² = r²+j²
Calculating j.
2² = 1²+j²
j² = 3
j = √(3) units.
It implies;
Area Purple is;
½*√(3)*1
= ½(√(3)) square units.
Therefore;
Area Green ÷ Area Purple is;
¼(3√(3)-√(15))÷ ½(√(3))
= (3√(3)-√(15))/(2√(3))
= ⅙(9-3√(5))
= ½(3-√(5))
Area Green ÷ Area Purple in terms of phi is;
Notice;
phi = ½(1+√(5))
It implies;
-½(-4+1+√(5))
-(-2+½(1+√(5)))
Where;
½(1+√(5)) is phi.
Therefore;
-(-2+phi)
= 2-phi
It implies;
Area Green ÷ Area Purple in terms of phi is;
= (2-phi)
