Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side length of the large square be 1 unit.
2a² = 1
a = ½√(2) unit.
b² = ½+1-√(2)cos(360-195)
b = 1.69293396321 units.
(1.69293396321/sin(360-195)) = (1/sinc)
c = 8.79397688726°
d = 180-165-8.79397688726
d = 6.20602311274°
e² = 1.69293396321²+1-2*1.69293396321cos(45+8.79397688726)
e = 1.36602540379 units.
(1.36602540379/sin(45+8.79397688726)) = (1/sinf)
f = 36.20602311308°
g = f-d
g = 30°
It implies;
Area Gold is;
Area triangle with height 1.36602540379 units and base sin30 unit.
½*1.36602540379sin30
= 0.34150635095 square units.
= 288714953/845416058 square units.
Calculating Area Magenta.
cos30 = 1/h
h = 1/cos30
h = ⅓(2√(3)) units.
tan30 = i/1
i = ⅓(√(3)) units.
(j/sin45) ÷ (1/sin105)
j = (√(3)-1) unit.
k = h-j
k = ⅓(3-√(3)) unit.
It implies;
Area Magenta is;
Area triangle with height and base 1 unit respectively - Area triangle with height ⅓(√(3)) unit and base ⅓(3-√(3))sin60 units.
= ½(1²) - ½*⅓(√(3))*⅓(3-√(3))sin60
= (3-√(3))/12 square units.
Therefore;
Area Gold ÷ Area Magenta exactly is;
(288714953/845416058) ÷ (3-√(3))/12
= (288714953(3+√(3)))/422708029
= 0.86602540379
= sin60
= cos30
= √(3)/2
