Mathematics Question and Solution
Let GD (inscribed square side) be a.
Calculating a.
2(a/(tan60))+a = 12
a = 12/((2/√(3))+1)
a = 5.56921938165 cm.
b = ½(12-5.56921938165)
b = 3.21539030918 cm.
Where b is BD = CE.
c = 3.21539030918+5.56921938165
c = 8.78460969083 cm.
Where c is CD.
d² = 3.21539030918²+12²-24*3.21539030918cos60
d = 10.75890566601 cm.
Where d is AD.
(10.75890566601/sin60) = (12/sine)
e = 105°
Where e is angle ADB.
f = 105-90
f = 15°
Where f is angle ADG.
cos15 = 5.56921938165/g
g = 5.7656801693 cm.
Where g is DP.
h = 45-15
h = 30°
Where h is angle PDF.
cos30 = i/5.7656801693
i = 4.99322549671 cm.
Where i is DQ.
j² = 8.78460969083²+4.99322549671²-2*4.99322549671*8.78460969083cos45
j = 6.33002837587 cm.
Where j is CQ.
(6.33002837587/sin45) = (4.99322549671/sink)
k = 33.90219461063°
Where k is angle ECR.
cos33.90219461063 = 3.21539030918/l
l = 3.87400660361 cm.
Where l is CR.
m = 6.33002837587-3.87400660361
m = 2.45602177226 cm.
Where m is QR.
It implies;
Length QR ÷ Length CR is;
2.45602177226÷3.87400660361
= 0.63397459622
