Mathematics Question and Solution
a² = 12²-6²
a = 6√(3) cm.
Where a is BD.
b = ½(a)
b = 3√(3) cm.
Where b is BG = DG.
c = atan(6/(3√(3))
c = 49.10660535087°
Where c is angle AGD = angle BGE.
d = 60-49.10660535087
d = 10.89339464913°
Where d is angle BAE.
e = 180-60-10.89339464913
e = 109.10660535087°
Where e is angle BEA.
(12/sin109.10660535087) = (f/sin60)
f = 10.99818166789 cm.
Where f is AE.
(12/sin109.10660535087) = (g/sin10.89339464913)
g = 2.4 cm.
Where g is BE.
h = 12-2.4
h = 9.6 cm.
Where h is CE.
i² = (6√(3))²+2.4²-12√(3)*2.4cos30
i = 8.4 cm.
Where i is DE.
(8.4/sin30) = (6√(3)/sinj)
j = 141.78678929826°
Where j is angle BED.
k = 180-141.78678929826
k = 38.21321070174°
Where k is angle DEC.
sin38.21321070174 = l/9.6
l = 5.93845991166 cm.
Where l is CF.
cos38.21321070174 = m/9.6
m = 7.54285714286 cm.
Where m is EF.
Therefore;
Shaded Triangle CEF is;
0.5*7.54285714286*5.93845991166
= 22.39647738113 cm²
